Title: Part IB / Groups, Rings and Modules
Tags: Rings Modules Integral_Domain_Factorisation Actions_and_Permutations p-Groups Group_Actions Groups Sylow_Theorems Alternating_Groups Finite_Abelian_Groups Matrix_Groups

Q: Theorem. Cayley's Theorem. Every group is isomorphic to some subgroup of some symmetric group
A: Take the left regular action of G on itself. This gives a group homomorphism \(\phi: G \to \text{Sym } G\)with \(\text{ker }\phi = \{e\}\) as the actino is faithful. By the isomorphism theorem, \(G \cong \text{im } \phi \leq \text{Sym }G\)

Q: Definition. Conjugation of element
A: Conjugation of \(a \in G\) by \(b \in G\) is given by \(bab^{-1} \in G\). Given any \(a,c\), if there exists some b such that \(c = bab^{-1}\) , then we say that a and c are conjugate

Q: Lemma. Conjugation action. Any group acts on itself by conjugation. \(g(x) = gxg^{-1}\)
A: 0. \(g(x) = gxg^{-1} \in G\) for all \(x,g \in G\)
1. \(e(x) = exe^{-1} = x\)
2. \(g(h(x)) = g(hxh^{-1}) = ghxh^{-1}g^{-1} = (gh)x(gh)^{-1} = (gh)(x)\)

Q: Definition. Conjugacy classes and centralizers
A: The conjugacy classes are the orbits of the conjugacy action. \[\text{ccl}(a) = \{ b \in G : (\exists g \in G)gag^{-1} = b\}\]
The centralizers are the stabilizers of this action. i.e. elements that commute with a \[C_{G}(a) = \{g \in G: gag^{-1} = a\} = \{g\in G: ga = ag\} \]

Q: Definition. Center of group
A: The elements that commute with all other elements. \[Z(G) = \{g\in G:(\forall a)gag^{-1} = a\} = \{ g\in G:(\forall a)ga = ag\}\]

Q: Theorem. First Isomorphism Theorem. If \(\phi : G \to H\), then \(\frac{G}{\text{ker }\phi} \cong \text{Im }\phi\)
A: Let \(K = \text{ker }(\phi)\). We know K is normal.
Define \(\Phi : \frac{G}{K} \to \text{Im }(\phi)\), so that \(gK \mapsto \phi(g)\).
\(\Phi\) is well defined and injective: \(g_1K = g_2K \iff g_2^{-1}g_1 \in K \iff \phi(g_2^{-1}g_2) = 1 \iff \phi(g_1) = \phi(g_2)\)
\(\Phi\) is a homomorphism: \[\Phi(g_1Kg_2K) = \Phi(g_1g_2K) = \phi(g_1g_2) = \phi(g_1)\phi(g_2) = \Phi(g_1K)\Phi(g_2K)\].
\(\Phi\) is surjective: \[x \in \text{Im} (\theta), \phi(g) = x, g \in G \implies x = \Phi(gK) \in \text{Im }(\Phi)\]

Q: Theorem. Second Isomorphism Theorem. If \(H \leq G\) and \(Κ \unlhd G\), then \(H \cap Κ \unlhd H, HK \leq G\) and \(\frac{H}{H \cap Κ} \cong \frac{HΚ}{Κ}\)
A: First to show that HK is a subgroup, take \(h_1k_1, h_2k_2 \in HK.\) Then \[h_1k_1(h_1k_1)^{-1} = h_1h_2^{-1}h_2k_1k_2^{-1}h_2^{-1} \in HK\].
Now let \(\phi: H \to \frac{G}{K}, h \mapsto hK\). This is the composite of \(H \to G\) and the quotient map \(G \to \frac{G}{K}\), hence it is a homomorphism.
\[\text{Ker }(\phi) = \{ h \in H | hK = K\} = H \cap K \unlhd H\]
\[\text{Im }(\phi) = \{hK | h \in H\} = \frac{HK}{K}\].
The result follows form the first isomorphism theorem.

Q: Theorem. Third Isomorphism Theorem. Let \(K \leq H \leq G\) be normal subgroups of G. Then \(\frac{\frac{G}{K}}{\frac{H}{K}} \cong \frac{G}{H}\)
A: Let \(\phi: \frac{G}{K} \to \frac{G}{H}, gK \mapsto gH\). If \(g_1K = g_2K\), then \(g_2^{-1}g_1 \in K \leq H \implies g_1H = g_2H\). Thus \(\phi\) is well-defined.
\(\phi\) is surjective group homomorphism with kernel \(\frac{H}{K}\) so the result follows form the first isomorphism theorem.

Q: Lemma. Let G be an abelian group. G is simple if and only if \(G \cong C_p\) for some prime p.
A: Backward direction: Let \(H \leq C_p\). Lagrange's theorem tells us this group is simple as the only subgroups are trivial.
Forward direction: Let \(1 \neq g \in G\). G contains the subgroup \(<g> = <\dots, g^{-1}, 1 , g, g^2, \dots>\)  which is normal in G since G is abelian. Since G is simple, \(<g> = G\). If G infinite, \(G \cong (\mathbb{X}, +)\) and \(2\mathbb{Z} \in \mathbb{Z}\) so this is a contradiction.
Otherwise \(G \cong C_n\) for some n, so \(g^n = 1\). If \(m | n\) then \(g^{n/m}\) generates a subgroup of order m in G. So G being simple means the only factors of n and 1 and n so n is prime.

Q: Theorem. If G is a finite group, then G has a composition series \[1 \cong G_0 \unlhd G_1 \unlhd \dots \unlhd G_n = G\]
where each quotient \(G_{i+1}/G_i\) is simple
A: Induction on \(|G|\). If \(|G|=1\), then G=1. Conversely, if \(|G| > 1\), let \(G_{n-1}\) be a normal subgroup of largest possible order not equal to \(|G|\). Then \(G/G_{n-1}\) exists and we must prove it is simple. If not is has a \(K \unlhd G / G_{n-1}\) proper. This corresponds to some \(G_{n-1} \leq K' \unlhd G\). Then this is a strongly larger subgroup so contradiction. Now do the same to \(G_{n-1}\), this ends as G is finite.

Q: Definition. \(\text{Aut}(G) = \{\text{group isomorphisms} f: G \to G\}\)
A: Note that \(\text{Aut}(G) \leq \text{Sym}(G)\), and \(\phi: G \to \text{Sym}(G)\) has image in \(\text{Aut }(G)\)

Q: Lemma. \(A_n\) is generated by 3-cycles.
A: All elements of \(A_n\) are generated by an even number of transpositions.
\[(a \; b )(c \; d) = (a \; c \; b)(a \; c \; d), \; \; (a \; b)(b \;c) = (a \;b \;c)\]

Q: Lemma. If \(n \geq 5\), all 3-cycles in \(A_n\) are conjugate.
A: If \((a\;b\;c)\) is a 3-cycle, we have \((a\;b\;c) = \sigma (1\;2\;3) \sigma^{-1}\) for some \(\sigma \in S_n\).
If \(\sigma \in A_n\), we are done.
Otherwise \(\sigma \mapsto \sigma(4\;5) \in A_n\) suffices since \((4\;5)\) commutes with \((1\;2\;3)\).

Q: Theorem. \(A_n\) is simple for \(n \geq 5\).
A: Suppose \(1 \neq N \unlhd A_n\). To disprove normality, if suffices to show that N contains a 3-cycle, since the normality of N would imply N contains all 3-cycles and hence all events of \(A_n\).
Let \(1 \neq \sigma \in N\), writing \(\sigma\) as a product of disjoint cycles.
i) Suppose it contains a cycle of length \(r \geq 4\). Let \(\sigma = (1\;2\;3\dots\;r)\tau\) where \(\tau\) fixes \(1,3, \dots r\).
Now let \(\delta = (1\;2\;3)\). We have \[(\sigma^{-1})(\delta^{-1}\sigma\delta^{-1}) = (2\;3\;r)\]with both the parenthesised parts being in N, so N contains a 3-cycle.
ii)Suppose \(\sigma\) contains two 3-cycles, which can be written as \((1\;2\;3)(4\;5\;6)\tau\). Let \(\delta = (1\;2\;4)\), and then \[\sigma^{-1}\delta^{-1}\sigma\delta = (1\;2\;4\;5\;6)\]therefore there exists an element of N which contains a cycle of length \(5 \geq 4\). This reduces the problem to case i)
iii) Finally, suppose \(\sigma\) contains two 2-cycles, which will be written \((1\;2)(3\;4)\tau\). Then let \(\delta = (1\;2\;3)\) and \[\sigma^{-1}\delta^{-1}\sigma\delta = (1\;4)(2\;3) = \pi\]
Let \(\epsilon = (2\;3\;5)\). Then \[\pi^{-1}\epsilon^{-1}\pi\epsilon = (1\;4)(2\;3)(2\;5\;3)(1\;4)(2\;3)(2\;3\;5) = (2\;5\;3)\]Thus N contains a cycle.
The reminding cases are that \(\sigma\) is a transposition, a 3-cycle, or a transposition composed with a 3-cycle. Only the 3-cycle case is possible as we need even order. In which case \(A_n\) contains a 3-cycle so it generates all elements of \(A_n\).

Q: Theorem. If G is a p-group, the centre \(Z(G)\) is non-trivial
A: For \(g \in G\), \(|\text{ccl}(g)||C(g)| = p^n\). In particular, \(|\text{ccl}(g)|\) divides \(p^n\), and they partition G.
Since G is a disjoint union of conjugate classes, module p we have \[|G| \equiv \text{number of conjugate classes of size 1} \equiv 0 \implies |Z(G)| \equiv 0\]
So \(Z(G)\) cannot be trivial. Check this by noting that \(g \in Z(G) \iff x^{-1}gx = g\) for all x, which is true if and only if \(\text{ccl}_G(g) = \{g\}\)

Q: Corollary. The only simple p-groups are the cycle groups of order p
A: Let G be a simple p-group, since \(Z(G)\) is a normal subgroup of G, we have \(Z(G) = 1\) or \(Z(G) = G\). But \(Z(G)\) is not trivial, so G is abelian. The only abelian simple groups are cyclic of prime order. So \(G \cong C_p\)

Q: Theorem. If \(|G| = p^n\), \(\exists\) subgroups of order \(p^a\) for any \(0 \leq a \leq n\).
A: If n = 1, then \(\{e\}\) and G are the required subgruops. Suppose all groups of order \(p^m, m < n\) have the property
We have that \(\{e\} \neq Z(G)\), containing \(x \neq e\)
The order of x is \(p^l\) for some l so \(\langle xp^{l-1}\rangle \) has order exactly p.
Then \(\langle xp^{l-1} \rangle\) is a normal subgroup of size p
So \(G / \langle xp^{l-1} \rangle\) is a group of order \(p^{n-1}\)
Now use correspondence between \[\{\text{subgroups of } G / \langle xp^{l-1} \rangle\} \leftrightarrow \{\text{subgroups of G containing } \langle xp^{l-1} \rangle\}\]Where \[H \mapsto \bar{H} := \{g \in G | g\langle xp^{l-1} \rangle \in H\}\]Observe that \(\bar{H} / \langle xp^{l-1} \rangle = H\), so \(|\bar{H}| = p |H|\)
Since \(G / \langle xp^{l-1} \rangle\) has subgroups of order \(p^0, \dots , p^{n-1}\) then G has subgroups of order \(p^1, \dots , p^n\) and also trivially has \(\{e\}\)

Q: Lemma. Let G be a group. If \(\frac{G}{Z(G)}\) is cyclic, then G is abelian. This then implies that \(Z(G) = G\), so in particular \(\frac{G}{Z(G)} = 1\).
A: Let \(gZ(G)\) be a generator for \(\frac{G}{Z(G)}\). Then each coset of \(Z(G)\) is of the form \(g^rZ(G)\) for some \(r \in \mathbb{Z}\). Thus, \(G = \{g^rz: r \in \mathbb{Z}, z \in Z(G)\}\). Now multiply two elements of this group and find \[g^{r_1}z_1 g^{r_2}z_2 = g^{r_1 + r_2}z_1z_2 = g^{r_1 + r_2}z_2z_1 = z_2z_1g^{r_1+r_2} = g^{r_2}z_2g^{r_1}z_1\].
So any two elements in G commute.

Q: Corollary. Any group of order \(p^2\) is abelian.
A: \(|Z(G)| \in \{1,p,p^2\}\). The centre cannot be trivial since G is a p-group.
If \(|Z(G)| = p\), then \(\frac{G}{Z(G)}\) is cyclic as it has order p. So G is abelian. This is a contradiction since the centre of an abelian group is the group itself.
If \(|Z(G)| = p^2\) then \(Z(G) = G\) and then G is clearly abelian.

