Rings and modules course of the University of Oxford (A3)

Solutions to all three questions of the University of Oxford Part A paper on rings and modules, sat on 23 June 2023. Source: Oxford-released examination PDF, parent PDF pages 229–231; paper pages 1–3.

Multiplication in a direct product of rings is coordinatewise. Thus $(r_1,\dots ,r_n)$ has a two-sided inverse exactly when each $r_j$ has a two-sided inverse in $R_j$, and thenConsequently its unit group is the direct product of groupsThe identification preserves multiplication because both sides use coordinatewise multiplication. This is the unit group of a direct product of rings formula.

Write $T={\prod }_{j=1}^n{R}_j$, and let ${\rho }_j:T\to R_j$ be the coordinate ring homomorphism. Set $I_j={\rho }_j(I)$. Each $I_j$ is an ideal: it is an additive subgroup, and multiplying a lift in $I$ on either side by a tuple with prescribed $j$th coordinate proves closure under left and right multiplication by $R_j$.

Certainly $I\subseteq {\prod }_j{I}_j$. Conversely, let $a=(a_1,\dots ,a_n)\in {\prod }_j{I}_j$. For each $j$, choose $b^{(j)}\in I$ with $j$th coordinate $a_j$, and let $e_j$ have $1_{R_j}$ in coordinate $j$ and zero elsewhere. Since $I$ is an ideal, $e_j{b}^{(j)}\in I$, andThereforeThe finite sum is essential to this proof of ideals of a finite product of rings.

The element $-1_R$ lies in the unit group and satisfies $(-1_R{)}^2=1_R$. Its order of a group element divides two and, by Lagrange's theorem, divides the odd number $|U(R)|$. Its order is therefore one, so $-1_R=1_R$ and $2\cdot 1_R=0$.

It follows thatis a well-defined unital ring homomorphism. For a nonzero ring, this says its characteristic of a ring is two. If the zero ring is permitted, the same map still exists. This is the odd unit group forces characteristic two argument, with that harmless exception to the characteristic wording.

Since the irreducible element $x$ is not a unit, the principal ideal $(x)$ is proper. Suppose $(x)\subseteq J\subseteq R$. As $R$ is a principal ideal domain, write $J=(d)$. Then $x=dr$ for some $r\in R$. The definition of an irreducible element says that $d$ or $r$ is a unit.

If $d$ is a unit, then $J=R$. If $r$ is a unit, then $d$ and $x$ are associate elements, so $J=(x)$. Thus $(x)$ is a maximal ideal. By the permitted criterion that a quotient ring by a maximal ideal is a field,

Every multiple of $x_1\cdots x_r$ is a multiple of each $x_j$, giving one inclusion. For the reverse inclusion, a principal ideal domain is a unique factorization domain, so each irreducible element $x_j$ is a prime element.

Let $a\in {\bigcap }_{j=1}^r(x_j)$. The case $a=0$ is immediate. Otherwise, write $a=x_{1}b$. Since $x_2\mid a$ and the prime element $x_2$ cannot divide the non-associate irreducible element $x_1$, it divides $b$. Write $a=x_1{x}_2{b}_2$, and repeat. At stage $j$, the prime element $x_j$ divides none of $x_1,\dots ,x_{j-1}$, so it divides the remaining factor. Hence $x_1\cdots x_r\mid a$, proving

The polynomial ring ${\mathbb{F}}_2[X]$ is a euclidean domain under polynomial degree, hence a principal ideal domain. By part (b)(i), eachis a field. By polynomial division, every residue class has exactly one representative of polynomial degree less than $n_j$. Thus $1,\overline{X},\dots ,{\overline{X}}^{n_j-1}$ form a basis over ${\mathbb{F}}_2$, and $|K_j|=2^{n_j}$.

For $j\ne k$, the ideals $(f_j)$ and $(f_k)$ are distinct maximal ideals, so they are comaximal ideals: their sum is an ideal properly containing $(f_j)$ and must be the whole ring. The Chinese remainder theorem for ideals and part (b)(ii) therefore give the ring isomorphismIn particular, this is a product of finite fields of the required sizes. The map preserves multiplication as well as addition and is surjective by the Chinese remainder theorem for ideals.

