1. 1. Let $R$ be a commutative ring.
      1. Show that if $R$ is an integral domain with the ACCP then $R$ is a factorisation domain.
      2. We say that $x ∈ R$ is irreducible if $⟨x⟩$ is maximal amongst proper principal ideals. State which of the following elements are irreducible in the given rings and briefly justify your answers: $$ 3+2 \sqrt{2} \text { in } ℤ[\sqrt{2}] ;   25 \text { in } ℤ_{10} ;   2 \text { in } ℤ[\sqrt{-5}] $$
      3. Show that if $R$ is a PID then every non-unit in $R$ has an irreducible factor.
      4. Show that $R$ is a field if and only if 0 is irreducible.
   2. Let $\overline{ℤ}$ be the set of $α ∈ ℂ$ for which there is a monic polynomial $p ∈ ℤ[X]$ such that $p(α)=0$.
      1. Show that $α ∈ \overline{ℤ}$ if and only if the ℤ-module $ℤ[α]$ is finitely generated.
      2. Hence show that $\overline{ℤ}$ is a subring of $ℂ$.
         [Hint: you may assume that a submodule of a finitely generated module over a PID is also finitely generated.]
      3. Show that $\overline{ℤ}$ does not contain any irreducible elements.
2. 1. Show that if $R$ is an integral domain with non-zero characteristic $p$ then $p$ is prime and $R$ is a vector space over $𝔽_p$ in such a way that multiplication on $R$ is bilinear.
   2. Show that if $p$ is a prime and $R$ is a ring of order $p^2$ then either $R ≅ ℤ_{p^2}$ or there is a polynomial $q ∈ 𝔽_p[X]$ such that $R ≅ 𝔽_p[X] /⟨q⟩$.
   3. Let $p ≡ 3\pmod4$ be prime.
      1. Show that if $d ∈ 𝔽_p$ is not a square then $d=-x^2$ for some $x ∈ 𝔽_p^*$.
      2. Show that if $q ∈ 𝔽_p[X]$ is a degree 2 irreducible polynomial then $$ 𝔽_p[X] /⟨q⟩ ≅ 𝔽_p[X] /\left\langle X^2+1\right\rangle $$
   4. 1. Show that if $R$ is a Euclidean domain then there is a prime $p ∈ R$ such that if $q: R→R /⟨p⟩$ is the quotient map then $U(q(R))=q(U(R))$. [Hint: consider the minimal values of the Euclidean function.]
      2. Show that $A:=ℝ[X, Y] /\left\langle X^2+Y^2+1\right\rangle$ is not a Euclidean domain. [You may assume that $U(A)=ℝ^*$, and also that the ℝ-vector space $A /⟨p⟩$ is finite-dimensional for any non-zero prime $p ∈ A$.]Quotient of polynomials, PID but not Euclidean domain?
3. 1. 1. Show that if $R$ is a Euclidean domain then every $A ∈ M_{n, m}(R)$ is equivalent by elementary operations to a diagonal matrix.
      2. Show that if $R$ is a commutative ring and $A, B ∈ M_{n, m}(R)$ are both in Smith Normal Form with $A$ equivalent to $B$ then $A_{i, i}$ is an associate of $B_{i, i}$ for all $i$. State clearly any results you use.
      3. Show that if $R=M_2(𝔽)$ for a field 𝔽 and $R^n ≅ R^m$ as $R$-modules then $n=m$.
   2. Let $U, V$ and $W$ be vector spaces over 𝔽 and let $R:=\operatorname{End}_{𝔽}(V)$.
      1. Show that the map $R×L(U, V)→L(U, V)$ which sends $(φ, ψ)$ to $φ ∘ ψ$ is well-defined and gives the commutative group $L(U, V)$ of 𝔽-linear maps $U→V$ the structure of an $R$-module.