Q: Theorem. Let G be a finite group of order \(p^am\) where p is prime and p does not divide m.
i) The set \(\text{Syl}_p(G) = \{P \leq G:|P| = p^a\}\) of Sylow p-subgroups is non-empty
A: Let \(\Omega\) be the set of all subsets of G of order \(p^a\). We can directly find \[\Omega| = \frac{p^am}{p^a} \cdot \frac{p^am-1}{p^a-1}\dots \frac{p^am -p^a+1}{1}\]
so \(|\Omega|\) is coprime to p.
Let G act on \(\Omega\) by left-multiplication, so \(g * X = \{gx:x\in X\}\). For any \(Z \in \Omega\), the orbit-stabiliser theorem can be applied to show that \[|G_X||\text{orb}_G(X)| = |G| = p^am\].
There must exists an orbit with size coprime to p, since orbits partition \(\Omega\). For such an X, \(p^a \big | |G_X|\)
On the other hand, if \(g \in G\) and \(x \in X\), then \(g \in (gx^{-1})*X\) and hence \[ G = \bigcup_{g \in G} g*X = \bigcup_{Y \in \text{orb}_G(X)}Y\]
Thus \( |G| \leq |\text{orb}_G(X)| \cdot |X| \),  \(\text{giving} (|G_X| = \frac{|G|}{|\text{orb}_G(X)|} \leq |X| = p^a\)
Combined with the result from before we get that \(|G_X| = p^a\). In other words, the stabiliser \(G_X\) is a Sylow p-subgroup of G.

Q: Theorem. Let G be a finite group of order \(p^am\) where p is prime and p does not divide m.
ii) All Sylow p-subgroups are conjugate
A: Can prove a stronger result, claim that if P is a Sylow p-subgroup and \(Q \leq G\) is a p-subgroup, then \(Q \leq gPg^{-1}\) for some \(g \in G\).
Let Q act on the left cosets of P in G by left multiplication.
Orbit-stabiliser tells us that each orbit has size which divides \(|Q| = p^k\) for some k.
So each orbit has size \(p^r\) for some r.
Since \(\left |\frac{G}{P} \right | = m\), which is coprime to p, there must exist an orbit of size 1.
So there exists \(g \in G\) such that \(q * gP = gP\) for all \(q \in Q\).
So \(g^{-1}qg \in P\) for all \(q \in Q\). This implies that \(Q \leq gPg^{-1}\) as required.

Q: Theorem. Let G be a finite group of order \(p^am\) where p is prime and p does not divide m.
iii)The amount of Sylow p-subgroups \(n_p = \left |\text{Syl}_p (G) \right|\) satisfies \[n_p \equiv 1 \mod p; \; \; \; n_p \Big | |G| \implies n_p | m\]
A: Let G act on \(\text{Syl}_p(G)\) by conjugation. Sylow theorem ii) tells us that this action is transitive. By the orbit-stabiliser theorem, \(n_p = |\text{Syl}p(G)| \Big | |G|\).
Now let \(P \in \text{Syl}_p(G)\). Then let P act on \(\text{Syl}_p(G)\) by conjugation.
The orbits of this action have size dividing \(p^a\), so the size is a power of p.
Need to show that \(\{P\}\) is the unique orbit of size 1.
Assume \(\{Q\}\) is another orbit of size 1, so Q is a Sylow p-subgroup which is preserved under conjugation by P.
P normalises Q, so \(P \leq N_G(Q)\). Both P and Q are Sylow p-subgroups of \(N_G(Q)\). So by Sylow theorem ii) they are conjugate inside \(N_G(Q)\).
Hence \(P=Q\) since \(Q \unlhd N_G(Q)\).
Thus \(\{P\}\) is the unique orbit of size 1, so \(n_p \equiv 1 \mod p\).

Q: Corollary. If \(n_p = 1\), there there is only one Sylow p-subgroup, and it is normal.
A: Let \(g \in G\) and \(P \in \text{Syl}_p(G)\). Then \(gPg^{-1}\) is a Sylow p-subgroup, hence \(gPg^{-1} = P\). P is normal in G.

Q: Lemma. The permutation representation \(PGL_2(\frac{\mathbb{Z}}{p\mathbb{Z}}) \to S_{p+1}\) is injection.
A: Suppose that \(\frac{az+b}{cz+d} = z\) for all \(z \in \frac{\mathbb{Z}}{p\mathbb{Z}} \cup \{\infty\}\).
Since \(z = 0\), we have \(b = 0\).
Since \(z = \infty\), we have \(c = 0\). Thus the matrix is diagonal. Finally, since \(z = 1\), \(\frac{a}{d} = 1\) hence \(a = d\). Thus the matrix is scalar.
So the permutation representation from \(PGL_2(\frac{\mathbb{Z}}{p\mathbb{Z}})\) has trivial kernel, giving us injectivity.

Q: Lemma. Let p be an odd prime. Then \[|PSL_2(\frac{\mathbb{Z}}{p\mathbb{Z}})| = \frac{(p-1)p(p+1)}{2}\]
A: \[|GL_2(\frac{\mathbb{Z}}{p\mathbb{Z}})| = p(p^2 - 1)(p-1)\]
The homomorphism \(GL_2(\frac{\mathbb{Z}}{p\mathbb{Z}}) \to (\frac{\mathbb{Z}}{p\mathbb{Z}})^{\times}\) given by the determinant is surjective. Since \(SL_2(\frac{\mathbb{Z}}{p\mathbb{Z}})\) is the kernel of this homomorphism, we have \[|SL_2(\frac{\mathbb{Z}}{p\mathbb{Z}})| = p(p-1)(p+1)\].
Now, if \(\left ( \begin{matrix} \lambda & 0 \\ 0 & \lambda\end{matrix}\right )\) is an element of the special linear group, then \(\lambda^2  \equiv 1 \mod p\). Then \(p | (\lambda-1)(\lambda+1)\) hence \(\lambda \equiv \pm 1 \mod p\).
Thus, \[Z \cap SL_2 (\frac{\mathbb{Z}}{p\mathbb{Z}}) = \{\pm 1\}\]
and the elements are distinct since \(p>2\). Hence the order of the projective special linear group is half the order of the special linear group as required.

Q: Definition. A ring is a triple \((R, + \cdot)\) where R is a set and \(+, \cdot\) are binary operations \(R \times R \to R\), satisfying the following axioms
A: i) \((R, +)\) is an a belgian group, and we will denote the identity element 0 and the inverse of x as -x
ii) \((R, \cdot)\) satisfies the group axioms except for the invertibility axiom, and we will denote the identity element 1 and the inverse of x as \(x^{-1}\) if it exists.
iii)for all \(x,y,z \in R\) we have \(x \cdot(y+z) = x \cdot y + x \cdot z\) and \((y+z) \cdot x = y \cdot x + z \cdot x\)

Q: Definition. A subset \(S \subseteq R\) is a subbing, denoted \(S \leq R\), if [...]
A: \((S, +, \cdot)\) is a ring with the same identity elements.

Q: Definition. Let R be a ring. A polynomial f over R is an expression [...]
A: \[f = a_0 + a_1X + \dots + a_nX^n\] for \(a_i \in R\)
Back Extra: We write \(R[X]\) for the set of all such polynomials over R.
Then \((R[X], + , \cdot)\) is a ring, where the operations are defined how you'd expect polynomials to be multiplied and added.

Q: Definition. An element \(r \in \R\) is a unit if [...]
A: r has a multiplicative inverse.
Back Extra: The units in a ring, denoted \(R^{\times}\), form an abelian group under multiplication.

Q: Definition. Field.
A: A ring where all nonzero elements are units and \(0 \neq 1\)

Q: Proposition. Let \(f,g '\in R[X]\) such that the leading coefficient of g is a unit. Then there exist polynomials \(q,r \in R[X]\) such that \(f = qg+r\), where the degree of r is less than the degree of g.
A: We carry out induction on the degree of f, n.
Let \[f = a_nX^n + \dots + a_0 \;\;\; g=  b_mX^m + \dots + b_0\]
By assumption, \(b_m \in R^{\times}\).
If \(n<m\), let \(q = 0, r = f\).
If \(n \geq m\), consider the polynomial \(f_1 = f - a_nb_m^{-1}gZ^{n-m}\).
This has degree at most n-1, hence we can use induction on n to decompose \(f_1\) as \(f_1 = q_1g+r\). Thus \(f = (q_1 + a_nb_m^{-1}X^{n-m})g + r\) as required.

Q: Definition. Unit in ring.
A: An element \(a \in R\) is a unit if it has a multiplicative inverse in R. Equivalently, an element a is a unit if and only if \((a) = R\), then \(1 \in (a)\) hence there exists a multiple of a equal to 1.

Q: Definition. An element \(a \in R\) divides \(b \in R\), written \(a|b\), if [...]
A: there exists \(c \in R\) such that b = ac
Back Extra: Equivalently \((b) \subseteq (a)\)

Q: Definition. Two elements \(a,b \in R\) are associates if [...]
A: \(a=bc\) where c is a unit.
Back Extra: Informally the two elements differ by multiplication by a unit.
Equivalently \((a) = (b)\)

Q: Definition. An element \(r \in R\) is a irreducible if [...].
An element \(r \in R\) is prime if [...]
A: r is not zero or a unit, and \(r= ab\) implies a is a unit or b is a unit; r is not zero or a unit, and \(r|ab\) implies \(r|a\) or \(r|b\)

Q: Lemma. \((r) \unlhd R\) is a prime ideal if and only if \(r=0\) or r is prime.
A: Suppose \((r)\) is a prime ideal with \(r \neq 0\). Since prime ideals are proper, r cannot be a unit.
Suppose \(r | ab\), or equivalently \(ab \in (r)\). By definition of a prime deal, \(a \in (r)\) or \(b \in (r)\). Hence, \(r | a\) or \(r | b\). By definition of a prime element, r is prime.
For the other direction we basically do the same in reverse.

Q: Lemma. Prime elements are irreducible.
A: Let r be a prime, then it is nonzero and not a unit. Suppose \(r = ab\). Then \(r|ab\) so \(r|a\) or \(r|b\) by primarily.
WLOG assume \(r | a\). Hence \(r = ac\) for some element \(c \in R\).
Then, \(r = ab = rcb\), so \(r(1-cb) = 0\). Since R is an integral domain, and \(r \neq 0\), we have \(cb = 1\), so b is a unit.

Q: Definition. An integral domain R is a principal ideal domain if [...]
A: all ideals are principal ideals.
Back Extra: In other words, for all ideals I, there exists an element r such that \(I = (r)\)

Q: Proposition. In a principal ideal domain, all irreducible elements are prime.
A: Let \(r \in R\) be irreducible and suppose \(r|ab\). If \(r|a\) the proof is complete, so suppose \(r \not | a\). Since R is a principal ideal domain, the ideal \((a,r)\) is generated by a single element \(d \in R\). In particular, since \(r \in (d)\), we have \(d | r\) so \(r = cd\) for some \(c \in R\).
Since r is irreducible, either c or d is a unit. If c is a unit, \((a,r) = (r)\), so in particular \(r | a\), which contradicts the assumption that \(r \not | a\), so c cannot be a unit. Thus d is a unit.
In this case, \((a,r) = R\). By definition of \((a,r)\), there exists \(s,t \in R\) such that \(1 = sa + tr\). Then, \(b = sab + trb\). We have \(r | sab\) since \(r|ab\), ad we know \(r | trb\). Hence \(r | b\) as required.

Q: Lemma. Let R be a principal ideal domain. Then an element r is irreducible if and only if (r) is maximal.
A: Suppose r is irreducible. Since r is not a unit, \((r) \neq R\). Suppose \((r) \subseteq J \subseteq R\) where J is an ideal in R. Since R is a principal ideal domain, \(J = (a)\) for some \(a \in R\). In particular, \(r = ab\) for some \(b \in R\), since \((r) \subseteq J\). Since r is irreducible, either a or b is a unit. If a is a unit we have \(J = R\), if b is a unit then a and r are associates so they generate the same ideal. Hence (r) is maximal.
Conversely suppose (r) is maximal. Note that r is not a unit, since \((r) \neq R\). Suppose \(r = ab\). Then \((r) \subseteq (a) \subseteq R\). But since (r) is maximal either \((a) = (r)\) or \((a) = R\). If \((a) = (r)\) then b is a unit. If \((a) = R\) then a is a unit. Hence r is irreducible. This direction of the proof only requires that R is an integral domain.

Q: Definition. Euclidean Domain. An integral domain is a Euclidean Domain if there exists a function \(\phi : R \setminus \{0\} \to \mathbb{Z}_{\geq 0}\) such that for all \(a,b \in R\):
[...]
[...]
A: i) if \(a | b\) then \(\phi(a) \leq \phi(b)\); ii) if \(b \neq 0\) then \(\exists q,r \in R\) such that \(a = bq+r\) and either \(r = 0\) or \(\phi(r) < \phi(b)\).

Q: Proposition. Euclidean domains are principal ideal domains.
A: Let R have Euclidean function \(\phi\). Let \(I \unlhd R\) be a nonzero ideal. Let \(b \in I \setminus \{0\}\) that minimises \(\phi(b)\). Then \((b) \subseteq I\). For any element \(a \in I\), we can use the Euclidean algorithm to show \(a = bq +r\) where \(r = 0\) or \(\phi(r) < \phi(b)\).  But since \(r = a - bq \in I, \phi(r)\) cannot be lower than the minimum l element \(\phi(b)\). Thus \(r = 0\), so \(a = bq\). Hence \( I = b()\), so all ideals are principal.