Choose a generator $u$ of the cyclic group $U(R)$, so $u^p=1$. Part (a)(iii) gives a unital ring homomorphism ${\mathbb{F}}_2\to R$. Its image lies in the centre of $R$, so even if $R$ is not a commutative ring, evaluating a polynomial at $u$ defines a ring homomorphismBecause $\varphi (X^p-1)=0$, it factors through the quotient ring to giveIts image is the subring $S={\mathbb{F}}_2[u]$. Every unit of $S$ is a unit of $R$, because an inverse in $S$ also lies in $R$. Conversely, every member of $U(R)$ is $u^j$ for some $j$, and both $u^j$ and its inverse $u^{p-j}$ lie in $S$. HenceThe inverse argument matters: merely belonging to a subring would not by itself make an ambient unit invertible there.

Using the assumed square-free polynomial property, factorinto distinct monic polynomials that are irreducible polynomials, with $n_j=\mathrm{deg}{f}_j$. Part (b)(iii) identifies $T={\mathbb{F}}_2[X]/(X^p-1)$ with ${\prod }_j{K}_j$, where $|K_j|=2^{n_j}$.

By the first isomorphism theorem for rings, the image $S$ from part (c)(i) is $T/J$, where $J$ is the kernel of a ring homomorphism. By part (a)(ii), $J={\prod }_j{J}_j$ for ideals $J_j\subseteq K_j$. A field has only the zero ideal and the whole field as ideals, so each $K_j/J_j$ is either $K_j$ or the zero ring. Consequently, for a subset $D\subseteq \{1,\dots ,r\}$,This is the quotient of a finite product of fields principle. The subset is nonempty because $|U(S)|=p>2$, whereas the zero ring has only one unit. By part (a)(i),Each factor is a positive integer. Since $p$ is a prime number, exactly one factor is $p$ and all remaining factors are one. Thusso $p$ is a Mersenne prime. If $n=ab$ with $a,b>1$, then $2^a-1$ would be a proper divisor of $2^{ab}-1$; thus $n$ is itself a prime number, although only the displayed form is needed.

The assumed square-free polynomial property can also be checked directly: in characteristic of a ring two, the formal derivative of $X^p-1$ is $X^{p-1}$ because $p$ is odd. These two polynomials have greatest common divisor one, since $X$ does not divide $X^p-1$, so no repeated irreducible polynomial factor occurs.

A subset $E\subseteq M$ is a basis of a module if every $m\in M$ has a unique expressionUniqueness concerns the entire family of coefficients, including the zero coefficients. Equivalently, $E$ generates $M$ and is linearly independent in a module: every finite relation ${\sum }_{e\in H}{r}_{e}e=0$, with $H\subseteq E$, has all $r_e=0$.

A module is a free module if it has a basis of a module. An infinite basis of a module still uses only finite sums in each expression; arbitrary infinite sums are not part of the module operations.

Yes. An $\mathbb{F}$-module over a field is a vector space. Start with a finite generating list. If it is linearly dependent, one vector has a nonzero coefficient in a linear relation; dividing by that coefficient expresses the vector using the others, so it can be removed without changing the span. Repeating terminates with a linearly independent generating list, which is a basis and therefore a basis of a module.

Thus every finitely generated module over a field is a free module. For the zero module, the empty list is the required basis of a module.

In the direct sum of modules $M=N\oplus P$, every $m\in M$ has a unique expression $m=n+p$, with $n\in N$ and $p\in P$. DefineThe uniqueness makes this well-defined. For $m=n+p$, $m^{\prime }=n^{\prime }+p^{\prime }$ and $r\in R$,so $\pi$ is a module homomorphism. Finally, $\pi (n+p)=0$ exactly when $p=0$, equivalently $m\in N$. Therefore $\boxed{\ker \pi =N}$. Its image is $P$, and ${\pi }^2=\pi$, so it is the projection onto that direct summand.

Put $N=⟨E^{\prime }⟩$ and $P=⟨E^{\prime \prime }⟩$, the submodules generated by the two parts of the basis of a module. Every finite expression in the basis of a module splits into its $E^{\prime }$ terms and its $E^{\prime \prime }$ terms. Thus $M=N+P$.