      2. Write down an $R$-linear isomorphism $α: L(U, V) ⊕ L(W, V)→L(U ⊕ W, V)$.
      3. Show that if $V=𝔽[X]$ considered as an 𝔽-vector space, then $V$ is 𝔽-linearly isomorphic to $V ⊕ V$.
      4. Deduce that $R ≅ R^2$ as $R$-modules.

Solution

1. 1. 1. Write $ℱ$ for the set of elements in $R^*$ that have factorisation into irreducibles so that all units and irreducible elements are in $ℱ . ℱ$ is closed under multiplication, by design and since $R$ is an integral domain. Were $ℱ$ not to be the whole of $R^*$ then there would be some $x_0 ∈ R^* \backslash ℱ$. Now create a chain iteratively: at step $i$ suppose we have $x_i ∈ R^* \backslash ℱ$. Since $x_i$ is not irreducible and not a unit there is $y_i|x_i$ with $y_i ≁ 1$ and $y_i ≁ x_i$; let $z_i ∈ R^*$ be such that $x_i=y_i z_i$. If $z_i ∼ x_i$, then $z_i ∼ y_i z_i$ and by cancellation $1 ∼ y_i$, a contradiction. We conclude $y_i, z_i ≁ x_i$. Since $ℱ$ is closed under multiplication we cannot have both $y_i$ and $z_i$ in $ℱ$. Let $x_{i+1} ∈\left\{y_i, z_i\right\}$ such that $x_{i+1} ∉ ℱ$; by design $x_{i+1}|x_i$ and $x_{i+1} ≁ x_i$. This process produces a sequence $…|x_2| x_1|x_0$ in which $x_i≁x_{i+1}$ for all $i ∈ ℕ_0$ contradicting the ACCP.
      2. $3+2 \sqrt{2}$ has $3-2 \sqrt{2}$ as a multiplicative inverse so is a unit in $ℤ[\sqrt{2}]$ and hence not irreducible.
         $25 ≡ 5\pmod{10}$ and $⟨5⟩=\{0,5\}$ is maximal amongst proper principal ideals in $ℤ_{10}$ and so irreducible.
         2 is irreducible in $ℤ[\sqrt{-5}]$, since if $2=(a+b \sqrt{-5})(c+d \sqrt{-5})$ then $4=\left(a^2+5 b^2\right)\left(c^2+5 d^2\right)$ and so $b=d=0$, and hence $a= ± 1$ or $c= ± 1$.
      3. (In the notes we prove that a PID has the ACCP, so they may choose to reproduce that and then apply the first part.) Let $x ∈ R \backslash U(R)$. Then $⟨x⟩$ is proper and so by Krull's Theorem it is contained in a maximal ideal $I$. Since $R$ is a PID $I=⟨d⟩$, and in particular $⟨d⟩$ is maximal amongst proper principal ideals so $d$ is irreducible, and since $x ∈ I=⟨d⟩$ we have $d|x$ as required.
      4. If $R$ is a field then the only ideals are $\{0\}$ and $R$ and so $\{0\}$ is maximal amongst proper principal ideals, and hence 0 is irreducible.
         On the other hand if $\{0\}$ is maximal amongst proper principal ideals and $x ∈ R^*$ then $\{0\} ⊊⟨x⟩$ and so by maximality $⟨x⟩=R$. Since $R$ is commutative there must be $y ∈ R$ such that $x y=1$, and again since $R$ is commutative it is a field.
   2. 1. For the first part, 'only if' follows since $1, α, α^2, …$ generate $ℤ[α]$ as a ℤ-module, but the degree $d$, say, monic $p$ of which α is a root gives an inductive way of writing $α^i$ as a ℤ-linear combination of $1, α, …, α^{d-1}$ for $i⩾d$.
         For 'if', suppose that $p_1, …, p_k ∈ ℤ[α]$ generate $ℤ[α]$ as a ℤ-module. Then $α^d$ is a ℤ-linear combination of $p_1, …, p_k$ for all $d ∈ ℕ_0$ and in particular for some $d>\max \left\{\deg p_1, …, \deg p_k\right\}$. This gives a monic satisfied by α as required.