Q: Proposition. If \(K \unlhd G\) then there is a bijection \[\text{subgroups of G/K} \leftrightarrow \text{subgroups of G containing K}\]
A: Take \(X \rightarrow \{g | gK \in X\}\) and \(L/K \leftarrow K \leq L \leq G\)

Q: Definition. A group G is a permutation group if [...]
A: it is a subgroup of Sym(X) for some X.
Back Extra: Say it is a permutation group of degree n if \(|X| = n\).

Q: Definition. An action of a group G on a set X is a function \(* : G \times X \to X\) s.t.
[...]
[...]
A: i) \(g_1 *(g_2 * x) = (g_1 \cdot g_2) * x \; \; \forall x \in X, \forall g_i \in G\); ii) \(e * x = x \; \; \forall x \in X\)

Q: Lemma. An action of G on X is the same as a homeomorphism \(\phi : G \to \text{Sym}(X)\). Either is called a permutation representation of G on X.
A: Let \(* : G \times X \to X\) be an action. Define \(\phi(g) = g * _ : X \to X\), a function. Note \(\phi(g^{-1}) \circ \phi(g)(x) = \phi(g^{-1})(g * x) = g^{-1} * (g * x) = (g^{-1} \cdot g) * x = e *x = x \; \forall x\) so \(\phi(g^{-1})\) is inverse to \(\phi(g)\) so \(\phi(g)\) is a bijection. This defines a function \(\phi : G \to \text{Sym}(X)\). This is a homomorphism.
Conversely, let \(\phi : G \to \text{Sym}(X)\) be a homomorphism. Let \(g * x = \phi(g)(x)\). The axioms follow: \(g * (h * x) = \phi(g)(\phi(h)(x)) = (\phi(g) \circ \phi(h))(x) = \phi(g \cdot h)(x)) = (g \cdot h) * x\) and we prove the other identity axiom so this is a group action.

Q: Definition. For an action of G on X, \[G_x = [...], \; \; G^x = [...] \]
A: \text{Im}(\phi : G \to \text{Sym}(X)); \text{Ker}(\phi : G \to \text{Sym}(X))

Q: Theorem. Let G be a finite group, \(H \leq G\) be a subgroup of index n. Then there is a \(K \unlhd G\) contained in H such that \(|G:K| \big | n!\) and \(n \big | |G:K|\)
A: Act by G on \(G / H\), set \(K = G_x \unlhd G\). Then \(G / G_x \cong G^x \leq \text{Sym}(G / H) \cong S_n\), so \(|G / G_x|\) divides n! , n divides it as \(K \leq H\).

Q: Corollary. Let G be a non-abelian simple group and H a subgroup of index \(n > 1\). Then G is isomorphic to a subgroup of \(A_n\), with \(n \geq 5\).
A: Act by G on the set \(G / H\), giving a homomorphism \(\phi : G \to \text{Sym} (G / H)\). Then \(G_x \unlhd G\) is normal, so as G is simple, \(G_x = \{e\}\) or \(G\).
As \(g * (eH) \neq eH\) for \(g \not \in H\) the kernel of the action is not trivial, so \(G_x \neq G\), \(G_x = \{e\}\), so \(\phi\) is injective.
So \(G = G/G_x \cong G^x \leq \text{Sym} (G / H) \cong S_n\).
Now \(A_n \unlhd S_n\), so \(G \cap A_n \unlhd G\), so as G is simple, this intersection is either trivial or the whole group. If \(G \cap A_n = \{e\}\), \(G = G / G \cap A_n \cong G \cdot A_n / A_n \leq S_n / A_n\). So G has at most 2 elements; this is a contraction as it is non-abelian.
So \(G \cap A_n = G\) so \(G \leq A_n\).
\(A_1, A_2, A_3, A_4\) have no non-abelian simple sub groups by observation.

Q: Definition. If G acts on X, the orbit of \(x \in X\) is [...] The stabiliser of \(x \in X\) is [...]
A: \(G \cdot x = \{x' \in X | x' = g*x \text{ for some } g \in G\}\).; \(G_x = \{g \in G | g * x  = x\}\).

Q: Theorem. Orbit-Stabiliser. There is a bijection \(G \cdot x \leftrightarrow G / G_x\) given by [...], in particular \(|G \cdot x| = |G : G_x |\)
A: \(g * x \leftrightarrow gG_x \ \)

Q: Definition. The normaliser \(H \subset G\) is [...]
A: \[N_G(H) = \{g \in G | gHg^{-1} = H\}\]
Back Extra: The largest subgroup inside which H is normal

Q: Theorem. Let G be a finite abelian group. Then \[G \cong C_{d_1} \times C_{d_2} \times \dots \times C_{d_k}\] with \(d_{i+1} | d_i\)
A: In section 3

Q: Lemma. If n,m coprime, \(C_n \times C_m \cong C_{nm}\)
A: Let \(g \in C_n\) be the generator, let \(h \in C_m\) be the generator.
Consider \((g,h) \in C_n \times C_m\)
Suppose this has order k, then \((g,h)^k = (e,e) \implies g^k = e, h^k = e, \implies o(g) = n | k, o(h) = m | k \implies nm|k\)
But since \((g,h) \in C_n \times C_m\) and \(|C_n \times C_m| = nm\), \(k | nm\)
Thus \(k = nm implies \langle (g,h) \rangle = C_n \times C_m\) so it is cyclic.

Q: Corollary. If G is a finite abelian group, then \[G \cong C_{n_1} \times C_{n_2} \times \dots \times C_{n_r}\] where each \(n_i\) is a prime power
A: Use that if \(d = n_1 n_2 \dots n_k\) with \(n_i\) prime powers, then \[C_d \cong C_{n_1} \times C_{n_2} \times \dots \times C_{n_k}\]We know that every finite abelian group is isomorphic to a product of cyclic groups so this gives us our result.

Q: Theorem. Let G be a finite group, p a prime number, and \(|G| = p^a m\) with \(p \not | m\).
[...]
[...]
[...]
A: i) The set of Sylow p-subgroups \[\text{Syl}_p(G) = \{P \leq G | |P| = p^a\}\] is not empty.; ii) All elements of \(\text{Syl}_p(G)\) are conjugate in G.; iii) The number \(n_p := |\text{Syl}_p(G)|\) of Sylow p-subgroups satisfies \[n_p \equiv 1 \mod p, \; n_p \big | |G|\]

Q: Corollary. Let G be a non-abelian simple group and \(p \big | |G|\). Then \(|G| \big | \frac{n_p!}{2}\) and \(n_p \geq 5\)
A: G acts on the set \(\text{Syl}_p(G)\) by conjugation, \[\implies \phi : G \to \text{Sym}(\text{Syl}_p(G)) \cong S_{n_p}\] Now \(\text{Ker}(\phi) \unlhd G\), so as G is simple, \(\text{Ker}(\phi) = \{e\} \text{ or } G\). If \(\text{Ker}(\phi) = G\) then all Sylow p-subgroups are normal. This contradicts G being simple \[\implies \text{Ker}(\phi) = \{e\}, \text{ so G is isomorphic to a subgroup of }S_{n_p}\] Now consider
If this is surjective, then the kernel is an index 2 normal subgroup of G, again contradicting G being simple.  \[\implies G = \text{Im}(\phi) \leq A_{n_p} \implies |G| \big | \frac{n_p!}{2}\]

Q: Lemma. If \(n_p = 1\) then the unique Sylow p-subgroup of G is normal.
A: Let P be the Sylow p-subgroup, \(g \in G\). Then \(g^{-1}Pg\) is again a Sylow p-subgroup, so is P so it is normal

Q: Definition. A quintuple \((R, +, \cdot, 0_R, 1_R)\) gives a ring if
[...]
A ring is commutative if [...]
A: i) \((R, +, 0_R)\) is an abelian group
ii) the equation \( - \cdot - : R \times R \to R\) is associative and satisfies \(1_R \cdot x = x = x \cdot 1_R\)
iii)multiplication distributes over addition \((r_1 + r_2) \cdot r_3 = r_1 \cdot r_3 + r_2 \cdot r_3\); \(a \cdot b = b \cdot a, \forall a,b \in R\)

Q: Definition. If \((R, + , \cdot, 0_R, 1_R)\) is a ring, a \(S \subseteq R\) is a subring if [...]
A: \(0_R, 1_R \in S\) and + and \(\cdot\) restrict to binary operations on S making it into a ring

Q: Definition. An \(r \in R\) is called a unit if [...]. If all non-zero elements of R are units, then R is called [...]
A: there is a \(s \in R\) s.t. \(r.s = 1_R\); a field

Q: Definition. Product of rings. Let R,S be rings, then the product \(R \times S\) is a ring via [...]
A: \[(r,s) + (r',s') = (r+r', s+ s'), \; (r,s) \cdot (r',s') = (r \cdot r', s \cdot s')\]
The zero element is \((0_R, 0_S)\) and the one element is \((1_R, 1_S)\)

Q: Definition. Polynomial. Let R be a ring. Then a polynomial with coefficients in R is an expression [...]
A: \[f = a_0 + a_1 X + \dots + a_nX^n\] with \(a_i \in R\)
Back Extra: The degree is largest m such that \(a_m \neq 0\), and a monic polynomial has \(a_m = 1\)

Q: Definition. Polynomial Ring. We write \(R[X]\) for the set of all polynomials with coefficients in R. [...]
A: The operations are performed in the obvious way. We identity R with the subring of constant polynomials, in particular \(0_R \in R\) and \(1_R \in R\) are the zero and one of \(R[X]\)

Q: Definition. Power Series. We write \(R[[x]]\) for the ring of power series with coefficients in R, i.e. [...]
A: \[f = a_0 + a_1 X + a_2 X^2 + \dots\] where each \(a_i \in R\). This has addition and multiplication the same as for polynomials, but without upper limits.

Q: Definition. Laurent Polynomials. We write \(R[X,X^{-1}]\) for the set of Laurent polynomials with coefficients in R, i.e. [...]
A: \[f = \sum_{i \in \mathbb{Z} } a_i X^i\] where \(a_i \in R\) and only finitely many \(a_i\) are non-zero. The operations of addition and multiplication are the obvious ones.

Q: Definition. Let R and S be rings. A function \(\phi: R \to S\) is a ring homomorphism if it satisfies
[...]
A: (i) \(\phi(r_1 + r_2) = \phi(r_1) + \phi(r_2)\)
(ii) \(\phi(0_R) = 0_S\)
(iii) \(\phi(r_1 \cdot r_2) = \phi(r_1) \cdot \phi(r_2)\)
(iv) \(\phi(1_R) = 1_S\)
Back Extra: If it's also a bijection we call it an isomorphism, in which case the inverse is also a ring homomorphism.

Q: Definition. The kernel of a homomorphism \(\phi : R \to S\) is [...] The image of a homomorphism is [...]
A: \[\ker (\phi) := \{r \in R : \phi(r) = 0_S\}\]; \[\text{im} (\phi) := \{s \in S : s = \phi(r) \text{ for some } r\in R\}\]

Q: Definition. A subset \(I \subseteq R\) is an ideal, written \(I \unlhd R\), if
[...]
A: (i) It is a subgroup of \((R, + , 0_R)\)      (additive closure)
(ii) If \(a \in I\) and \(b \in R\), then \(a \cdot b \in I\)      (strong closure)
Back Extra: We say that I is a proper ideal if \(I \neq R\)

Q: Lemma. If \(\phi : R \to S\) is a homomorphism, then \(\ker (\phi) \unlhd R\)
A: Since \(\phi : (R, + , 0_R) \to (S, + , 0_R)\) is a group homomorphism, \(\ker (\phi)\) is a subgroup of \((R, + , 0_R)\). For the second part, let \(a \in \ker (\phi), b \in R\). We need to show that their product is in the kernel. We have \[\phi (a \cdot b) = \phi(a) \cdot \phi(b) = 0 \cdot \phi(b) = 0\] so \(a \cdot b \in \ker (\phi)\)

Q: Definition. Generator of ideal. For an element \(a \in R\), we write [...] and call it the ideal generated by a. For \(A \subseteq R\) a subset, the ideal generated by A is [...]
A: \[(a) := aR := \{a \cdot r : r \in R\} \unlhd R\]; \[(A) = \left \{ \sum_{a \in A} r_a \cdot a : r_a \in R , \text{ only finitely many } r_a \text{ non-zero}\right \}\]
Back Extra: In general, for \(a_1 , a_2 , \dots , a_k \in R\), we write \[(a_1 , \dots , a_k) = \{a_1 r_1 + \dots + a_k r_k : r_1 , \dots , r_k \in R\}\] This is the ideal generated by \(a_1, \dots , a_k\)

Q: Definition. An ideal I is a principal ideal if [...]
A: \(I = (a)\) for some \(a \in R\)

Q: Proposition. Let I be an ideal of R. The quotient ring \(R / I\) is the set of cosets \(r + I\) of I in R with \[(r_1 + I) + (r_2 + I) = r_1 + r_2 + I\] \[(r_1 + I)\cdot (r_2 \cdot I) = r_1 r_2 \cdot I\] and \(0_{R/ I} = 0_R + I\), \(1_{R/ I} = 1_R + I\). This is a ring, and \(q : R \to R / I\) s.t. \(r \mapsto r + I\) is a ring homomorphism.
A: Already have \((R/I, +, 0_{R/I})\) is an abelian group. If \(r_1 + I = r_1' + I\) and \(r_2 + I = r_2' + I\) then \(r_1 = r_1' + a_1, r_2 = r_2' + a_2\) with \(a_1, a_2 \in I\). So \(r_1 r_2 = (r_1' + a_1)(r_2' + a_2) = r_1' r_2' + r_1' a_2 + r_2' a_1 + a_1 a_2\) so \(r_1 r_2 + I = r_1' r_2' + I\). So multiplication of cosets is well-defined. The other axioms are inherited from R.