If $m\in N\cap P$, there are finite expressionsSubtracting gives a finite relation among distinct members of $E$, because $E^{\prime }\cap E^{\prime \prime }=\varnothing$. Their linear independence in a module forces every coefficient to vanish, so $m=0$. HenceThis proves the basis partition gives a module direct sum lemma, also when one part is empty.

Choose a countable generating family $(n_k)$ for $N$. For each $k$, the expression of $n_k$ in the basis of a module $E$ has a finite set $H_k$ of nonzero coefficients, its coordinate support in a free module. LetBy the countable union of countable sets property, $E^{\prime }$ is a countable set. Every $n_k$ belongs to $⟨E^{\prime }⟩$, so all finite $R$-linear combinations of the $n_k$ do too. ThereforeHere and below, countable includes finite. The same construction applies to a finite or empty generating family. This is the countable basis support lemma.

For $j\in {\mathbb{N}}_0$, let $e_j\in B$ have $e_j(j)=1$ and $e_j(k)=0$ for $k\ne j$. If $f,g\in F$ and $z\in \mathbb{Z}$, then $f+g$ can be nonzero only where $f$ or $g$ is nonzero, and $zf$ can be nonzero only where $f$ is nonzero. These are finite sets. Since $0\in F$, this proves that $F$ is a submodule.

Every $f\in F$ has the finite expressionEvaluating at coordinate $k$ recovers the coefficient of $e_k$, so the expression is unique. Thus $\{e_j:j\in {\mathbb{N}}_0\}$ is a basis of a module for $F$, andis a free module. By contrast, the ambient module of integer sequences $B={\prod }_{j\in {\mathbb{N}}_0}\mathbb{Z}$ allows arbitrary integer sequences.

Under the temporary assumption that $E$ is a basis of a module for $B$, part (a)(v) applied to the countably generated submodule $F$ gives a countable set $E^{\prime }\subseteq E$ such that $F\subseteq C=⟨E^{\prime }⟩$. By part (a)(iv),Use part (a)(iii) to project onto the second direct summand. This gives a module homomorphism $\pi :B\to B$ withIt remains to check the required size of this kernel. Enumerate $E^{\prime }$ as $b_1,b_2,\dots$, with the finite case treated the same way. Every element of $C$ is a finite sum of these elements with integer coefficients. For each $m$, the set $⟨b_1,\dots ,b_m⟩$ is the image of the countable set ${\mathbb{Z}}^m$, so it is countable. Taking their countable union provesThis is the countability of countably generated abelian groups argument; it uses the countability of the scalar ring $\mathbb{Z}$.

ConsiderThe map sending a sequence of signs $({\epsilon }_j)$ to $({\epsilon }_j{2}^j)$ is a bijection from $\{-1,1{\}}^{{\mathbb{N}}_0}$ onto $W$. This is an uncountable set by the Cantor diagonal argument: given a proposed list $w_0,w_1,\dots$ of its elements, define $w(j)=-w_j(j)$. Then $w\in W$ differs from $w_j$ at coordinate $j$ for every $j$.

Since $\ker \pi$ is countable, $W$ cannot be contained in it. Consequently there exists

For every $i\in {\mathbb{N}}_0$, truncate $w$ before coordinate $i$:This $f_i$ lies in $F$. For $j<i$, $(w-f_i)(j)=0$. For $j\ge i$, $(w-f_i)(j)=w(j)=\pm 2^j$, which is divisible by $2^i$. In particular the integer sequencebelongs to $B$ and satisfies $w=f_i+2^i{g}_i$.

Apply the module homomorphism $\pi$. Since $f_i\in F\subseteq \ker \pi$,Fixing any coordinate $j$, the integer $\pi (w)(j)$ is therefore divisible by $2^i$ for every $i$. The only such integer is zero: if $a\ne 0$, choose $i$ with $2^i>|a|$, which rules out $a=2^{i}b$ for any nonzero integer $b$. Thus every coordinate of $\pi (w)$ vanishes, contradicting part (b)(iii).

It follows thatThe decisive property is vanishing of infinitely divisible integer sequences, namely ${\bigcap }_{i\ge 0}{2}^{i}B=\{0\}$.