      2. Suppose that $α, β ∈ \overline{ℤ}$. Then there are generators $p_1, …, p_k ∈ ℤ[α]$ and $q_1, …, q_m ∈ ℤ[β]$. But $ℤ[α+β]$ and $ℤ[α β]$ are both contained in the ℤ-module generated by $α^i β^j$ for $i, j ∈ ℕ_0$, which in turn is generated by the finite set $\left\{p_i q_j: 1 ⩽ i ⩽ k, 1 ⩽ j ⩽ m\right\}$. Hence $ℤ[α+β]$ and $ℤ[α β]$ are submodules of a finitely generated ℤ-module, and so themselves finitely generated. Finally, $ℤ[α]=ℤ[-α]$ and 1 is a root of $X-1$ and so $\overline{ℤ}$ is a ring by the subring test.
      3. If $α ∈ \overline{ℤ}^*$ is irreducible, then α is a root of some monic $p(X)$, so $\sqrt{α}$ is a root of $p\left(X^2\right)$. Since α is irreducible, $α≁1$, but since $α=\sqrt{α} \sqrt{α}$ we have $\sqrt{α} ∼ 1$ or $\sqrt{α} ∼ α$. If $\sqrt{α} ∼ 1$ then $α ∼ 1$, a contradiction; if $\sqrt{α} ∼ α$, then $\sqrt{α} ∼ 1$ again contradiction.
         If 0 is irreducible, by a(iv) we conclude that $\overline{ℤ}$ is a field. However, 2 does not have an inverse in ℤ since if $2 α=1$ and $p ∈ ℤ[X]$ is a monic then\begin{align*}2^d p(α)&=(2 α)^d+2 q(2 α)\text{ for some }q ∈ ℤ[X]\\&=1+2 q(1)\text{ is odd integer}\end{align*}so $p(α) ≠ 0$. Hence 0 is not irreducible.
2. 1. Let $χ_R: ℤ→R$ be the unique homomorphism from the integers, and suppose that $R$ has characteristic $p$. If $p=a b$ for $a, b⩾1$ then $0_R=χ_R(p)=χ_R(a) χ_R(b)$, and since $R$ is an integral domain we conclude that $χ_R(a)=0$ or $χ_R(b)=0$; say the former. Then by definition of characterstic, $a⩾p$ and so $a=p$ and $b=1$. We conclude that $p$ is prime. The kernel of $χ_R$ contains $p$ and is an ideal in ℤ. Since ℤ is a PID it has the form $⟨N⟩$ for some $N ∈ ℕ_0$, but then $N|p$, whence $N=1$ or $N=p$. If $N=1$ then $1_R=χ_R(1)=χ_R(0)=0_R$ contradicting the non-triviality of $R$. We conclude that $N=p$ and the ring $ℤ /⟨p⟩$ is the field $𝔽_p$ which is a field. By the First Isomorphism Theorem there is then an injective ring homomorphism $𝔽_p→R$ which induces an $𝔽_p$-vector space structure on the additive group of $R$ in such a way that right multiplication is 𝔽-linear. Since multiplication is commutative, it is 𝔽-bilinear.
   2. The additive order of 1 must divide $p^2$. It cannot be 1 since the ring is not trivial. If it is $p^2$ then $R ≅ ℤ_{p^2}$. The characteristic $p$ case is what remains. In this case $R$ is a vector space over $𝔽_p$, and for reasons of size must have a basis of size 2. Take $1 ∈ R$ which is non-zero and extend this to an $𝔽_p$-basis by some element $x$. Let $a, b ∈ 𝔽_p$ be such that $x^2=a x+b$, then the map $𝔽_p[X]→R ; f↦f(x)$ is a surjective ring homomorphism. The kernel contains $\left\langle X^2-a X-b\right\rangle$, and since $𝔽_p[X] /\left\langle X^2-a X-b\right\rangle$ is 2-dimensional over $𝔽_p, R$ is 2-dimensional, and the given homomorphism is $𝔽_p$-linear.