Q: Proposition. Euclidean Algorithm for Polynomials. Let F be a field, \(f(x), g(x) \in F[X]\). Then we can write \[f(x) = q(x) g(x) + r(x)\] with \(\deg r < \deg q\)
A: Let \(\deg f = n\) so \(f(X) = \sum_{i=0}^n a_i X^i\) with \(a_n \neq 0\). \(\deg g = m\) so \(g(X) = \sum_{i=0}^m b_j X^j\) with \(b_m \neq 0\)
If \(n < m\) then can take \(g(X) = 0\), \(r(X) = f(X)\). Suppose \(n \geq m\), proceed by induction on n.
Let \(f_1(X) = f(X) - a_n b_m^{-1} X^{n-m} g(X)\). The coefficient of \(X^n = 0\), so \(f_1(X)\) has degree < n. If we had n=m then \[f(x) = g(x)(a_n b_m^{-1} X^{n-m}) + f_1(X)\] If \(n > m\) then by induction \[f_1(X) = q_1(X)g(X) + r_1(X)\] with \(\deg r_1 < m\). So \[f(X) = (a_n b_m^{-1} X^{n-m} + q(x))g(x) + r_1(X)\] as required.

Q: Theorem. First Isomorphism Theorem for Rings. Let \(\phi : R \to S\) be a ring homomorphism. Then \(\ker (\phi) \unlhd R\), \(\text{Im} (\phi) \leq S\), and \[R / \ker \phi \cong \text{Im} \phi\] as rings
A: Have seen that \(\ker \phi\) is an ideal. Also, \(\text{Im} \phi\) is a subgroup of \((S, +, 0_S)\)(1st iso theorem for groups), and \(\phi(r_1) \cdot \phi (r_2) = \phi(r_1 r_2) \in \text{Im } \phi\), and \(1_S = \phi(1_R), 0_S = \phi(0_R)\). So \(\text{Im} \phi\) is a subring of S. Let \[\phi : R / \ker \phi \to \text{Im }\phi, r + \ker \phi \mapsto \phi(r)\] This is a bijection by 1st iso theorem for groups, just need to check it is multiplicative.

Q: Theorem. Second Isomorphism Theorem. Let \(R \leq S\) and \(J \unlhd S\). Then \(J \cap R \unlhd R \) and \[\frac{R+J}{J} = \{r + J : r \in R\} \leq \frac{S}{J}\] is a subring, and \[\frac{R}{R \cap J} \cong \frac{R+J}{J}\]
A: Define \[\phi : R \to S / J\] by \(r \mapsto r + j\). Since this is the quotient map, it is a ring homomorphism. The kernel is \[\ker \phi = \{r \in R : r + J = 0 \implies r \in J\} = R \cap J\]
Then the image is \[\text{im} \phi = \{r + J : r \in R\} = \frac{R+J}{J}\]
Then by the first isomorphism theorem, we know \(R \cap J \unlhd R\), and \(\frac{R+J}{J} \leq S\), and \[\frac{R}{R \cap J} \cong \frac{R+J}{J}\]

Q: Theorem. Third Isomorphism Theorem. Let \(I \unlhd R\) and \(J \unlhd R\), and \(I \subseteq J\). Then \(J / I \unlhd R / I\) and \[\left ( \frac{R}{I} \right ) / \left ( \frac{J}{I} \right ) \cong \frac{R}{J}\]
A: We define the map \(\phi : R / I \to R / J\) by \[r + I \mapsto r + J\]
This is well defined and surjective by the groups case. Also it is a ring homomorphism since the multiplication in both \(R / I\) and \(R / J\) is given by multiplication of coset representatives. The kernel is \[\ker \phi = \{r + I : r + J = 0 \implies r \in J\} = \frac{J}{I}\]
So the result follows from the first isomorphism theorem.

Q: Definition. Let R be a ring, and \(\iota : \mathbb{Z} \to R\) be the unique ring homomorphism. The characteristic of R is [...]
A: the unique non-negative n such that \(\ker (\iota) = n \mathbb{Z}\)

Q: Definition. Integral Domain. A non-zero ring R is an integral domain if [...]
A: for all \(a,b \in R\), if \(a \cdot b = 0_R\), then \(a = 0_R\) or \(b = 0_R\)
Back Extra: An element that violates this property is known as a zero divisor.

Q: Definition. Zero divisor. An element \(x \in R\) is a zero divisor if [...]
A: \(x \neq 0\) and there is a \(y \neq 0\) such that \(xy = 0\in R\)
Back Extra: A ring is an integral domain if it has no zero divisors

Q: Lemma. Let R be a finite ring which is an integral domain. Then R is a field.
A: Let \(a \in R\) be non-zero, and consider the group homomorphism \[a \cdot - : (R, +, 0_R) \to (R, + , 0_R)\] where \[b \mapsto a \cdot b\]
We want to show this is injective, for which it suffices to show that the kernel is trivial. This follows from R being an integral domain. So the kernel is trivial.
Since R is finite, \(a \cdot -\) must also be surjective. In particular, there is an element \(b \in R\) such that \(a \cdot b = 1_R\), so a has an inverse. Since a was arbitrary, R is a field.

Q: Lemma. If R is an integral domain, then \(R[X]\) is too.
A: We need to show the product of two non-zero elements are non-zero. Let \(f,g \in R[X]\) be non-zero, say \[f = a_0 + a_1 X + \dots + a_n X^n \in R[X]\]
\[g = b_0 + b_1 X + \dots + b_m X^m \in R[X]\] with \(a_n,b_m \neq 0\). Then the coefficient of \(X^{n+m}\) in fg is \(a_n b_m\). This is non-zero since R is an integral domain, so fg is non-zero. So \(R[X]\) is an integral domain.

Q: Definition. Field of fractions. Let R be an integral domain. A field of fractions F of R is a field with the following properties
[...]
A: i) \(R \leq F\)
ii) Every element of F may be written as \(a \cdot b^{-1}\) for \(a,b \in R\), where \(b^{-1}\) means the multiplicative inverse to \(b \neq 0\) in F.
Back Extra: \(\mathbb{Q}\) is a field of fractions of \(\mathbb{Z}\)

Q: Theorem. Every integral domain has a field of fractions.
A: The construction is exactly how we construct the rationals from the integers - as equivalence classes of pairs of integers. Let \[S = \{(a,b) \in R \times R : b \neq 0\}\] We think of \((a,b) \in S\) as the fraction \(\frac{a}{b}\). We define an equivalence relation \(~\) on S by \[(a,b) ~ (c,d) \iff ad = bc\] We can show this is an equivalence relation, using the fact that R is an integral domain.
Now let \(F = S / ~\) be the set of equivalence classes. We write \(\frac{a}{b} = [(a,b)] \in F\), and define addition and multiplication operations by \[\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}\] \[\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}\] This is well-defined, and makes \((F, +, \cdot, \frac{0}{1} , \frac{1}{1})\) into a ring. There are many things to check, but they are straightforward. Finally, we wish to show F is a field so need to show every non-zero element has an inverse. Let \(\frac{a}{b} \neq 0_F\), so \(\frac{a}{b} \neq \frac{0}{1}\) or \(a \cdot 1 \neq b \cdot 0 \in R\) so \(a \neq 0\). Then \(\frac{b}{a} \in F\) is defined, and is the inverse to \(\frac{a}{b}\) and hence F is a field.
We now need to construct a subring of F that is isomorphic to R. To do so, we need to define an injective homomorphism \(\phi : R \to F\). This is given by \[r \mapsto \frac{r}{1}\] This is a ring homomorphism, the kernel is trivial and so \(\phi\) is injective. By the first isomorphism theorem, \(R \cong \text{im}(\phi) \subseteq F\).
Finally, we need to show everything is a quotient of two things in R. We have \[\frac{a}{b} = \frac{a}{1} \cdot \frac{1}{b} = \frac{a}{1} \cdot \left ( \frac{b}{1} \right )^{-1}\] as required.

Q: Lemma. A non-zero ring R is a field if and only if its only ideals are \(\{0\}\) and R
A: Let \(I \unlhd R\) and R be a field. Suppose \(x \neq 0 \in I\). Then as x is a unit, \(I = R\).
Suppose \(x \neq 0 \in R\). Then \((x)\) is an ideal of R which is not \(\{0\}\) since it contains x. So \((x) = R\) and so \(1_R \in (x)\). Thus there is some \(u \in R\) such that \(x \cdot u = 1_R\), so x is a unit. Since x was arbitrary, R is a field.

Q: Definition. An ideal I of a ring R is maximal if [...]
A: \(I \neq R\) and for any ideal J with \(I \leq J \leq R\), either \(J = I\) or \(J = R\)

Q: Lemma. An ideal \(I \unlhd R\) is maximal if and only if \(R / I\) is a field.
A: \(R / I\) is a field if and only if \(\{0\}\) and \(R / I\) are the only ideals of \(R / I\). By the ideal correspondence, this is equivalent to saying I and R are the only ideals of R which contain I, so I is maximal.

Q: Definition. An ideal I of a ring R is prime if [...]
A: \(I \neq R\) and whenever \(a,b \in R\) are such that \(a \cdot b \in I\), then \(a \in I\) or \(b \in I\)

Q: Lemma. An ideal \(I \unlhd R\) is prime if and only if \(R / I\) is an integral domain.
A: Let I be prime. Let \(a + I, b + I \in R / I\), and suppose that \((a+I)(b+I) = 0_{R/ I}\). By definition \((a+I)(b+I) = ab + I\). So we must have \(ab \in I\). As I is prime, either \(a \in I\) or \(b \in I\). So \(a + I = 0_{R / I}\) or \(b + I = 0_{R/I}\), and hence \(R / I\) is an integral domain.
Conversely, suppose \(R / I\) is an integral domain. Let \(a,b \in R\) be such that \(ab \in I\). Then \((a + I)(b + I) = ab + I = 0_{R/I} \in R / I\). Since \(R/I\) is an integral domain, either \(a+I = 0_{R/I}\) or \(b + I = 0_{R/I}\), i.e. \(a \in I\) or \(b \in I\). So I is a prime ideal.

Q: Proposition. Every maximal ideal is a prime ideal.
A: \(I \unlhd R\) is maximal implies \(R / I\) is a field implies \(R / I\) is an integral domain implies I is prime.

Q: Lemma. Let R be an integral domain. Then it's characteristic is either 0 or a prime number.
A: Consider the unique map \(\phi : \mathbb{Z} \to R\), and \(\ker (\phi) = n \mathbb{Z}\). Then n is the characteristic of R by definition. By the first isomorphism theorem, \(\mathbb{Z} / n \mathbb{Z} = \text{im}(\phi) \leq R\). So \(\mathbb{Z} / n \mathbb{Z}\) is an integral domain. So \(n \mathbb{Z} \unlhd \mathbb{Z}\) is a prime. So \(n = 0\) or a prime number.

Q: Definition. An element \(a \in R\) is a unit if [...]
A: there is a \(b \in R\) such that \(ab = 1_R\).
Back Extra: Equivalently, if \((a) = R\).

Q: Definition. For elements \(a,b \in R\), we say a divides b, written \(a | b\), if [...]
A: there is a \(c \in R\) such that \(b = ac\)
Back Extra: Equivalently, if \((b) \subseteq (a)\)

Q: Definition. We say \(a,b \in R\) are associates if [...]
A: \(a = bc\) for some unit c.
Back Extra: Equivalently, if \((a) = (b)\). Equivalently if \(a | b\) and \(b | a\)

Q: Definition. We say \(a \in R\) is irreducible if [...]
A: \(a \neq 0\), a is not a unit, and if \(a = xy\) then x or y is a unit.
Back Extra: For integers, being irreducible is the same as being a prime. However it means something different in general rings.