Since $A^2+I=0$, its minimal polynomial $m_A$ divides $X^2+1$. The polynomial $X^2+1$ is an irreducible polynomial over the rational numbers: a reducible quadratic would have a rational root, whereas $r^2+1>0$ for every $r\in \mathbb{Q}$.

Because $n>0$, the constant polynomial one evaluates to the nonzero identity matrix $I$, so $m_A\ne 1$. Both $m_A$ and $X^2+1$ are monic polynomials; hence the only remaining divisor is

The nonconstant invariant factors of a linear operator satisfy $d_1\mid \cdots \mid d_k=m_A$. By part (a)(i), $m_A=X^2+1$ is an irreducible polynomial, so every $d_j$ must equal $X^2+1$. Their polynomial degrees sum to $n$, giving $n=2k$.

The companion matrix of $X^2+1$, in the basis $(1,X)$ of $\mathbb{Q}[X]/(X^2+1)$, isTherefore, for each fixed even $n$, every such $A$ has rational canonical formThere are no such matrices when $n$ is odd. This is an instance of the rational canonical form with irreducible minimal polynomial principle; equality of minimal polynomials alone does not determine rational canonical form in general.

Let $v\in {\mathbb{Q}}^2$ be nonzero. If the columns $v,Av$ were linearly dependent over the rational numbers, then $Av=\lambda v$ for some $\lambda \in \mathbb{Q}$. Applying $A$ once more would giveso $({\lambda }^2+1)v=0$. Since $v\ne 0$, this forces ${\lambda }^2=-1$, impossible in $\mathbb{Q}$. Thus the two columns are linearly independent, and the square matrixis an invertible matrix.

In the basis $(v,Av)$, the linear map given by $A$ sends $v$ to the second basis vector and sends $Av$ to $A^{2}v=-v$. Equivalently,Multiplying on the left by the matrix inverse $E^{-1}$ givesThis change of basis produces the companion matrix found in part (a)(ii).

The Gaussian integers form an integral domain, since they are a subring of the complex numbers. Use the Gaussian integer normwhich is a positive integer for nonzero $a+bi$ and satisfies $N(\alpha \beta )=N(\alpha )N(\beta )$.

For $\alpha ,\beta \in \mathbb{Z}[i]$ with $\beta \ne 0$, write $\alpha /\beta =x+iy$. Choose integers $m,n$ with $|x-m|\le 1/2$ and $|y-n|\le 1/2$, put $q=m+in$, and set $r=\alpha -\beta q$. ThenThus division with remainder has either $r=0$ or a strictly smaller Gaussian integer norm. This proves that $N$ is a euclidean function and

Over a principal ideal domain $D$, an $m$-by-$n$ matrix is in Smith normal form if all entries off the main diagonal vanish and its nonzero diagonal entries $d_1,\dots ,d_r$ occur first and satisfyAll remaining diagonal entries are zero. The zero matrix is allowed, with $r=0$. For an arbitrary matrix $M$, the Smith normal form theorem gives invertible matrices $P\in {\mathrm{GL}}_m(D)$ and $Q\in {\mathrm{GL}}_n(D)$ withwhere the right side is interpreted as a rectangular diagonal matrix when necessary. The $d_j$ are determined up to multiplication by units. Positivity can be required over $\mathbb{Z}$, but is not a condition over the Gaussian integers.

Regard $A$ as a matrix over the rational numbers. Part (a)(ii), with $n=2$, gives its characteristic polynomial $X^2+1$. Therefore, writing $i$ for the imaginary unit,The matrix $M=iI-A$ is not zero: otherwise $A=iI$ would have noninteger diagonal entries. It consequently has rank of a matrix one over the fraction field $\mathbb{Q}(i)$ of the Gaussian integers.