   3. 1. The map $𝔽_p^*→𝔽_p^* ; x↦x^2$ is a homomorphism of the multiplicative group, and its image has index at most 2 since degree 2 polynomials over an integral domain have at most 2 roots, and at least 2 since $-1$ is not a square modulo $p$ for congruence reasons. Since cosets partition a group, if $Q$ are the quadratic residues in $𝔽_p^*$ then $-Q$ is the set of non-residues as required.
      2. Since $q$ is a quadratic there are $a, b, c ∈ 𝔽_p$ with $a ≠ 0$ such that $q(X)=a X^2+b X+c$ by completing the square (since $p$ is odd) $q(X)=a\left((X-b / 2 a)^2+Δ\right)$ for $Δ=c-b^2 / 4 a^2$. Since $q$ is irreducible it has no root so $-Δ$ is not a square, so by $\mathrm{c}(\mathrm{i})$ we have $Δ=d^2$ for $d ≠ 0$. Hence $q(X)=a d^2\left((X / d-b / 2 a d)^2+1\right)$. Dilating ideals by a unit does not change them so the map $𝔽_p[X] /⟨q⟩→𝔽_p[X] /\left\langle X^2+1\right\rangle;X↦d X+b / 2 a$ is an isomorphism.
   4. 1. Let $f$ be a Euclidean function on $R$ and $p ∈ R$ have $f(p)$ minimal over all nonzero non-units. Then if $x ∈ R^* \backslash U(R)$, then either $p|x$ or there is $r ∈ R^*$ with $x=b p+r$ and $f(r)< f(p)$. By minimality of $f(p)$ we have $r ∈ U(R)$ and hence $x+⟨p⟩ ∈ U(q(R))$. It follows that $U(q(R)) ∩ q\left(R^* \backslash U(R)\right) ⊂ q(U(R))$, and hence $U(q(R)) ⊂ q(U(R))$. Units remain units under quotienting which is the other direction. Finally, $p$ is prime, because it is not a unit, and if $p|x y$ and $p≁x$ then by the above $p ∣(b p+r) y$ for $r ∈ U(R)$, but then $p|r y| y$ as required.
      2. Let $p$ be as in $\mathrm{d}(\mathrm{i})$. Then $A /⟨p⟩$ contains elements $x, y$ with $x^2+y^2+1=0$. It is also an integral domain that is finite dimensional over $ℝ$ and so $A /⟨p⟩$ is a field and hence $A /⟨p⟩=U(A /⟨p⟩) ∪\{0\}=ℝ^* ∪\{0\}=ℝ$, but there are no $X, Y ∈ ℝ$ with $X^2+Y^2+1=0$.
3. 1. 1. Let $𝒜_k$ be those matrices $B ∼_ℰ A$ with the additional property that whenever $i< k$ and $j ≠ i$, or $j< k$ and $i ≠ j$, we have $B_{i, j}=0$. We shall show by induction that $𝒜_k$ is non-empty for $k ⩽ \min \{m, n\} ; 𝒜_1$ contains $A$ and so is certainly non-empty. Let $f$ be a Euclidean function for $R$, and suppose that $𝒜_k ≠ ∅$ and $k<\min \{m, n\}$. Let $B ∈ 𝒜_k$ be a matrix with $f\left(B_{k, k}\right)$ minimal. First we show that $B_{k, k}|B_{k, i}$ for all $i>k$ : if not, there is some $i>k$ with $B_{k, i}=q B_{k, k}+r$ with $f(r)< f\left(B_{k, k}\right)$ and we apply the elementary operations $c_i↦c_i-c_k q$ and $c_k ↔ c_i$ to get a matrix $B'∈ 𝒜_k$ with $B_{k, k}'=B_{k, i}-q B_{k, k}=r$, but $f\left(B_{k, k}'\right)=f(r)< f\left(B_{k, k}\right)$ which contradicts the minimality in our choice of $B$. Similarly, but with row operations in place of column operations, $B_{k, k}|B_{i, k}$ for all $i>k$. For $k< i⩽ m$ let $q_i$ be such that $B_{k, i}=B_{k, k} q_i$. Apply elementary column operations $c_{k+1}↦c_{k+1}-c_k q_{k+1}, …, c_m↦c_m-c_k q_m$ to get a matrix $B'$. For $k< i⩽ n$ let $p_i$ be such that $B_{i, k}=p_i B_{k, k}$. Apply elementary row operations $r_{k+1}↦r_{k+1}-p_{k+1} r_k, …, r_n↦r_n-p_n r_k$ to $B'$ to get a matrix $B^{\prime \prime}$. Then $B^{\prime \prime} ∼_{ℰ} B'∼_{ℰ} B ∼_{ℰ} A$ and $B^{\prime \prime} ∈ 𝒜_{k+1}$. The inductive step is complete. It follows that $𝒜_{\min \{m, n\}} ≠ ∅$; any $B$ in this set is diagonal and equivalent to $A$.