Q: Definition. We say \(a \in R\) is prime if [...]
A: a is non-zero, not a unit, and whenever \(a | xy\), either \(a|x\) or \(a | y\)

Q: Lemma. A principal ideal \((r)\) is a prime ideal in R if and only if \(r = 0\) or r is prime.
A: Let \((r)\) be a prime ideal. If \(r = 0\), then done. Otherwise, as prime ideals are proper, r is not a unit. Now suppose \(r | a \cdot b\). Then \(a \cdot b \in (r)\). But \((r)\) is prime. So \(a \in (r)\) or \(b \in (r)\). So \(r | a\) or \(r | b\). So r is prime.
If \(r = 0\), then \((0) = \{0\} \unlhd R\), which is prime since R is an integral domain. Otherwise, let \(r \neq 0\) be prime. Suppose \(a \cdot b \in (r)\). This means \(r | a \cdot b\). So \(r | a\) or \(r | b\). So \(a \in (r)\) and \(b \in (r)\). So \((r)\) is prime.

Q: Lemma. Let \(r \in R\) be prime. Then it is irreducible.
A: Let \(r \in R\) be prime, and suppose \(r = ab\). Since \(r | r = ab\), and r is prime, we must have \(r | a\) or \(r | b\), WLOG \(r | a\). So \(a = rc\) for some \(c \in R\). So \(r = ab = rcb\). Since we are in an integral domain, we must have \(1 = cb\). So b is a unit.

Q: Definition. Euclidean domain. An integral domain R is a Euclidean domain if [...]
A: there is a Euclidean function \(\phi : R \setminus \{0\} \to \mathbb{Z}_{\geq 0}\) such that
i) \(\phi(a \cdot b) \geq \phi(b)\) for all \(a,b \neq 0\)
ii) If \(a,b \in R\), with \(b \neq 0\), then there are \(q,r \in R\) such that \[a = b \cdot q + r\] and either \(r=0\) or \(\phi(r) < \phi(b)\)
Back Extra: Examples are integers with Euclidean function norm

Q: Definition. Principal Ideal Domain. A ring R is a principle ideal domain if [...]
A: it is an integral domain, and every ideal is a principal ideal
Back Extra: For all \(I \unlhd R\), there is some a such that \(I = (a)\)

Q: Proposition. Let R be a Euclidean domain. Then R is a principal ideal domain.
A: Let R have a Euclidean function \(\phi : R \setminus \{0\} \to \mathbb{Z}_{\geq 0}\). We let \(I \unlhd R\) be a non-zero ideal, and let \(b \in I \setminus \{0\}\) be an element with \(\phi(b)\) minimal. Then for any \(a \in I\), we write \[a = bq + r\] with \(r = 0\) or \(\phi(r) < \phi(b)\). However any such r must be in I since \(r = a - bq \in I\). So we cannot have \(\phi(r) < \phi (b)\). So we must have \(r = 0\). So \(a = bq\). So \(a \in (b)\). Since this is true for all \(a \in I\), we must have \(I \subseteq (b)\). On the other hand, since \(b \in I\), we must have \((b) \subseteq I\). So we must have \(I = (b)\).

Q: Definition. Unique Factorisation domain. An integral domain R is a unique factorisation domain if
[...]
A: i) Every non-unit may be written as a product of irreducible
ii) If \(p_1 \dots p_n = q_1 \dots q_m\) with \(p_i, q_j\) irreducibles, then n = m, and they can be reordered such that \(p_i\) is an associate of \(q_i\)

Q: Lemma. Let R be a principal ideal domain. If \(p \in R\) is irreducible, then it is prime.
A: Let \(p \in R\) be irreducible, and suppose \( p | a \cdot b\). Also suppose \(p \not | a\). We need to show \(p | b\).
Consider the ideal \((p,a) \unlhd R\). Since R is a principal ideal domain, there is some \(d \in R\) such that \((p,a) = (d)\). So \(d | p\) and \(d | a\).
Since \(d | p\) there is some \(q_1\) such that \( p =q_1 d\). As p is irreducible, either \(q_1\) or d is a unit.
If \(q_1\) is a unit, then \(d = q_1^{-1}p\), and this divides a. So \(a = q_1^{-1} p x\) for some x. This is a contradiction, since \(p \not | a\).
Therefore d is a unit. So \((p,a) = (d) = R\). In particular, \(1_R \in (p,a)\). So suppose \(1_R = rp + sa\), for some \(r,s \in R\). We now take the whole thing and multiply by b. Then \[b = rpb + sab\] We observe that \(ab\) is divisible by p, and so is p. So b is divisible by p. So done.

Q: Lemma. Let R be a principal ideal domain. Let \(I_1 \subseteq I_2 \subseteq I_3 \dots\) be a chain of ideals. Then there is some \(N \in \mathbb{N}\) such that \(I_n = I_{n+1}\) for some \(n \geq N\)
A: The obvious thing to do when we have an infinite chain of ideals is to take then union of them. We let \[I = \bigcup_{n \geq 1}^{\infty} I_n\] which is again an ideal. Since R is a principal ideal domain, \(I = (a)\) for some \(a \in R\). We know \(a \in I = \bigcup_{n=0}^{\infty} I_n\). So \(a \in I_N\) for some N. Then we have \((a) \subseteq I_N \subseteq I = (a)\). So we must have \(I_n = I\). So \(I_n = I_N = I\) for all \(n \geq N\)

Q: Proposition. Let R be a principal ideal domain. Then R is a unique factorisation domain.
A: We first need to show any (non-unit) \(r \in R\) is a product of irreducibles. Suppose \(r \in R\) cannot be factored as a product of irreducibles. Then it is certainly not irreducible. So we can write \(r = r_1 s_1\), with \(r_1 s_1\) both non-units. Since r cannot be factored as a product of irreducibles, WLOG suppose that \(r_1\) cannot be factored as a product of irreducibles (if both \(r_1\) and \(s_1\) can, then r would be a product of irreducibles). So we can write \(r_1 = r_2 s_2\) with \(r_2 ,s_2\) not units. Again, WLOG suppose that \(r_2\) can be factored as a product of irreducibles. We continue in this way.
By assumption, the process does not end, and then we have the following chain of ideals \[(r) \subseteq (r_1) \subseteq (r_2) \subseteq \dots \subseteq (r_n) \dots\] But then we have an ascending chain of ideals. By the previous lemma, these are all eventually equal. In particular, since \((r_n) = (r_{n+1})\), and \(r_n = r_{n+1}s_{n+1}\), then \(s_{n+1}\) is a unit. But this is a contradiction since it is not a unit. So r must be a product of irreducibles.
To show uniqueness, we let \(p_1 p_2 \dots p_n = q_1 q_2 \dots q_m\) with \(p_i, q_i\) irreducible. So in particular \(p_1 | q_1 \dots q_m\). Since \(p_1\) is irreducible, it is prime. So \(p_1\) divides some \(q_i\). We reorder and suppose \(p_1 | q_1\). So \(q_1 = p_1 \cdot a\) for some a. But since \(q_1\) is irreducible, a must be a unit. So \(p_1, q_1\) are associates. Since R is a principal ideal domain, hence integral domain, we can cancel \(p_1\) to obtain \[p_2 p_3 \dots p_n = (aq_2)q_3 \dots q_m\]. We now remain \(aq_2\) as \(q_2\). We can then continue to show that \(p_i\) and \(q_i\) are associates for all i. This also shows that \(n=m\), to else if \(n = m + k\), say, then \(p_{k+1} \dots p_n = 1\), which is a contradiction.

Q: Definition. d is a greatest common divisor of \(a_1, a_2 , \dots , a_n\) if [...]. m is a least common multiple of \(a_1, \dots a_n\) if [...]
A: \(d | a_i\) for all i, and if any other \(d'\) satisfies \(d' | a_i\) for all i, then \(d' | d\); \(a_i | m\) for all i, and if any other \(m'\) satisfies \(a_i | m'\) for all i, then \(m | m'\).

Q: Lemma. Let R be a unique factorisation domain. Then gcd's and lcm's exist, and are unique up to associates.
A: We construct the greatest common divisor using prime factorisation. We let \(p_1 , p_2 \dots , p_m\) be a list of all irreducible factors of \(a_i\), such that no two of these are associates of each other. We now write \[a_i = u_i \prod_{j=1}^m p_j^{n_{ij}}\] where \(n_{ij} \in \mathbb{N}\) and \(u_i\) are units. We let \[m_j = \min_i \{n_{ij} \}\] and choose \[d = \prod_{j=1}^m p_j^{m_j}\] As by definition, \(m_j \leq n_{ij}\) for all i, we know \(d | a_i\) for all i. Finally if \(d' | a_i\) for all i, then we let \[d' = v \prod_{j=1}^m p_j^{t_j}\] then we must have \(t_j \leq n_{ij}\)for all i,j. So we must have \(t_j \leq m_j\) for all j. So \(d' | d\).
Uniqueness is immediate since any two greatest common divisors have to divide each other. The argument for least common multiples is similar.

Q: Definition. Let R be a UFD and \(f = a_0 + a_1 X + \dots + a_n X^n \in R[X]\). The content \(c(f)\) of f is [...]
A: \[c(f) = \gcd (a_0 , a_1 , \dots, a_n) \in R\]
Back Extra: Since gcd is only defined up to a unit, so is the content.

Q: Definition. A polynomial is primitive if [...]
A: \(c(f)\) is a unit, i.e. the \(a_i\) are coprime.

Q: Lemma. Gauss' Lemma. Let R be a UFD, and \(f \in R[X]\) be a primitive polynomial. Then f is irreducible in \(R[X]\) if and only if f is irreducible in \(F[X]\), where F is the field of fractions of R.
A: We will show that a primitive \(f \in R[X]\) is reducible in \(R[X]\) if and only if it is redouble in \(F[X]\). Let \(f = gh\) be a product in \(R[X]\) with \(g,h\) not units. As f is primitive, so are g and h. So both have degree > 0. So \(g,h\) are not units in \(F[X]\). So f is reducible in \(F[X]\).
The other direction is less obvious. We let \(f = gh\) in \(F[X]\), with g,h not units. So g and h have degree > 0, since F is a field. So we can clear denominators by finding \(a,b \in R\) such that \((ag), (bh) \in R[X]\)(eg. let a be the product of denominators of coefficients of g). Then we get \[abf = (ag)(bh)\] and this is a factorisation in \(R[X]\). Here we have to be careful - \((ag)\) is an element of \(R[X]\), and is not necessarily a product in \(R[X]\), since g might not be in \(R[X]\). So we should just treat it as a single symbol. We now write \[(ag) = c(ag)g_1\] \[(bh) = c(bh)h_1\] where \(g_1, h_1\) primitive. So we have \[ab = c(abf) = c((ag)(bh)) = u \cdot c(ag) c(bh)\] where \(u \in R\) is a unit, by the previous corollary. But also we have \[abf = c(ag)c(gh)g_1h_1 = u^{-1} ab g_1 h_1\] So cancelling ab gives \[f = u^{-1} g_1 h_1 \in R[X]\] So f is reducible in \(R[X]\)

Q: Lemma. Let R be a UFD. If \(f,g \in R[X]\) are primitive, then so is fg.
A: We let \[f = a_0 + a_1 X + \dots + a_n X^n\] \[g = b_0 + b_1 X + \dots + b_m X^m\]
where \(a_n, b_m \neq 0\) and f,g are primitive. We want to show that the content of fg is a unit.
If fg is not primitive then \(c(fg)\) is not a unit. Since R is a UFD, we can find an irreducible \(p \in R\) which divides \(c(fg)\)
By assumption, \(c(f)\) and \(c(g)\) are units, so \(p \not | c(f)\) and \(p \not | c(g)\). So suppose \(p | a_0, p | a_1 , \dots , p|a_{k-1}\) but \(p \not | a_k\). Note it is possible that \(k = 0\). Similarly suppose \(p | b_0 , p| b_1 , \dots , p| b_{l-1} , p \not | b_l\).
We look at the coefficient of \(X^{k+l}\) in fg. It is given by \[\sum_{i+j = k+l} a_i b_j = a_{k+l}b_0 + \dots + a_{k+1}b_{l-1} + a_k b_l + a_{k-1} b_{l+1} + \dots + a_0 b_{l+k}\]
By assumption, this is divisible by p so \[p | \sum_{i+j = k+l} a_i b_j\]
However the terms, \(a_{k+l} b_0 + \dots + a_{k+1}b_{l-1}\) is divisible by p, and the same for \(a_{k-1}b_{l+1} + \dots + a_0 b_{l+k}\). So we must have \(p | a_k b_l\). As p is irreducible, and hence prime, we must have \(p | a_k\) or \(p | b_l\). This is a contradiction. So \(c(fg)\) must be a unit.

Q: Corollary. Let R be a UFD. Then for \(f,g \in R[X]\), we must have that \(c(fg)\) is an associate of \(c(f)c(g)\).
A: We can write \(f = c(f)f_1\) and \(g = c(g)g_1\), with \(f_1, g_1\) primitive. Then \[fg = c(f)c(g)f_1 g_1\] Since \(f_1 g_1\) is primitive, \(c(f)c(g)\) is a gcd of the coefficients of fg; so is \(c(fg)\), by definition. Thus they are associates.

