Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 1 (Commutative Algebra)

A ring is Artinian when its ideals satisfy the descending chain condition: every chain $I_1\supseteq I_2\supseteq \cdots$ eventually becomes constant. It is Noetherian when its ideals satisfy the ascending chain condition, equivalently when every ideal is finitely generated. We first prove finite length of a commutative Artinian ring; this gives the stronger structural reason for its Noetherian property.

If $\mathfrak{p}$ is a prime ideal of an Artinian ring, the quotient $R/\mathfrak{p}$ is an Artinian integral domain. For a nonzero element $x$ of that domain, the chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes. Thus $x^n=b{x}^{n+1}$ for some $b$, and cancellation gives $bx=1$. The quotient is a field, so every prime ideal is a maximal ideal.

There are only finitely many maximal ideals. Otherwise, choose distinct ones ${\mathfrak{m}}_1,{\mathfrak{m}}_2,\dots$. The intersections ${\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_n$ form a strictly descending chain. To see strictness, for each $i\le n$ choose $a_i\in {\mathfrak{m}}_i\setminus {\mathfrak{m}}_{n+1}$; their product belongs to the first $n$ ideals but not to the next one, since ${\mathfrak{m}}_{n+1}$ is prime. This contradicts the descending chain condition.

Put $J={\bigcap }_i{\mathfrak{m}}_i$. This Jacobson radical is also the nilradical, since all primes are maximal. We need the stronger conclusion that $J$ is nilpotent, without assuming Noetherianity. Its powers stabilize, say $K=J^n=J^{n+1}=JK$. Suppose $K\ne 0$. By the descending chain condition, choose an ideal $L$ minimal subject to $KL\ne 0$. Some $x\in L$ has $Kx\ne 0$, so minimality gives $L=Rx$. Moreover $K(JL)=KL\ne 0$, and minimality gives $JL=L$. Hence $x=jx$ for some $j\in J$. But $1-j$ is a unit: it cannot lie in any maximal ideal, because $j$ lies in all of them. Thus $x=0$, a contradiction. Therefore $J^n=0$.

The Chinese remainder theorem gives $R/J\cong {\prod }_{i}R/{\mathfrak{m}}_i$, a finite product of fields. Each quotient $J^i/J^{i+1}$ is an Artinian module over this product, and each field component must be a finite-dimensional vector space; an infinite-dimensional vector space admits a strictly descending chain of subspaces. Consequently every layer has finite composition length. The finite filtrationshows that $R$ itself has finite composition length. A strict inclusion of submodules strictly increases length, so an ascending chain cannot continue indefinitely. Every commutative Artinian ring is therefore Noetherian. This is the Artinian rings are Noetherian result.

For the formal power series ring, let $B=R[[X]]$ with $R$ Noetherian, and let $H$ be any ideal of $B$. For $n\ge 0$ define a coefficient idealThese coefficient ideals of a formal power series ideal satisfy $A_n\subseteq A_{n+1}$, by multiplication by $X$. The ascending chain condition gives $A_n=A_N$ for all $n\ge N$. For each $0\le n\le N$, choose finitely many series $f_{n,j}\in H\cap X^{n}B$ whose coefficients at $X^n$ generate $A_n$.

We claim that these finitely many series generate $H$ as an ordinary ideal. Given $f\in H$, cancel its coefficient at $X^k$ successively. After coefficients below $k$ have vanished, its coefficient at $X^k$ lies in $A_k$. If $k\le N$, use an $R$-linear combination of the $f_{k,j}$. If $k>N$, use a combination of $X^{k-N}{f}_{N,j}$, since $A_k=A_N$. The remainder then belongs to $H\cap X^{k+1}B$.

Collect all the cancellations against each fixed generator. For $n<N$ its multiplier is a polynomial, while the multipliers of the $f_{N,j}$ are well-defined formal power series: at any fixed degree, only finitely many cancellation steps contribute. The remainder has every coefficient zero. ThusThis is a finite sum of ideal generators, rather than merely a topological closure assertion. Since $H$ was arbitrary,This coefficient-cancellation argument proves Noetherianity of a formal power series ring.

The corresponding Artinian assertion is false. For any nonzero ring $R$, the idealsin $R[[X]]$ are strictly decreasing, since $X^n$ has a nonzero coefficient in degree $n$ and no multiple of $X^{n+1}$ does. In particular, a field $k$ is Artinian, but $k[[X]]$ is not. The zero ring is the harmless exception.