By part (b)(i), $\mathbb{Z}[i]$ is a euclidean domain, hence a principal ideal domain, so apply Smith normal form. There exist $P,Q\in {\mathrm{GL}}_2(\mathbb{Z}[i])$ such thatIndeed, invertible multiplication preserves rank of a matrix, so exactly one diagonal entry is nonzero. For $e_2=(0,1{)}^{\mathsf{T}}$, multiplication on the right gives $P(iI-A)Q{e}_2=0$. Since $P$ is an invertible matrix,The Smith normal form supplies a kernel basis lemma ensures that the null vector $Q{e}_2$ is a column of an invertible matrix over $\mathbb{Z}[i]$; an arbitrary nonzero solution over $\mathbb{Q}(i)$ would not itself establish that requirement.

Each entry of the matrix $Q$ from part (b)(iii) has unique real and imaginary integer coefficients. WriteIt already belongs to the general linear group ${\mathrm{GL}}_2(\mathbb{Z}[i])$. Expanding the relation from part (b)(iii) givesThe real and imaginary parts must both vanish, soThese are the required identities, with $e_2=(0,1{)}^{\mathsf{T}}$.

Solutions to all three questions of the University of Oxford Part A paper on rings and modules, sat on 23 June 2021. Source: Oxford-released examination PDF, parent PDF pages 223–225; paper pages 1–3.

We prove the stated Eisenstein criterion directly. Assume $n\ge 1$. Suppose $f=gh$ in the polynomial ring $R[X]$. Since the principal ideal domain $R$ is an integral domain, polynomial degree is additive under polynomial multiplication, and the product of the leading coefficients of $g,h$ is the leading coefficient $1$ of $f$. These two coefficients are therefore units. Multiplying $g$ and $h$ by inverse units, without changing their product, makes both factors monic polynomials.

If either factor is a constant polynomial, that factor is now $1$, so the original factor was a unit. It remains to rule out $r=\mathrm{deg}g>0$ and $s=\mathrm{deg}h>0$.

The prime element $p$ generates a maximal ideal in the principal ideal domain $R$. Indeed, if $(p)\subseteq (d)$, then $p=de$, and the irreducible element property of $p$ forces either $(d)=R$ or $(d)=(p)$. Consequently $k=R/(p)$ is a field. Apply reduction modulo an ideal to every coefficient. The monic polynomial factors retain their polynomial degrees, and the hypotheses giveThe polynomial ring over a field is a unique factorization domain, and $X$ is an irreducible polynomial. Thus its only monic polynomial divisors in $X^n$ are powers of $X$. Hence $\overline{g}=X^r$ and $\overline{h}=X^s$. Both have zero constant term, so $p\mid g(0)$ and $p\mid h(0)$. This gives $p^2\mid g(0)h(0)=a_0$, a contradiction.

The monic polynomial $f$ has positive polynomial degree, so it is a nonzero nonunit. Every factorization therefore has a unit factor, proving

Regard $\mathbb{R}[X,Y]$ as the polynomial ring $(\mathbb{R}[X])[Y]$. The coefficient ring $\mathbb{R}[X]$ is a Euclidean domain, hence a principal ideal domain. The element $X$ is a prime element: the quotient ring $\mathbb{R}[X]/(X)\cong \mathbb{R}$ is an integral domain.

The monic polynomial $Y^2+X$ has its other coefficients divisible by $X$, while $X^2\nmid X$. Part (a)(i), with $X$ playing the role of $p$ and $Y$ the indeterminate, proves that $Y^2+X$ is an irreducible polynomial.

Now use the automorphism of a polynomial ringIts inverse fixes $X$ and sends $Y$ to $Y-1$. A ring isomorphism preserves units and factorizations, hence preserves irreducible elements. Sincewe obtainApplying the Eisenstein criterion directly to the second expression with the prime element $X$ would fail, since $X\nmid 2$; the change of indeterminate supplies the required argument.

All principal ideals in this part are taken inside the subring $T$, rather than inside $\mathbb{R}[X]$. First, $T$ is a subring: sums and additive inverses retain zero coefficient of $X$, and the coefficient of $X$ in a product $uv$ is $u(0)[X]v+v(0)[X]u=0$. It contains $1$.

The principal ideal $I=(X^3{)}_T=X^{3}T$ is preciselyIndeed, multiplying an element of $T$ by $X^3$ gives a possible term of polynomial degree $3$ and terms of polynomial degrees at least $5$, with no $X^4$ term. Conversely, dividing any displayed polynomial by $X^3$ gives a polynomial with no $X$ term, hence an element of $T$.