Q: Proposition. Eisenstein's Criterion. Let R be a UFD, and let \[f=a_0 + a_1X + \dots + a_nX^n \in R[X]\] be primitive with \(a_n \neq 0\). Let \(p \in R\) be an irreducible(hence a prime) such that
(i) \(p \not | a_n\)
(ii)\(p|a_i\) for all \(0 \leq < n\)
(iii)\(p^2 \not | a_0\)
Then f is irreducible in \(R[X]\) and hence in \(F[X]\)
A: Suppose we have a factorisation \(f = gh\) with \[g = r_0 + r_1 X + \dots r_k X^k\]
\[h = s_0 + s_1 X + \dots s_l X^l\]
for \(r_k, s_l \neq 0\)
We know that \(r_k s_l = a_n\). Since \(p \not | a_n\), so \(p \not | r_k\) and \(p \not | s_l\). We also know that \(r_0 s_0 = a_0\). We know \(p | a_0\) but \(p^2 not | a_0\). So p divides exactly one of \(r_0,s_0\). Let us suppose \(p | r_0\) and \(p \not | s_0\).
Now let j be such that \[p | r_0, p| r_1, \dots , p|r_{j-1}, p \not | r_j\]
We now look at \(a_j\). This is, by definition, \[a_j = r_0 s_j + r_1 s_{j-1} + \dots + r_{j-1}s_1 + r_j s_0\] We know \(r_0, \dots , r_{j-1}\) are all divisible by p. Also, since \(p \not | r_j\) and \(p \not | s_0\), we know \(p \not | r_j s_0\), using the fact that p is prime. So \(p \not | a_j\). So we must have \(j = n\). We also know that \(j \leq k \leq n\). So we must have \(j = k = n\). So \(\text{deg} g = n\) and \(\deg h = 0\). But we also know that f is primitive, so h must be a unit. So this was not a proper factorisation.

Q: Proposition. A prime number \(p \in \mathbb{Z}\) is a prime in \(\mathbb{Z}[i]\) iff \(p \neq a^2 + b^2\) for \(a,b \in \mathbb{Z} \setminus \{0\}\)
A: If \(p = a^2 + b^2\), then \(p = (a+ib)(a-ib)\). So p is not irreducible
Now suppose \(p = uv\) with \(u,v\) not units. Taking norms, we get \(p^2 = N(u)N(v)\). So if u and v are not units, then \(N(u) = N(v) = p\). Writing \(u = a+ib\), then this says \(a^2 + b^2 = p\)

Q: Definition. The Gaussian integers is [...]
A: the subring \[\mathbb{Z}[i] = \{a+bi : a,b \in \mathbb{Z}\} \leq \mathbb{C}\]
Back Extra: The norm, which is multiplicative, is a Euclidean function for \(\mathbb{Z}[i]\), so this is a Euclidean domain, hence a principal ideal domain, hence a unique factorisation domain.

Q: Definition. Algebraic Integer. An \(\alpha \in \mathbb{C}\) is called an algebraic integer if [...]
A: it is a root of a monic polynomial in \(\mathbb{Z}[X]\), i.e. there is a monic \(f \in \mathbb{Z}[X]\) such that \(f(\alpha) = 0\).

Q: Definition. Minimal polynomial. Let \(\alpha \in \mathbb{C}\) be an algebraic integer. Then the minimal polynomial of \(\alpha\) is [...]
A: the irreducible monic polynomial \(f_{\alpha}\) such that \(I = \ker (\phi) = (f_{\alpha})\)

Q: Proposition. Let \(\alpha \in \mathbb{C}\) be an algebraic integer. Then the ideal \[I = \ker(\phi : \mathbb{Z}[X] \to \mathbb{C}, f \mapsto f(\alpha))\] is principal, and is equal to \((f_{\alpha})\) for some irreducible monic \(f_{\alpha}\)
A: By definition, there is a monic \(f \in \mathbb{Z}[X]\) such that \(f(a) = 0\). So \(f \in I\). So \(I \neq 0\). Now let \(f_{\alpha} \in I\) be such a polynomial of minimal degree. We may suppose that \(f_{\alpha}\) is primitive. We want to show that \(I = (f_{\alpha})\), and that \(f_{\alpha}\) is irreducible.
Let \(h \in I\). In \(\mathbb{Q}[X]\) we have the Euclidean algorithm, so we can write \[h = f_{\alpha} q + r\] with \(r = 0\) or \(\text{deg}(r) < \text{deg}(f_{\alpha})\). This was done over \(\mathbb{Z}[X]\), not \(\mathbb{Z}[X]\). We now clear denominators: multiplying by some non-zero \(a \in \mathbb{Z}\) we get \[ah = f_{\alpha}(aq) + (ar)\] where now \((aq), (ar) \in \mathbb{Z}[X]\). We now evaluate these polynomials at \(\alpha\). Then we have \[ah(\alpha) = f_{\alpha}(\alpha)aq(\alpha) + ar(\alpha)\] We know \(f_{\alpha}(\alpha) = h(\alpha) = 0\), since \(f_{\alpha}\) and h are both in I. So \(ar(\alpha) = 0\). So \((ar) \in I\). As \(f_{\alpha} \in I\) has minimal degree, we cannot have \(\text{deg}(r) = \text{deg}(ar) < \text{deg}(f_a)\). So we must have \(r = 0\).
Hence we know \[ah = f_{\alpha} \cdot (aq)\] is a factorisation in \(\mathbb{Z}[X]\). This is almost right, but we want to factor h, not ah. Again, by taking contents of everything, we get \[ac(h) = c(ah) = c(f_{\alpha}(aq)) = c(aq)\] as \(f_{\alpha}\) is primitive. In particular, \(a | c(aq)\). This, by definition of content, means that \((aq)\) can be written as \(a\hat{q}\), where \(\hat{q} \in \mathbb{Z}[X]\). Cancelling a, we get \(q = \hat{q} \in \mathbb{Z}[X]\). So we know \[h = f_{\alpha}q \in (f_{\alpha})\] So we know \(I = (f_{\alpha})\).
To show that \(f_{\alpha}\) is irreducible, note that \[\frac{\mathbb{Z}[X]}{(f_{\alpha})} \cong \frac{\mathbb{Z}[X]}{\ker \phi} \cong \text{im}\phi = \mathbb{Z}[\alpha] \leq \mathbb{C}\] Since \(\mathbb{C}\) is an integral domain, so is \(\text{im}(\phi)\). So we know \(\mathbb{Z}[X]/(f_{\alpha})\) is an integral domain. So \((f_{\alpha})\) is prime. So \(f_{\alpha}\) is prime, hence it is irreducible.

Q: Lemma. Let \(\alpha \in \mathbb{Q}\) be an algebraic integer. Then \(\alpha \in \mathbb{X}\)
A: Let \(f_{\alpha} \in \mathbb{Z}[X]\) be the minimal polynomial, which is irreducible. In \(\mathbb{Q}[X]\), the polynomial \(X - \alpha\) must divide \(f_{\alpha}\). However, by Gauss' lemma, we know \(f \in \mathbb{Q}[X]\) is irreducible. So we must have \(f_{\alpha} = X - \alpha \in \mathbb{Z}[X]\). So \(\alpha\) is an integer.

Q: Definition. Noetherian ring. A ring is Noetherian if [...]
A: for any chain of ideals \[I_1 \subseteq I_2 \subseteq \dots\] then there is some N such that \(I_N = I_{N+1} = \dots\)
Back Extra: This condition is known as the ascending chain condition (ACC)

Q: Definition. Finitely generated ideal. In ideal I is finitely generated if [...]
A: it can be written as \(I = (r_1, \dots , r_n)\) for some \(r_1, \dots , r_n \in R\)

Q: Proposition. A ring is Noetherian if and only if every ideal is finitely generated.
A: Suppose every ideal of R is finitely generated. Given the chain \(I_1 \subseteq I_2 \subseteq \dots\), consider the ideal \[I = I_1 \cup I_2 \dots\] This is an ideal. We know I is finitely generated, say \(I = (r_1, \dots , r_n)\), with \(r_i \in I_{k_i}\). Let \[K = \max_{i = 1,\dots , n} \{k_i\}\] Then \(r_1, \dots, r_n \in I_K\). So \(I_K = I\). So \(I_K = I_{K+1} = \dots\)
To prove the other direction, suppose there is an ideal \(I \unlhd R\) that is not finitely generated. We pick \(r_1 \in I\). Since I is not finitely generated, we know \((r_1) \neq I\). So we can find some \(r_2 \in I \setminus (r_1)\). Again \((r_1, r_2) \neq I\). So we can find \(r_3 \in I \setminus (r_1, r_2)\). We continue on, and then can find an infinite strictly ascending chain \[(r_1) \subseteq (r_1, r_2) \subseteq (r_1, r_2 , r_3) \subseteq \dots\] So R is not Noetherian

Q: Theorem. Hilbert Basis Theorem. Let R be a Noetherian ring. Then so is \(R[X]\).
A: Let \(J \unlhd R[X]\) be an ideal. Let \(f_1 \in J\) be a polynomial of minimal degree. If \(J \neq (f_1)\) then let \(f_2 \in J \setminus (f_1)\) be a polynomial of minimal degree. If \(J \neq (f_1, f_2)\) let \(f_3 \in J \setminus (f_1, f_2)\) be a polynomial of minimal degree. Continuing in this way, if \(J = (f_1, \dots, f_i)\) then we are done, so suppose note.
Let \(a_i \in R\) be the non-zero coefficient of the largest power of X in \(f_i\), and consider the ideals \((a_1) \subseteq (a_1, a_2) \subseteq \dots \subseteq (a_1, \dots , a_i) \subseteq \dots\) of R. As R is Noetherian these stabilise, so \((a_1, a_2, \dots ) = (a_1,a_2, \dots a_m)\) for some m.
Now \(a_{m+1} \in (a_1 , \dots, a_m)\), so \[a_{m+1} = \sum_{i=1}^m a_i \cdot b_i\] Thus the polynomial \[g = \sum_{i=1}^m b_i f_i X^{\text{deg} f_{m+1} - \text{deg} f_i}\] has the same degree and top coefficient as \(f_{m+1}\). (Note that \(\text{deg}f_{m+1} \geq \text{deg}f_i\) for \(i \leq m\)) Thus \(f_{m+1} - g\) has degree strictly smaller than \(f_{m+1}\). But \(g \in (f_1, \dots , f_m)\)  and \(f_{m+1} \not \in (f_1, \dots , f_m)\), so \[f_{m+1} - g \not \in (f_1, \dots , f_m)\] which contradicts the fact that we chose \(f_{m+1}\) to have minimal degree among polynomials in J but not in \((f_1 , \dots , f_m)\).
Hence the process we started with must terminate at some point, so J is finitely-generated

Q: Corollary. \(\mathbb{Z}[X_1, X_2, \dots , X_n]\) is Noetherian, and for F a field \(F[X_1, \dots, X_n]\) is Noetherian.
A: Since if R is a Noetherian ring so is \(R[X]\).

Q: Proposition. Let R be a Noetherian ring and I be an ideal of R. Then \(R / I\) is Noetherian.
A: Let \(J \unlhd R / I\) be an ideal. We want to show that J is finitely generated. By the ideal correspondence, it corresponds to some ideal \(J' \unlhd R\) containing I. This is an ideal of R, and is hence finitely generated since R is Noetherian. So \(J' = (r_1, \dots r_n)\) for some \(r_1, \dots ,r_n \in R\). Then J may be finitely generated by \(r_1 + I, \dots , r_n + I\)

Q: Corollary. Any finitely-generated ring is Noetherian.
A: A finitely-generated ring is a quotient of some \(\mathbb{Z}[X_1, \dots , X_n]\) which is Noetherian.

Q: Definition. Module. Let R be a commutative ring. We say a quadruple \((M, + , 0_M, \cdot)\) is an R-module if
[...]
A: (i) \((M, + , 0_M)\) is an abelian group
(ii) The operation \(\cdot : R \times M \to M\) satisfies
(a) \((r_1 + r_2) \cdot m = (r_1 \cdot m) + (r_2 \cdot m)\);
(b) \(r \cdot (m_1 + m_2) = (r\cdot m_1) + (r \cdot m_2)\);
(c) \(r_1 \cdot (r_2 \cdot m) = (r_1 \cdot r_2) \cdot m\);
(d) \(1_R \cdot M = m\)
Back Extra: We can think of modules as rings acting on abelian groups, just as groups can act on sets.