A prime ideal $\mathfrak{p}$ is minimal over $I$ if $I\subseteq \mathfrak{p}$ and no strictly smaller prime contains $I$. An associated prime of a module is an annihilator of an individual nonzero element which happens to be prime. For the quotient module this readswhere ${\mathrm{Ann}}_R(\bar{a})=(I:a)=\{r\in R:ra\in I\}$. This is the annihilator of an individual element, rather than necessarily the annihilator of the whole quotient.

Minimal primes exist because $I$ is proper. First choose a maximal ideal containing $I$. Within the primes contained in it and containing $I$, an intersection of any decreasing chain is again prime. Indeed, if $ab$ is in the intersection and $a$ is absent from one member, then $b$ belongs to that member and to every smaller member; it also belongs to every larger member. The intersection is still proper and contains $I$. Zorn's lemma, applied with reverse inclusion, produces a minimal prime. This proves existence of minimal primes over a proper ideal, even without the Noetherian hypothesis.

Now fix a minimal prime $\mathfrak{p}$ and put $A=R/I$. The localization at a prime ideal $A_{\mathfrak{p}}$ is nonzero and is a Noetherian local ring. By prime ideal correspondence for localization, its only prime is $\mathfrak{p}{R}_{\mathfrak{p}}/I{R}_{\mathfrak{p}}$. Write this maximal ideal as $\mathfrak{n}$. It is the nilradical; since it is finitely generated and each generator is nilpotent, some power of $\mathfrak{n}$ is zero. Explicitly, if its generators have nilpotence exponents $e_1,\dots ,e_r$, every product of degree $1+{\sum }_i(e_i-1)$ vanishes.

Choose the smallest $s\ge 1$ such that ${\mathfrak{n}}^s=0$, and choose a nonzero element in ${\mathfrak{n}}^{s-1}$; when $\mathfrak{n}=0$, choose $1$. Its annihilator over $R_{\mathfrak{p}}$ is exactly $\mathfrak{p}{R}_{\mathfrak{p}}$. Represent it as $a/u$ with $u\notin \mathfrak{p}$. Multiplication by the unit $u$ shows that $a/1$ has the same annihilator and remains nonzero.

Let $p_1,\dots ,p_r$ generate $\mathfrak{p}$ in $R$. For each $i$, the equality $p_{i}a/1=0$ supplies $u_i\notin \mathfrak{p}$ such that $u_i{p}_i\bar{a}=0$ in $A$. Put $u={\prod }_i{u}_i$ and $\bar{b}=u\bar{a}$. Its localization is nonzero, so $\bar{b}\ne 0$, and every element of $\mathfrak{p}$ kills it. Conversely, an element outside $\mathfrak{p}$ becomes a unit and cannot kill the nonzero element $\bar{b}/1$. ThereforeThe key step in minimal primes are associated primes is clearing denominators for a finite generating set of $\mathfrak{p}$; that is where Noetherianity is used.

For an embedded associated prime, take $R=k[x,y]$ and $I=(x^2,xy)$. Its radical is $(x)$, so $(x)$ is its unique minimal prime. The nonzero class of $x$ satisfiesIndeed, $rx\in x(x,y)$ is equivalent to $r\in (x,y)$ by cancellation in the polynomial domain. Thus $(x,y)$ is associated but is not minimal, since $(x)⊊(x,y)$. For comparison, ${\mathrm{Ann}}_R(\bar{y})=(x)$, exhibiting the minimal associated prime as well.

An integral extension $R\subseteq T$ means that every element $t\in T$ satisfies a monic polynomial with coefficients in $R$:The Krull dimension is the supremum of the lengths of strict chains of primes:There need not be a finite bound on these lengths.

Here are the prime-ideal facts behind dimension preservation, with their relevant proofs. If a domain $B$ is integral over a subdomain $A$ and $B$ is a field, then $A$ is a field: for $0\ne a\in A$, a monic equation for $a^{-1}\in B$, multiplied by a suitable power of $a$, expresses $a^{-1}$ as an element of $A$. Conversely, an integral domain integral over a field is itself a field: a nonzero element has a polynomial equation with nonzero constant term after removing any factor of the indeterminate, and that equation expresses its inverse. Applied to quotients, these observations show that a prime in an integral extension is maximal if and only if its contraction is maximal.

For the Lying-over theorem, localize $R\subseteq T$ at $S=R\setminus \mathfrak{p}$. The inclusion remains injective and integral, and $S^{-1}T$ is nonzero. Any maximal ideal of $S^{-1}T$ contracts to the unique maximal ideal of $R_{\mathfrak{p}}$, by the field criterion just proved. The prime ideal correspondence for localization then gives a prime of $T$ contracting to $\mathfrak{p}$.

