Past exam of the rings and modules course of the University of Oxford / 2021 (A3, A15291W1)

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.

Let $e_1,\dots ,e_n$ denote the standard basis of a module of the free module $R^n$. Since the module homomorphism $\varphi :R^m\to M$ is surjective, choose $v_j\in R^m$ such that $\varphi (v_j)=\psi (e_j)$ for each $j$. DefineUniqueness of coordinates in a basis of a module makes this well-defined, and the formula preserves addition and scalar multiplication, so $q$ is a module homomorphism. For every $x={\sum }_j{r}_j{e}_j$,Thus $\boxed{\varphi \circ q=\psi }$. This is the lifting property of a free module; only the surjectivity of $\varphi$ is needed for this part.

Put $K=\ker \varphi$, $H=\ker \psi$, and $Q=\mathrm{Im}q$. The kernel of a module homomorphism $H$ is a submodule, so its image $L=q(H)$ is a submodule. For $x\in H$, $\varphi (q(x))=\psi (x)=0$, showing $L\subseteq K$. Thus the quotient module $K/L$ exists. Also $L\subseteq Q$.

Consider the restriction to $K$ of the natural module homomorphism $R^m\to R^m/Q$:Its kernel of a module homomorphism is $K\cap Q$. We have $L\subseteq K\cap Q$. Conversely, if $y\in K\cap Q$, choose $x\in R^n$ with $y=q(x)$. Thenso $x\in H$ and $y\in L$. Hence $\boxed{K\cap Q=L}$.

The map $\alpha$ is surjective. Given $v+Q\in R^m/Q$, use the surjectivity of $\psi$ to choose $x\in R^n$ with $\psi (x)=\varphi (v)$. Set $y=v-q(x)$. ThenThus $y\in K$ is a preimage of the arbitrary class $v+Q$.

By the first isomorphism theorem for modules, $\alpha$ induces the module isomorphismExplicitly, replacing $y$ by $y+\ell$ with $\ell \in L$ leaves its class modulo $Q$ unchanged, proving well-definedness. Its formula preserves addition and scalar multiplication. If its value is zero, $y\in K\cap Q=L$, so $y+L=0$, proving injectivity; surjectivity was proved above. These checks give all the requested properties of the induced map.

Assume $H=\ker \psi$ is a finitely generated module. The restriction $q:H\to L$ is a surjective module homomorphism, so part (a)(i) makes $L$ a finitely generated module.

The free module $R^m$ has its finite standard basis of a module. Its quotient module $R^m/\mathrm{Im}q$ is therefore a finitely generated module by part (a)(i). Part (c) transfers this property to the isomorphic quotient module $\ker \varphi /L$. Finally apply part (a)(ii) to the submodule $L\subseteq \ker \varphi$. It follows thatMore concretely, take the images under $q$ of finitely many generators of $H$, and adjoin lifts in $\ker \varphi$ of finitely many generators of $\ker \varphi /L$. This finite list generates $\ker \varphi$. The argument shows why finite presentation of a module is independent of the chosen finite generating set of a module.

The exact orders printed in the question are inconsistent. The source PDF says that the orders of group elements $a,b,c$ are $4,12,24$, respectively, and also imposes $3a+2b=0$. Multiplying this relation by $6$ and using $12b=0$ gives $18a=0$. Since $4a=0$ also holds, subtracting $4(4a)=0$ gives $2a=0$, contradicting the stated order of a group element $4$. Thus

The natural abelian group presentation interpretation is to impose the relations $4a=12b=24c=0$ and $3a+2b=0$, with no further relations, without requiring the orders of the images of the generators to stay $4,12,24$. Under this interpretation the calculation is as follows. Since $3a=-a$ follows from $4a=0$, the mixed relation gives $a=2b$. Consequently $8b=0$, and together with $12b=0$ this gives $4b=0$. Conversely, $a=2b$, $4b=0$, and $24c=0$ imply every original relation. Eliminating $a$ leaves two independent cyclic groups:To verify that these relations force no further identifications, map $a,b,c$ to $(2,0),(1,0),(0,1)$ in ${\mathbb{Z}}_4\oplus {\mathbb{Z}}_{24}$. All defining relations hold, and the inverse group homomorphism sends the two standard generators to $b,c$. These group homomorphisms are mutually inverse. The divisibility condition $4\mid 24$ gives the requested invariant factors of a finitely generated module over $\mathbb{Z}$. In this presented group, $a,b,c$ actually have orders $2,4,24$.