Q: Definition. Module. Let R be a commutative ring. We say a quadruple \((M, + , 0_M, \cdot)\) is an R-module if
[...]
A: (i) \((M, + , 0_M)\) is an abelian group
(ii) The operation \(\cdot : R \times M \to M\) satisfies
(a) \((r_1 + r_2) \cdot m = (r_1 \cdot m) + (r_2 \cdot m)\);
(b) \(r \cdot (m_1 + m_2) = (r\cdot m_1) + (r \cdot m_2)\);
(c) \(r_1 \cdot (r_2 \cdot m) = (r_1 \cdot r_2) \cdot m\);
(d) \(1_R \cdot M = m\)

Q: Definition. Submodule. Let M be an R-module. A subset \(N \subseteq M\) is an R-submodule if [...] We write \(N \leq M\).
A: it is a subgroup of \((M, + , 0_M)\), and if \(n \in N\) and \(r \in R\), then \(rn \in N\).

Q: Definition. Quotient Module. Let \(N \leq M\) be an R-submodule. The quotient module \(M / N\) is [...]
A: the set of N-cosets in \((M, + , 0_M)\), with the R action given by \[r \cdot (m + N) = (r \cdot m) + N\]
Back Extra: This is well-defined and is indeed a module. We can in modules, quotient by arbitrary submodules. There is no notion of "normal submodules"

Q: Definition. R-module homomorphism and isomorphism. A function \(f : M \to N\) between R-modules is a R-module homomorphism if [...] An isomorphism is a bijective homomorphism.
A: it is a homomorphism of abelian groups, and satisfies \[f(r \cdot m) = r \cdot f(m)\] for all \(r \in R, m \in M\).

Q: Theorem. First Isomorphism Theorem. Let \(f : M \to N\) be an R-module homomorphism. Then \[\ker f = \{m \in M : f(m) = 0\}\] is an R-submodule of M. Similarly, \[\text{im} \{f(m) : m \in M\}\] is an R-submodule of N. Furthermore there is an R-module isomorphism \[\frac{M}{\ker f} \cong \text{im} f\]
A: Proof exactly the same as before

Q: Theorem. Second Isomorphism Theorem. Let \(A,B \leq M\). Then \[A + B = \{m \in M : m = a + b\} \text{ for some } a\in A, b \in B\]is a submodule of M and \(A \cap B\) is a submodule of M. Furthermore there is an R-module isomorphism \[\frac{A+B}{A} \cong \frac{B}{A \cap B}\]
A: Proof same as before.

Q: Theorem. Third Isomorphism Theorem. Let \(N \leq L \leq M\). Furthermore there is an R-module isomorphism \[\frac{M}{L} \cong \left ( \frac{M}{N} \right) / \left ( \frac{L}{N} \right)\]
A: Proof same as before. As usual, we have a correspondence between submodules of M/N and submodules of M containing N.

Q: Definition. Submodule generated by an element. Let M be an R-module, and \(m \in M\). The submodule generated by m is \[[...]\]
A: Rm = \{r \cdot m \in M : r \in R\}
Back Extra: Considering the R-module homomorphism \(\phi : R \to M\) with \(r \mapsto rm\) and using the first isomorphism theorem we get \[Rm \cong R / \text{Ann}(m)\]

Q: Definition. Finitely-generated module. An R-Module M is finitely-generated if there is a finite list of elements \(m_1, \dots , m_k\) such that \[[...]\]
A: M = Rm_1 + \dots + Rm_k = \{r_1 m_1 + \dots + r_k m_k : r_i \in R\}

Q: Lemma. An R-module M is finitely-generated if and only if there is a surjective R-module homomorphism \(f : R^k \to M \) for some finite k.
A: If \(M = Rm_1 + \dots + Rm_k\) then we define \(f : R^k \to M\) by \[(r_1 , \dots , r_kl) \mapsto r_1 m_1 + \dots + r_k m_k\]It is clear that this is an R-module homomorphism. It is by definition surjective, so we are done.
Conversely, given a surjection \(f : R^k \to M\), we let \[m_i = f(0,0,\dots , 0,1,0 \dots , 0)\] where the 1 appears in the ith position. We now claim that \[M = Rm_1 + \dots + Rm_k\] So let \(m \in M\). As f is surjective, we know \[m = f(r_1, \dots , r_k)\] for some \(r_i\). We then have \[f(r_1, \dots r_k) = f((r_1, 0,0 \dots ,0) + (0, r_2, 0, \dots , 0) + \dots + (0,0, \dots , 0, r_k)) = r_1 m_1 + \dots + r_k m_k\] using definitions of modules. So the \(m_i\) generate M.

Q: Corollary. Let M be finitely-generated and \(N \leq M\). Then \(M / N\) is also finitely generated.
A: Since m is finitely-generated, we have some surjection \(f : R^k \to M\). Moreover, we have the surjective quotient map \(q : M \to M / N\). Then we get the following composition \[R^k \to^f M \to^q M / N\] which is a surjection, since it is a composition of surjections. So \(M / N\) is finitely-generated.

Q: Definiton. Annihilator. Let M be a R-module, and \(m \in M\). The annihilator of m is \[[...]\] For any set \(S \subseteq M\), we define \[[...]\]In particular, for the module M itself, we have \[[...]\]
A: \text{Ann}(m) = \{r \in R : r \cdot m = 0 \}; \text{Ann}(S) = \{r \in R : r \cdot m = 0 \text{ for all } m \in S \} = \bigcap_{m \in S} \text{Ann}(m); \text{Ann}(M) =\{r \in R : r\cdot m = 0 \text{ for all } m \in M\} = \bigcap_{m \in M} \text{Ann}(m)
Back Extra: Note that the annihilator is a subset of R. Moreover it is an ideal - if \(r \cdot m = 0\) and \(s \cdot m = 0\), then \((r + s) \cdot m = 0\). So \(r + s\) is in the annihilator. Similarly for multiplication.

Q: Definition. Direct Sum of Modules. Let \(M_1, \dots , M_k\) be R-modules. The direct sum is the R-module \[M_1 \bigoplus M_2 \bigoplus \dots \bigoplus M_k\] which is the set [...]
A: \(M_1 \times M_2 \times \dots \times M_k\), with addition given by \[(m_1, \dots , m_k) + (m_1', \dots , m_k') = (m_1 + m_1' , \dots , m_k +m_k')\] and the R action is given by \[r \cdot (m_1, \dots , m_k) = (rm_1, \dots , rm_k)\]

Q: Definition. Linear Independence. Let \(m_1 , \dots , m_k \in M\). Then \(\{m_1, \dots , m_k\}\) is linearly independent if [...]
A: \[\sum_{i=1}^k r_i m_i = 0 \implies r_1 = r_2 = \dots = r_k = 0\]

Q: Definition. Freely generate. A subset \(S \subseteq M\) generates M freely if:
[...]
A: (i) S generates M
(ii) Any set function \(\varphi : S \to N\) to an R-module N extends to an R-module map \(\theta : M \to N\)

Q: Definition. Free module and basis. An R-module is free if [...], and [...] is called a basis.
A: it is freely generated by some subset \(S \subseteq M\); S

Q: Proposition. For a subset \(S = \{m_1 , \dots , m_k\} \subseteq M\), the following are equivalent:
(i)S generates M freely
(ii)S generates M and the set S is independent
(iii)Every element of M is uniquely expressive as \[r_1 m_1 + \dots + r_k m_k\] for some \(r_i \in R\)
A: The fact that (ii) and (iii) are equivalent is something we would expect from what we know from linear algebra, and in fact the proof is the same. So we only show that (i) and (ii) are equivalent.
Let S generate M freely. If S is not independent, then we can write \[r_1 m_1 + \dots + r_k m_k = 0\] with \(r_i \in R\) and say, \(r_1\) non-zero. We define the set function \(\varphi : S \to R\) by sending \(m_1 \mapsto 1_R\) and \(m_i \mapsto 0\) for all \(i \neq 1\). As S generates M freely, this extends to an R-module homomorphism \(\theta : M \to R\). By definition of a homomorphism, we can compute \[0 = \theta(0) = \theta(r_1 m_1 + \dots + r_k m_k) = r_1\] This is a contradiction, so S must be independent.
To prove the other direction, suppose every element can be uniquely written as \(r_1 m_1 + \dots + r_km_k\). Given any set function \(\varphi : S \to N\), we define \(\theta : M \to N\) by \[\theta(r_1 m_1 + \dots + r_k m_k) = r_1 \varphi(m_1) + \dots + r_k \varphi(m_k)\] This is well-defined by uniqueness, and is clearly a homomorphism. So it follows that S generates M.

Q: Definition. Relations. If M is a finitely-generated R-module, we have shown that there is a surjective R-module homomorphism \(\phi: R^k \to M\). We call [...] the relation module for those generators.
A: \(\ker \phi\)

Q: Definition. Finitely presented module. A finitely-generated module is finitely-presented if [...]
A: we have a surjective homomorphism \(\phi : R^k \to M\) and \(\ker \phi\) is finitely-generated.

Q: Definition. Elementary Row Operations. On an \(m \times n\) maitre A with entires in R are operations of the form:
[...]
A: (ER1) Add \(c \in R\) times the ith row to the jth row. This is done by multiplying by the following matrix on the left
where c appears in the ith column of the jth row
(ER2) Swap the ith and jth rows. This can be done by left multiplication by the permutation matrix
where the off-diagonal entries are in positions (i,j) and (j,i)
(ER3) Multiplying the ith row by a unit \(c \in R\). We do this via the following matrix
Back Extra: Clearly these are all invertible. Column operations are defined by doing right multiplication with the same matrices.

Q: Definition. Two matrices are equivalent if [...]
A: we can get from one to the other via a sequence of elementary row and column operations.
Back Extra: Note that if A and B are equivalent we can write \[B = QAT^{-1}\] for invertible \(Q,T^{-1}\)

Q: Theorem. Smith Normal Form. An \(m \times n\) matrix over a Euclidean domain R is equivalent to a diagonal matrix
with the \(d_i\) all non-zero and \[d_1 | d_2 | d_3 | \dots | d_r\]
A: If A = 0 then done. So suppose \(A \neq 0\). So entry is not zero, say, \(A_{ij} \neq 0\). Swapping the ith and first row, then jth and first column, we arrange that \(A_{11} \neq 0\). We now try to reduce \(A_{11}\) as much as possible. We have the following two possible moves.
(i) If there is an \(A_{1j}\) not divisible by \(A_{11}\), then we can use the Euclidean algorithm to write \[A_{1j} = qA_{11} + r\]
By assumption, \(r \neq 0\). So \(\phi(r) < \phi(A_{11})\). So we do a column operation to subtract q copies of the first column from the jth column. Then in position (1,j), we now have r. We swap the first and jth column such that r is in position (1,1), and we have strictly reduced the value of \(\phi\) at the first entry.
(ii) If there is an \(A_{i1}\) not divisible by \(A_{11}\) we do the same thing, and this again reduces \(\phi (A_{11})\).
We keep performing these until no move is possible. Since the value of \(\phi(A_{11})\) strictly decreases every move, we must finish after finitely many applications. Then we know that we must have \(A_{11}\) dividing all \(A_{1j}\) and \(A_{i1}\). Now we can just subtract appropriate multiples of the first column from other so that \(A_{ij} = 0\) for \(j \neq 1\). We do the same thing with rows so that the first row is cleared. Then we have a matrix of the form
(iii) Suppose there is an entry of C not divisible by d, say \(A_{ij}\) with \(i,j > 1\).
We suppose \[A_{ij} = qd + r\] with \(r \neq 0\) and \(\phi(r) < \phi(d)\). We add column 1 to column j, and subtract q times row 1 from row i. Now we get r in the (i,j)th entry, and we want to send it back to the (1,1) position. We swap row i with row 1, swap column j with row 1, so that r is in the (1,1)th entry, and \(\phi(r) < \phi(d)\).
Now we have messed up the first row and column, so we go back and do (i) and (ii) again until the first row and columns are cleared. Then we get
where \[\phi(d') \leq \phi(r) < \phi(d)\]
As this strictly decreases the value of \(\phi(A_{11})\), we can only repeat this finitely many times. When we stop, we will end up with a matrix
and d divides every entry of C. Now we apply the entire process to C. When we do this process, notice all allowed operations don't change the fact that d divides every entry of C. So applying this recursively, we obtain a diagonal matrix with the claimed divisibility property.

Q: Definition. The [...] are called the invariant factors of A.
A: \(d_k\) obtained in the Smith normal form

Q: Definition. Minor. A \(k \times k\) minor of a matrix A is [...]
A: the determinant of a k x k sub-matrix of A.

Q: Definition. Fitting Ideal. For a matrix A, the kth fitting ideal \(\text{Fit}_k (A) \unlhd R\) is [...]
A: the ideal generated by the set of all \(k \times k\) minors of A.