      2. This is on problem sheet 4. I am expecting them to quote the uniqueness theorem and the fact that equivalent matrices produce isomorphic presentations, so $R^m / \operatorname{Im} L_A ≅ R^m / \operatorname{Im} L_B$, and since both $A$ and $B$ are diagonal with, say, entries $a_1, …, a_k$ and $b_1, …, b_k$ (where $k=\min \{n, m\}$) we have $\operatorname{Im} L_A=\left\langle a_1\right\rangle×⋯×\left\langle a_k\right\rangle×\{0\}×⋯×\{0\}$ where there are $m-k$ copies of $\{0\}$. Moreover, $a_1|a_2|⋯|a_k$ so $\left\langle a_1\right\rangle ⊃ ⋯ ⊃\left\langle a_k\right\rangle ⊃⟨0⟩ ⊃ ⋯ ⊃⟨0⟩$. Similarly for $\operatorname{Im} L_B$, and it follows by the uniqueness theorem that $a_i ∼ b_i$ for all $i$ as required.
      3. If $R=M_2(𝔽)$ then $R$ is $𝔽$-linearly isomorphic to $𝔽^4$, and if $R^n$ is $R$-linearly isomorphic to $R^m$ then the underlying $𝔽$-vector spaces are $𝔽$-linearly isomorphic, and so $𝔽^{4 n}$ is $𝔽$-linearly isomorphic to $𝔽^{4 m}$ and hence $4 n=4 m$ and so $n=m$.
   2. 1. The map is well-defined because the composition of linear maps is linear. The multiplicative identity in $\operatorname{End}_{𝔽}(V)$ is the identity function and so $1_R ⋅ ψ=ψ ; φ ∈ \operatorname{End}_{𝔽}(V)$ is additive and so $(φ ⋅(ψ+π))(x)=φ(ψ(x)+π(x))=φ(ψ(x))+φ(π(x))=(φ . ψ)(x)+(φ . π)(x)$ for all $x ∈ V ;\left(\left(φ+φ'\right) . ψ\right)(x)=\left(φ+φ'\right)(ψ(x))=φ(ψ(x))+φ'(ψ(x))=(φ ⋅ ψ)(x)+\left(φ'. ψ\right)(x)$ for all $x ∈ V$; and finally $\left(φ ∘ φ'\right) . ψ=\left(φ ∘ φ'\right) ∘ ψ=φ ∘\left(φ'∘ ψ\right)=φ ⋅\left(φ'. ψ\right)$ since functional composition is associative.
      2. In the first case, the map $α: L(U, V) ⊕ L(W, V)→L(U ⊕ W, V) ;(φ, ψ)↦(u+w↦φ(u)+ψ(w))$ is a well-defined $R$-linear isomorphism.
      3. The map $β: 𝔽[X] ⊕ 𝔽[X]→𝔽[X] ; p↦p\left(X^2\right)+q\left(X^2\right) X$ is an 𝔽-linear bijection.
      4. Let $U=W=V$ in the isomorphism α and note that the map $R→R^2 ; φ↦α^{-1}(φ ∘ β)$ is well-defined and an $R$-linear bijection and the claim is proved.