For the Going-up theorem, suppose $\mathfrak{q}$ lies over $\mathfrak{p}$ and $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$. The quotient inclusion $R/\mathfrak{p}\subseteq T/\mathfrak{q}$ is integral. Apply Lying-over theorem to the prime ${\mathfrak{p}}^{\prime }/\mathfrak{p}$; lifting back gives ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ contracting to ${\mathfrak{p}}^{\prime }$.

For the incomparability theorem for integral extensions, suppose $\mathfrak{q}\subseteq {\mathfrak{q}}^{\prime }$ contract to the same $\mathfrak{p}$. After localizing at $R\setminus \mathfrak{p}$, both are maximal ideals, because they lie over the maximal ideal of $R_{\mathfrak{p}}$. Their inclusion is therefore equality. The bijection between primes under localization gives $\mathfrak{q}={\mathfrak{q}}^{\prime }$.

Now contract a strict chain of primes in $T$. Incomparability theorem for integral extensions ensures that every contraction remains strict, so $\dim T\le \dim R$. Conversely, Lying-over theorem lifts the first member of any finite chain in $R$, and repeated Going-up theorem lifts the remaining members; different contractions ensure a strict chain in $T$. Taking suprema, including the possibility of infinity, givesThis proves that integral extensions preserve Krull dimension.

For the given quotient, put $B=k[Y][X]/(X^2+YX+Y^3)$. The relation is monic in $X$, so monic polynomial division gives a unique representative $a(Y)+b(Y)X$. Thus $k[Y]\hookrightarrow B$ is injective and $B$ is free of rank two as a $k[Y]$-module. In particular, $B$ is integral over $k[Y]$. The one-variable polynomial ring has dimension one: its zero prime is strictly below $(Y)$, and every nonzero prime is maximal because $k[Y]$ is a principal ideal domain. HenceThis is an instance of dimension of a monic plane hypersurface. No irreducibility or algebraic-closure assumption on $k$ is needed; monicity supplies the integral extension in every characteristic.

Use the following finite-length version of the Hilbert-Serre theorem. Let $S={⨁}_{n\ge 0}{S}_n$ be a graded algebra generated over an Artinian ring $A=S_0$ by finitely many homogeneous elements $x_1,\dots ,x_r$ of positive degrees $e_1,\dots ,e_r$. For a finitely generated nonnegatively graded $S$-module $M$, define its Poincare series of a graded module byEvery component has finite length, and the theorem saysFor a module whose grading is merely bounded below, the same statement holds with a Laurent-polynomial numerator. When the generators all have degree one, the denominator is $(1-t{)}^r$.

Here is an induction proof. An Artinian ring is Noetherian by question 1, so the Hilbert basis theorem makes $S$ Noetherian. If there are no positive-degree generators, $S=A$ and a finite homogeneous generating set for $M$ occupies only finitely many degrees. Each component is a finite module over the Artinian ring $A$, hence has finite length of a module, and $P_M$ is a polynomial.

For $r>0$, put $x=x_r$, $e=e_r$, $N=\{m\in M:xm=0\}$ and $C=M/xM$. Both are finitely generated graded modules annihilated by $x$, hence modules over $S/(x)$, which is generated by the first $r-1$ homogeneous elements. Their multiplication exact sequence isHere $M(-e{)}_n=M_{n-e}$. Additivity of length degree by degree givesThis is the Hilbert series multiplication exact sequence. By induction, the right-hand side has denominator ${\prod }_{i<r}(1-t^{e_i})$. Division by $1-t^e$ completes the proof of the Hilbert-Serre theorem. The finite-component and finite-generation claims also follow from the finite set of positive-degree algebra and module generators; no analytic convergence of a series is involved.

For the local invariant, use the usual Noetherian local ring hypothesis of Hilbert–Samuel growth dimension. Locality alone does not guarantee finite lengths or polynomial growth; the printed question leaves this finiteness assumption implicit. For instance, in the localization at a prime ideal of the polynomial ring $k[x_1,x_2,\dots ]$ at $(x_1,x_2,\dots )$, the vector space $\mathfrak{m}/{\mathfrak{m}}^2$ has the infinitely many independent classes of the variables, so its length is not finite. Write $\mathfrak{m}=P$ and $\kappa =R/\mathfrak{m}$. Its associated graded ringis generated over $\kappa$ in degree one, because $\mathfrak{m}$ is finitely generated. Thus its Hilbert series is rational with a denominator that is a power of $1-t$. After cancelling factors, writeDefine $d(R)=d$, the pole order at $t=1$, with $d=0$ when $P_G$ is a polynomial.