Q: Lemma. Let A and B be equivalent matrices. Then for all k we have \[\text{Fit}_k(A) = \text{Fit}_k (B)\]
A: It suffices to show that changing A by a row or column operation does not change the fitting ideal. Since taking the transpose doesn't change the determinant, it suffices to consider just row operations.
The most difficult one is taking linear combinations. Let B be the result of adding c times the ith row to the jth row, and fix C a \(k \times k\) minor of A. Suppose that the corresponding minor of B is C'. We then want to show that \(\det C' \in \text{Fit}_k(A)\).
If the jth row is outside of C, then the minor \(\det C\) is unchanged. If both the ith and jth rows are in C, then C' is obtained from C by a row operations, which does not change the determinant.
Suppose the jth row is in C and the ith row is not. Suppose the ith row is \(f_1, dots , f_k\). Then C is changed to C', differing only in the jth row, which is \[(C_{j1} + cf_1, C_{j2} + cf_2, \dots, C_{jk} + cf_k)\] We compute \(\det C'\) by expanding along this row, giving \[\det C' = \det C + c \det D\] where D is the matrix obtained by replacing the jth row of C with \((f_1, \dots , f_k)\). Now \(\det C\) is definitely a minor of A, and \(\det D\) is still a minor of A, just another one. Since ideals are closed under addition and multiplications, it follows that \(\det C' \in \text{Fit}_k (A)\).
The other operations are much simpler, they just follow by standard properties of the effect of swapping rows or multiplying rows on determinants.
So after any row operation the resultant submatrix C' satisfies \(\det C' \in \text{Fit}_k (A)\). Since this is true for all minors, we must have \(\text{Fit}_k (B) \subseteq \text{Fit}_k(A)\). But row operations are invertible. So we must have \(\text{Fit}_k (A) \subseteq \text{Fit}_k (B)\) as well. So they must be equal.

Q: Corollary. If A has Smith normal form
then \[\text{Fit}_k (A) = (d_1 d_2 \dots d_k)\] In particular, \(d_k\) is unique up to associates.
A: We have \(\text{Fit}_k (B) = (d_1 d_2 \dots d_k)\) as the only possible contributing minors are from the diagonal sub matrices, and the minor from the top left square submatrix divides all other diagonal ones. By the previous lemma this is also \(\text{Fit}_k(A)\). The final claim follows since we can find \(d_k\) by dividing a generator of \(\text{Fit}_k(A)\) by a generator of \(\text{Fit}_{k-1}(A)\)

Q: Lemma. Let R be a principal ideal domain. Then any submodule of \(R^m\) is generated by at most m elements.
A: Let \(N \leq R^m\) be a submodule. Consider the ideal \[I = \{r \in R : (r,r_2, r_3, \dots , r_m) \in N \text{ for some } r_2, \dots , r_m \in R\}\] It is clear this is an ideal. Since R is a principal ideal domain, we must have \(I = (a)\) for some \(a \in R\). We now choose an element \[n = (a, a_2, \dots , a_m) \in N\] Now for any \( (r_1 , r_2 , \dots , r_m) \in N\) we know that \(r_1 \in I\), so \(a | r_1\). Thus we can write \(r_1 = ra\). Then we can form \[(r_1 , r_2 , \dots, r_m) - r(a,a_2 , \dots , a_m) = (0, r_2 - ra_2, \dots, r_m - ra_m) \in N\] This lies in \(N' = N \cap (\{0\} \times R^{m-1}) \leq R^{m-1}\). Thus everything in N can be written as a multiple of n plus something in N'. But by induction, since \(N' \leq R^{m-1}\), we know N' is generated by at most \(m-1\) elements. So there are \(n_2 , \dots , n_m \in N'\) generating N'. So \(n, n_2, \dots , n_m\) generate N.

Q: Theorem. Let R be a Euclidean domain, and let \(N \leq R^m\) be a submodule. Then there exists a basis \(v_1 , \dots , v_m\) of \(R^m\) such that N is generated by \(d_1 v_1, d_2 v_2, \dots , d_r v_r\) for some \(0 \leq r \leq m\) and some \(d_i \in R\) such that \(d_1 | d_2 | \dots | d_r\).
A: By the previous lemma, N is generated by some elements \(x_1, \dots , x_n\) with \(n \leq m\). Each \(x_i \) is an element of \(R^m\). So we can think of it as a column vector of length m, and we can form a \(m \times n\) matrix
We can put this in Smith normal form, since there are no more columns than there are rows, this is of the form
Recall that we got to the Smith normal form by row and column operations. Performing row operations is just changing the basis of \(R^m\), while each column operation changes the generators of N.
So what this tells us is that there is a new basis \(v_1, \dots ,v_m\) of \(R^m\) such that N is generated by \(d_1 v_1, \dots ,d_r v_r\). By definition of Smith normal form, the divisibility condition holds.

Q: Corollary. Let R be a Euclidean domain. A submodule of \(R^m\) is free of rank at most m. In other words, the submodule of a free module is free, and of a smaller (or equal) rank.
A: Continuing with the notation in the argument above, we claim that the set \(d_1 v_1, \dots , d_r v_r\) freely generate N. This is because a linear dependence between them would give a linear dependence between the \(v_1 , \dots , v_m\).
A submodule of a free module is free over a Euclidean domain.

Q: Theorem. Classification of finitely-generated modules over a Euclidean domain. Let R be a Euclidean domain, and M be a finitely-generated R modules. Then \[M \cong \frac{R}{(d_1)} \bigoplus \frac{R}{(d_2)} \bigoplus \dots \bigoplus \frac{R}{(d_r)} \bigoplus R \bigoplus R \bigoplus \dots \bigoplus R\]
for some \(d_i \neq 0\) and \(d_1 | d_2 | \dots | d_r\)
A: Since M is finitely generated, there is a surjection \(\phi : R^m \to M\). So by the first isomorphism theorem we have \[M \cong \frac{R^m}{\ker \phi}\]
Since \(\ker \phi\) is a submodule of \(R^m\), by the previous theorem, there is a basis \(v_1 , \dots , v_m\) of \(R^m\) such that \(\ker \phi\) is generated by \(d_1 v_1 , \dots , d_r v_r\) for \(0 \leq r \leq m\) and \(d_1 | d_2 | \dots | d_r\). So \[M \cong \frac{R^m}{((d_1, 0 , \dots, 0), (0, d_2, 0 , \dots , 0), \dots , (0,0, \dots , 0, d_r , 0 , \dots , 0))}\] This is \[\frac{R}{(d_1)} \bigoplus \frac{R}{(d_2)} \bigoplus \dots \bigoplus \frac{R}{(d_r)} \bigoplus R \bigoplus \dots \bigoplus R\] with \(m - r\) copies of R

Q: Corollary. Classification of finitely-generated abelian groups. Any finitely-generated abelian group is isomorphic to \[C_{d_1} \times \dots \times C_{d_r} \times C_{\infty} \times \dots \times C_{\infty}\] where \(C_{\infty} \cong \mathbb{Z}\) is the infinite cyclic group, with \(d_1 | d_2 | \dots | d_r\)
A: Let \(R = \mathbb{Z}\), and apply the classification of finitely-generated R-modules.

Q: Lemma. Chinese Remainder Theorem. Let R be a Euclidean domain, and \(a,b \in R\) be such that \(\text{gcd}(a,b) = 1\). Then \[\frac{R}{(ab)} \cong \frac{R}{(a)} \bigoplus \frac{R}{(b)}\] as R-modules
A: Consider the R-module homomorphism \[\phi : \frac{R}{(a)} \bigoplus \frac{R}{(b)} \to \frac{R}{(ab)}\] \[(r_1 + (a), r_2 + (b)) \mapsto br_1 + ar_2 + (ab)\] To show this is well-defined, suppose \[(r_1 + (a), r_2 + (b)) = (r_1 ' + (a), r_2' + (b))\] Then \[r_1 = r_1' + xa, \; r_2 = r_2' + yb\] So \[br_1 + ar_2 + (ab) = br_1' + xab + ar_2' + yab + (ab) = br_1' + ar_2' + (ab)\] So this is indeed well-defined. It is clear that this is a module map.
We now have to show it is surjective and injective. So far, we have not used the hypothesis that \(\text{gcd}(a,b) = 1\). As \(\text{gcd}(a,b) = 1\), by the Euclidean algorithm we can write \[1 = ax + by\] for some \(x,y \in R\). So we have \[\phi(y + (a), x + (b)) = by + ax + (ab) = 1 + (ab)\] So \(1 \in \text{im} \phi\). Since this is an R-module map, we get \[\phi(r(y + (a), x + (b))) = r \cdot (1 + (ab)) = r+(ab)\] The key fact is that \(R / (ab)\) as an R-module is generated by 1. Thus \(\phi\) is surjective.
Finally, we have to show it is injective, i.e. that the kernel is trivial. Suppose \[\phi(r_1 + (a), r_2 + (b)) = 0 + (ab)\] Then \[br_1 + ar_2 \in (ab)\] So we can write \[br_1 + ar_2 = abx\] for some \(x \in R\). Since \(a | ar_2\) and \(a | abx\), we know \(a | br_1\). Since an and b are coprime, unique factorisation implies \(a | r_1\). Similarly we know \(b | r_2\) \[(r_1 + (a), r_2 + (b)) = (0 + (a), 0 + (b))\] So the kernel is trivial.

Q: Theorem. Primary decomposition theorem. Let R be a Euclidean domain, and M be a finitely-generated R-module. Then \[M \cong N_1 \bigoplus N_2 \bigoplus \dots \bigoplus N_t\] where each \(N_i\) is either R or is R / (p^n) for some prime \(p \in R\) and some \(n \geq 1\).
A: We already know that \[M \cong \frac{R}{(d_1)} \bigoplus \dots \bigoplus \frac{R}{(d_r)} \bigoplus R \bigoplus \dots \bigoplus R\] so it suffices to show that each \(R / (d_i)\) can be written in that form. We let \[d = p_1^{n_1} p_2^{n_2} \dots p_k^{n_k}\] with \(p_i\) distinct primes. So each \(p_i^{n_i}\) is coprime to each other. So by the previous Lemma iterated a few times, we have \[\frac{R}{(d_1)} \cong \frac{R}{(p_1^{n_1})} \bigoplus \dots \bigoplus \frac{R}{(p_k^{n_k})}\]

Q: Lemma. If V is a finite-dimensional vector space, then \(V_{\alpha}\) is a finitely generated \(\mathbb{F}[X]\)-module.
A: If \(v_1, \dots , v_n\) generate V as an \(\mathbb{F}\)-module, i.e. span V as a vector space over \(\mathbb{F}\), then they also generate \(V_{\alpha}\) as an \(\mathbb{F}[X]\)-module since \(\mathbb{F} \leq \mathbb{F} [X]\)

Q: Theorem. Rational Canonical Form. Let \(\alpha : V \to V\) be a linear endmorphism of a finite-dimensional vector space over \(\mathbb{F}\), and \(V_{\alpha}\) be the associated \(\mathbb{F}[X]\)-module. Then \[V_{\alpha} \cong \frac{\mathbb{F}[X]}{(f_1)} \bigoplus \frac{\mathbb{F}[X]}{(f_2)} \bigoplus \dots \bigoplus \frac{\mathbb{F}[X]}{(f_r)}\] with \(f_1 | f_2 | \dots | f_r\). Thus there is a basis for V in which the matrix for \(\alpha\) is the block diagonal
Where \(c(f)\), the companion matrix for the monic polynomial \(f = a_0 + a_1X + \dots + X^r\) is given by
A: We know that \(V_{\alpha}\) is a finitely-generated \(\mathbb{F}[X]\)-module. By the structure theorem for these, we know \[V_{\alpha} \cong \frac{\mathbb{F}[X]}{(f_1)} \bigoplus \frac{\mathbb{F}[X]}{(f_2)} \bigoplus \dots \bigoplus \frac{\mathbb{F}[X]}{(f_r)} \bigoplus 0\]
(There can be no copies of \(\mathbb{F}[X]\) since \(V_{\alpha} = V\) is finite-dimensional over \(\mathbb{F}\) but \(\mathbb{F}[X]\) is not). The divisibility criterion also follows from the structure theorem, then the form of the matrix is immediate.

Q: Lemma. The prim elements of \(\mathbb{C}[X]\) are the \(X - \lambda\) for \(\lambda \in \mathbb{C}\)(up to multiplication by units.)
A: Let \(f \in \mathbb{C}[X]\). If f is constant, then it is either a unit of 0. Otherwise, by the fundamental theorem of algebra, it has a root \(\lambda\). So it is divisible by \(X - \lambda\). So if f is irreducible, it must have degree 1. And clearly everything of degree 1 is prime.

Q: Theorem. Jordan Normal Form. Let \(\alpha : V \to V\) be an endomorphism of a vector space V over \(\mathbb{C}\), and \(V_{\alpha}\) be the associated \(\mathbb{C}[X]\) module. Then \[V_{\alpha} \cong \frac{\mathbb{C}[X]}{((X - \lambda_1)^{a_1})} \bigoplus \frac{\mathbb{C}[X]}{((X-\lambda_2)^{a_2})} \bigoplus \dots \bigoplus \frac{\mathbb{C}[X]}{((X-\lambda_t)^{a_t})} \]
where \(\lambda_i \in \mathbb{C}\) do not have to be distinct. Thus there is a basis of V in which \(\alpha\) has matrix
where
is an m x m matrix.
A: Apply the prime decomposition theorem to \(V_{\alpha}\). Then all primes are of the form \(X - \lambda\). We then use our second example at the beginning of the chapter to get the form of the matrix.