Equivalently, the relation matrixhas nonzero Smith normal form entries $1,4,24$, with the trivial ${\mathbb{Z}}_1$ factor omitted. The displayed direct sum of modules is the answer to the presentation interpretation; the inconsistency of the literal exact-order formulation remains.

Every multiple of $dh$ is a multiple of both $d$ and $h$, so $dh\mathbb{Z}\subseteq d\mathbb{Z}\cap h\mathbb{Z}$.

For the reverse inclusion, let $z=du=hv$ lie in this intersection of ideals. Since $d,h$ are coprime integers, Bézout's identity supplies $r,s\in \mathbb{Z}$ with $rd+sh=1$. Multiplying by $u$ givesThus $h\mid u$, and $dh\mid du=z$. ConsequentlyHere Bézout's identity follows from the allowed principal ideal domain property of $\mathbb{Z}$: the ideal $d\mathbb{Z}+h\mathbb{Z}$ is generated by their greatest common divisor, which is $1$.

Define the group homomorphismBy part (b)(i), its kernel of a group homomorphism isIt is surjective. Indeed, choose $r,s\in \mathbb{Z}$ with $rd+sh=1$ by Bézout's identity. For any prescribed classes $u+d\mathbb{Z}$ and $v+h\mathbb{Z}$, the integersatisfies $t\equiv u(\mathrm{mod}d)$ and $t\equiv v(\mathrm{mod}h)$. The first isomorphism theorem for groups therefore givesThis proves the required isomorphism and is the Chinese remainder theorem for coprime integers. It also shows that $(1+d\mathbb{Z},1+h\mathbb{Z})$ generates the direct sum of modules when these additive groups are viewed as $\mathbb{Z}$-modules.

Let $N=|G|$ be a square-free integer. If $N=1$, then $G$ is the trivial cyclic group. Otherwise, finiteness means that a direct sum of modules decomposition into cyclic groups has only finitely many nontrivial factors, each of finite group order. Omit trivial factors and writeThe integers $n_i,n_j$ are coprime integers whenever $i\ne j$. If a prime number $p$ divided both, then $p^2\mid N$, contradicting the square-free integer hypothesis.

Moreover, $n_1\cdots n_{j-1}$ and $n_j$ are coprime integers: any common prime divisor would divide one of the earlier factors as well as $n_j$. Thus part (b)(ii) can be applied repeatedly to combine factors, givingIn particular $G$ is a cyclic group. Under the displayed identifications, the element whose coordinate in every factor is the class of $1$ is a generator of a group.

Write $q=|\mathbb{F}|$. The unit group $U(\mathbb{F})=\mathbb{F}\setminus \{0\}$ is a finite abelian group of group order $q-1$, since multiplication in a field satisfies commutativity. The structure theorem for finitely generated abelian groups, applied to this finite group, gives invariant factors of a finitely generated moduleafter omitting trivial factors. If there are no factors, the unit group is trivial and hence a cyclic group.

Otherwise put $D=d_r$. Every element $u\in U(\mathbb{F})$ satisfies $u^D=1$, because the order of a group element in each coordinate divides $D$. Hence all $q-1$ distinct nonzero elements of $\mathbb{F}$ are roots of a polynomial $X^D-1$. The root bound for a polynomial over a field givesFor completeness, that bound follows by induction from the factor theorem: if a nonzero polynomial $f$ vanishes at $a$, write $f=(X-a)g$; every other root of a polynomial is a root of $g$, because a field has no zero divisors. Each new distinct root therefore uses one unit of polynomial degree.

On the other hand, the group order in the decomposition isEquality is forced, so $d_1\cdots d_{r-1}=1$. As every retained factor is greater than $1$, there can be only one. ConsequentlyThe same proof establishes the more general result finite multiplicative subgroup of a field is cyclic. The square-free integer result in part (c)(i) alone would not suffice, since $q-1$ need not be a square-free integer; the root bound for a polynomial rules out multiple invariant factors of a finitely generated module even then.