Equivalently, the Hilbert–Samuel functionhas generating series $P_G(t)/(1-t)$ and eventually agrees with a polynomial of degree $d$. Indeed, coefficients of $(1-t{)}^{-d-1}$ are $\left(\genfrac{}{}{0ex}{}{n+d}{d}\right)$, and multiplication by $q(t)$ leaves leading term $q(1)n^d/d!$. Consequently $d(R)$ is the degree of the cumulative Hilbert-Samuel polynomial, rather than the degree of the individual graded-component function; the latter has degree $d-1$ when $d>0$. This pole/growth invariant also equals Krull dimension by the local dimension theorem, although that theorem is not required to define it here.

For an example, take $R=k[x,y{]}_{(x,y)}$. Its associated graded ring is $k[x,y]$ with the ordinary degree grading: degree $n$ has the $n+1$ monomials $x^i{y}^{n-i}$ as a basis. ThereforeThe length formula also follows directly by counting monomials of total degree at most $n$ in $R/(x,y{)}^{n+1}$.

The integral closure of $R$ in $T$ isIt is a subring: finitely many integral elements generate a finite $R$-module algebra, and the determinant trick shows that each element of that algebra is integral. In particular, sums and products of integral elements remain integral.

A valuation ring is an integral domain $V$ such that, for every nonzero $z$ in its fraction field, either $z\in V$ or $z^{-1}\in V$. Equivalently, its ideals are totally ordered by inclusion. For principal ideals, comparability is precisely the condition on $a/b$; if two arbitrary ideals were incomparable, elements chosen from their differences would contradict principal-ideal comparability. Such a ring is local. Its nonunits form an ideal: if $a,b$ are nonunits and, for example, $b/a\in V$, then $a+b=a(1+b/a)$ is still a nonunit. The unique maximal ideal consists of those nonunits.

First, valuation rings are integrally closed. If $z\notin V$, then $y=z^{-1}\in V$ is a nonunit and lies in its maximal ideal $\mathfrak{m}$. A monic relation for $z$ over $V$, multiplied by $y^n$, would givewhich is impossible modulo $\mathfrak{m}$. Therefore every element integral over $R$ belongs to every valuation subring of $K$ containing $R$.

For the reverse inclusion, we will construct a valuation overring that excludes any chosen nonintegral element. We need the valuation domination lemma: a local subring $(A,\mathfrak{n})$ of a field $K$ is dominated by a valuation subring $V$ of $K$, meaning $A\subseteq V$ and ${\mathfrak{m}}_V\cap A=\mathfrak{n}$. Here is a proof, including the crucial maximality step.

Order the local subrings of $K$ dominating $A$ by domination. For a chain, take the union of the rings and of their maximal ideals. The union is a local ring: an element outside the union ideal is already a unit in a member of the chain, while an element in that ideal cannot become a unit in a later dominating member. The union still dominates $A$. Thus Zorn's lemma supplies a maximal pair $(V,\mathfrak{m})$.

For any $z\in K^{\times }$, at least one of $\mathfrak{m}V[z]$ and $\mathfrak{m}V[z^{-1}]$ is proper. Suppose otherwise. There would be relationswith $n,s$ chosen minimal. Both are positive. Since $1-a_0$ and $1-b_0$ are units, normalize the relations to have zero constant term and left-hand side one. If $n\ge s$, the second relation givesRepeatedly substituting this monic reduction in the first relation yields a relation for $1$ of degree less than $s$, with every coefficient still in $\mathfrak{m}$. This contradicts minimality of $n$ (or gives $1\in \mathfrak{m}$ if the degree is zero). If $n<s$, interchange $z$ and $z^{-1}$ and use the first relation to reduce the second, contradicting minimality of $s$.

Choose whichever extension has a proper extended ideal, then a maximal ideal containing it. Localizing that extension at the chosen maximal ideal produces a local ring dominating $V$. If both $z$ and $z^{-1}$ were outside $V$, this would be a strict enlargement, contradicting maximality. Thus $V$ has the valuation property. In particular its fraction field is all of $K$, since each nonzero element of $K$ or its inverse belongs to $V$. This proves the valuation domination lemma.