Given $p\in T$, let $a,b,c$ be its coefficients of $1,X^2,X^4$, respectively. Subtracting these terms leaves an element of $I$, soFor uniqueness, if two such representatives differ by an element of $I$, their difference has possible monomials only $1,X^2,X^4$, whereas every element of $I$ has zero coefficients at all three. Comparing coefficients gives $a=a^{\prime }$, $b=b^{\prime }$, $c=c^{\prime }$.

In particular, $X^4\notin I$. Treating $I$ as the principal ideal of $\mathbb{R}[X]$ instead would incorrectly discard this surviving term.

Set $I=(X^3{)}_T$ and $J=(Y^3{)}_{\mathbb{R}[Y]}$. Part (b)(i) gives a unique representative $a+b{X}^2+c{X}^4$ for each element of the quotient ring $T/I$. By polynomial division, each element of the quotient ring $\mathbb{R}[Y]/J$ likewise has a unique representative $a+bY+c{Y}^2$. Thus the proposed map $\Phi$ is well-defined and a bijection.

It preserves addition and $1$. To verify multiplication, all terms of polynomial degree at least $5$ lie in $I$. HenceIn the other quotient ring, $Y^3=0$, giving exactly the same coefficient formula:Thereforeis a ring isomorphism. Both sides are examples of a truncated polynomial ring; the generator on the left is the class of $X^2$, whose square is nonzero and whose cube vanishes.

No such property exists. The ring isomorphism in part (b)(ii) identifies quotient rings by elements having different irreducible element behaviour.

The element $X^3$ is an irreducible element of $T$. It is nonzero and a nonunit. Every nonzero nonunit of $T$ has polynomial degree at least $2$: the only possible polynomials of smaller polynomial degree are constant polynomials, and every nonzero constant polynomial is a unit. If $X^3=uv$ with $u,v\in T$ both nonunits, additivity of polynomial degree in the integral domain $\mathbb{R}[X]$ would givewhich is impossible. Thus one factor must be a unit.

In $\mathbb{R}[Y]$, however, $Y^3=Y\cdot Y^2$ is a product of two nonunits, so $Y^3$ is not an irreducible element. If a ring isomorphism invariant $P$ characterized irreducible elements by the indicated quotient ring, the irreducible element $X^3\in T$ would imply that $T/(X^3{)}_T$ has $P$. Its isomorphic quotient ring $\mathbb{R}[Y]/(Y^3)$ would then have $P$, incorrectly making $Y^3\in \mathbb{R}[Y]$ an irreducible element.

Consequently,

The required property isFor a nonzero element $x$ of an integral domain $R$, this follows fromIndeed, $\overline{a}\overline{b}=0$ in the quotient ring exactly when $x\mid ab$. The prime element condition says this forces $x\mid a$ or $x\mid b$, exactly the absence of zero divisors in the quotient ring. The nonunit condition makes the quotient ring nonzero. Being an integral domain is preserved by every ring isomorphism.

With the usual definition of a prime element, $x\ne 0$ is understood. The prime ideal formulation works for every ideal of any commutative ring, including the zero ideal; it avoids this convention about elements.

Choose a finite generating set of a module $n_1,\dots ,n_t$ for the finitely generated module $N$. For any $m\in M$, the surjectivity of the module homomorphism $\varphi$ gives $n\in N$ with $\varphi (n)=m$. Write $n={\sum }_{j=1}^t{r}_j{n}_j$. Since $\varphi$ is a module homomorphism,Thusis a finitely generated module. The argument includes $N=0$, in which case surjectivity forces $M=0$ and the empty generating set of a module suffices.

Let $n_1,\dots ,n_t$ generate the submodule $N$, and let $m_1+N,\dots ,m_s+N$ generate the quotient module $M/N$. Choose each $m_i\in M$ as a representative of its class. For any $m\in M$, there are $r_1,\dots ,r_s\in R$ such thatTherefore $m-{\sum }_i{r}_i{m}_i\in N$. Expressing that difference using the generating set of a module $n_1,\dots ,n_t$ givesHenceis a finitely generated module. This proves that finite generation is preserved by module extensions.