Now let $x\in K$ be nonintegral over $R$, so $x\ne 0$, and set $y=x^{-1}$, $B=R[y]$. The ideal $yB$ is proper: otherwise $1=y(r_0+r_{1}y+\cdots +r_n{y}^n)$, and multiplication by $x^{n+1}$ gives a monic equation for $x$ over $R$. Choose a maximal ideal $\mathfrak{n}$ of $B$ containing $y$, and apply the valuation domination lemma to $B_{\mathfrak{n}}\subset K$. Its dominating valuation ring $V$ contains $R$ and has $y\in {\mathfrak{m}}_V$. Hence $y$ is not invertible in $V$, so $x\notin V$.

We have excluded every nonintegral element from at least one valuation overring, while every integral element belongs to all of them. Therefore the integral closure as an intersection of valuation rings isNeither Noetherianity nor a discrete valuation is required for this separation argument.

Set $S=R\setminus P$, a multiplicative subset. The localization at a prime ideal is $R_P=S^{-1}R$, whose elements are fractions $r/s$. Equality of two fractions means that $u(r{s}^{\prime }-r^{\prime }s)=0$ for some $u\in S$. Likewise the localization of a module is $M_P=S^{-1}M$, withAddition uses a common denominator, and the module action isThe equivalence relations make these operations well-defined, so $M_P$ is an $R_P$-module.

If $M\ne 0$, choose $0\ne m\in M$. Its annihilator is proper, so it lies in a maximal ideal $P$. The element $m/1$ cannot vanish in $M_P$: vanishing would mean $sm=0$ for some $s\notin P$, contrary to ${\mathrm{Ann}}_R(m)\subseteq P$. The converse is immediate by localizing the zero module. ThusThis proof of localization detects zero elements actually applies to arbitrary modules, without finite-generation or Noetherian assumptions. We will use that extra generality for an Ext functor module below.

An injective module $E$ has the extension property: for each inclusion $A\hookrightarrow B$, every map $A\to E$ extends to a map $B\to E$. Equivalently, ${\mathrm{Hom}}_R(-,E)$ is exact. A projective module $Q$ has the lifting property: for each surjection $B\twoheadrightarrow C$, every map $Q\to C$ lifts to $Q\to B$. Equivalently, ${\mathrm{Hom}}_R(Q,-)$ is exact, or $Q$ is a direct summand of a free module.

For the local criterion for injectivity over a Noetherian ring, recall the Baer criterion: $E$ is injective precisely when every map from an ideal $J\subseteq R$ into $E$ extends to $R$. Through the short exact sequence $0\to J\to R\to R/J\to 0$, this is equivalent toWe also need localization of Ext over a Noetherian ring. Because $R$ is Noetherian, the module $R/J$ has a free resolution with every term finitely generated: all successive kernels are finitely generated, so this can be built recursively. For a finite free term $F$, the natural mapis an isomorphism, as is clear from a finite basis. Exactness of localization lets us pass to cohomology of the Hom complex. ThereforeThis explains the finiteness hypothesis needed for the localization argument, rather than assuming that localization preserves injectivity automatically.

If $M$ is injective, the left-hand side vanishes for every $J$ and $P$. Every ideal $L$ of $R_P$ is $J{R}_P$, where $J$ is its contraction to $R$: if $r/s\in L$, then $r/1=s(r/s)\in L$, and conversely localization of a member of the contraction stays in $L$. Hence all ideal tests for $M_P$ vanish, and the Baer criterion over $R_P$ makes $M_P$ injective.

Conversely, suppose every $M_P$ is injective. For each ideal $J$, the displayed Ext functor localization is zero at every prime. The zero-detection argument above gives ${\mathrm{Ext}}_R^1(R/J,M)=0$, without needing this Ext module to be finitely generated. Applying the Baer criterion over $R$ provesIn fact, under the Noetherian ring hypothesis this equivalence holds for arbitrary $M$; the printed finite-generation assumption is more than is needed.

The global dimension iswhere projective dimension is the smallest length of a projective resolution, or infinity if there is no finite one. If $\mathrm{gldim}R=0$, every module is projective. Given an inclusion $A\hookrightarrow B$, the quotient $B/A$ is then projective, so the exact sequence $0\to A\to B\to B/A\to 0$ splits. A retraction $\rho :B\to A$ exists. For any module $E$ and any map $f:A\to E$, the composition $f\rho :B\to E$ extends $f$. Thus every module is also injective. Global dimension zero makes all modules both projective and injective, as recorded by global dimension zero and split exact sequences.