Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 (Commutative Algebra)

Suppose every submodule of $M$ is finitely generated. For an ascending chain $N_1\subseteq N_2\subseteq \cdots$, its union $N$ is a submodule. Finitely many generators of $N$ all lie in one $N_r$, so $N=N_r$ and the chain stabilizes. Thus the ascending chain condition holds.

If the ascending chain condition holds, any nonempty collection of submodules has a maximal member: otherwise, starting from one member and repeatedly choosing a strictly larger one constructs a nonstationary ascending chain.

Finally, assume the maximal condition. Among the finitely generated submodules of a given $N\subseteq M$, choose a maximal one $F$. If $F\ne N$, then $F+Rx$ is a larger finitely generated submodule for any $x\in N\setminus F$, a contradiction. Hence every $N$ is finitely generated. These are the three equivalent characterizations of a Noetherian module.

Let $n_1,\dots ,n_r$ generate $N$, and let the images of $m_1,\dots ,m_s$ generate $M/N$. For any $m\in M$, subtracting a suitable linear combination of the $m_j$ leaves an element of $N$, which is a combination of the $n_i$. Thus

If $M$ is Noetherian, every submodule of $N$ is a submodule of $M$, and submodules of $M/N$ correspond to submodules of $M$ containing $N$; hence both are Noetherian.

Conversely, suppose $N$ and $M/N$ are Noetherian. For any $P\subseteq M$, the intersection $P\cap N$ is finitely generated and the image $(P+N)/N$ is finitely generated. Lifting generators of the image and applying part i toshows that $P$ is finitely generated. Thus $M$ is Noetherian.

The exact sequenceand part ii show that a direct sum of two modules is Noetherian exactly when both summands are. Induction proves the assertion for every finite direct sum.

Each $R/I_j$ is Noetherian as an $R$-module because its $R$-submodules are precisely its ideals as a Noetherian ring. The diagonal maphas kernel ${\bigcap }_j{I}_j=0$. Hence $R$ is an $R$-submodule of a finite direct sum of Noetherian modules, so $R$ is a Noetherian $R$-module; equivalently, $R$ is a Noetherian ring.

DefineThis makes $M[X]$ an $R[X]$-module.

Let $N\subseteq M[X]$ be an $R[X]$-submodule. For $n\ge 0$, let $L_n\subseteq M$ consist of zero and the leading coefficients of elements of $N$ of degree $n$. Each $L_n$ is an $R$-submodule, and multiplication by $X$ givesBecause $M$ is Noetherian, this chain stabilizes at some $n_0$, and each $L_n$ for $n\le n_0$ is finitely generated. Choose finitely many polynomials $f_{nj}\in N$ of degree $n$ whose leading coefficients generate $L_n$.

For $f\in N$ of degree $d$, if $d\le n_0$, subtract an $R$-linear combination of the $f_{dj}$ to lower its degree. If $d>n_0$, use $L_d=L_{n_0}$ and subtract a combination of $X^{d-n_0}{f}_{n_{0}j}$. Induction on degree expresses $f$ in terms of the finite collection $\{f_{nj}\}$. Therefore every submodule $N$ is finitely generated andThis is the module form of the Hilbert basis theorem.

The localization of a ring $S^{-1}R$ consists of pairs $(r,s)\in R\times S$ modulofor some $u\in S$. Addition and multiplication are

The localization is the zero ring exactly when $1/1=0/1$, which means that some $u\in S$ is zero. Thus $0\in S$. This holds exactly when $S$ contains a nilpotent element: if $s^n=0$ and $s\in S$, multiplicative closure gives $0\in S$, while zero itself is nilpotent.

Let $e=(1,0)$. Since $e$ becomes invertible in $S^{-1}(A\times B)$ and $e(0,b)=0$, every $(0,b)$ becomes zero. The mapis therefore a well-defined ring isomorphism, with inverse $a\mapsto (a,0)/1$.

For a subring $\mathbb{Z}\subseteq R\subseteq \mathbb{Q}$, letIf $a/b\in R$ is in lowest terms, Bézout gives integers $u,v$ with $ua+vb=1$, soEvery prime divisor $p$ of $b$ then satisfies $1/p=(b/p)/b\in R$. Conversely, every rational whose denominator uses only primes in $\mathcal{P}$ lies in $R$. Hence all intermediate rings are exactlywhere $S$ is generated by an arbitrary set $\mathcal{P}$ of primes. For example, $1/3=1-2/3$, so $\mathbb{Z}[2/3]=\mathbb{Z}[1/3]=\{3^n:n\ge 0{\}}^{-1}\mathbb{Z}$.

For $x\in L$, choose a monic equation over $K$,Choose nonzero $a\in R$ with $a{c}_i\in R$ for every $i$. Multiplying by $a^n$ shows that $b=ax$ satisfieswhose coefficients lie in $R$. Thus $b\in \overline{R}$ and

Extension and contraction give the prime ideal correspondence for localizationExplicitly, $\mathfrak{p}$ maps to $S^{-1}\mathfrak{p}$, while $\mathfrak{q}\subseteq S^{-1}R$ maps to $\{r:r/1\in \mathfrak{q}\}$. Primality follows by clearing denominators, and the two operations are inverse because an ideal in a localization contains $r/s$ exactly when it contains $r/1$.

The weak Hilbert Nullstellensatz says that if a field is a finitely generated algebra over $k$, then it is a finite algebraic extension of $k$. Since every maximal ideal is prime, the nilradical $\mathcal{N}(R)$ is contained in the Jacobson radical $J(R)$.

Conversely, let $f$ be nonnilpotent. Then $R_f\ne 0$, and it is a finitely generated $k$-algebra, so it has a maximal ideal $\mathfrak{n}$. Its contraction $\mathfrak{p}$ to $R$ avoids $f$. Moreover,is a finitely generated $k$-domain inside a finite algebraic extension of $k$, hence is itself a field. Thus $\mathfrak{p}$ is maximal and does not contain $f$. Therefore $f\notin J(R)$, proving

The tensor product $M{\otimes }_{R}N$ is the free abelian group on symbols $(m,n)$ modulo additivity in each variable and the balancing relation $(rm,n)=(m,rn)$. It has the universal property that balanced bilinear maps $M\times N\to P$ correspond uniquely to homomorphisms $M{\otimes }_{R}N\to P$.

The mapis well defined and surjective. Its inverse sends $m+IM$ to $(1+I)\otimes m$; elements of $IM$ map to zero because $(1+I)\otimes am=(a+I)\otimes m=0$ for $a\in I$. Hence

Let $k=R/\mathfrak{m}$. Associativity and part i giveIf $M{\otimes }_{R}N=0$, the tensor product on the right is zero. Two nonzero vector spaces over a field have nonzero tensor product, so one factor vanishes. Nakayama lemma then gives $M=0$ or $N=0$.

An $R$-module $F$ is flat when $-{\otimes }_{R}F$ preserves injections, equivalently all finite exact sequences. Since $-{\otimes }_{R}R$ is naturally the identity functor, $R$ is flat. A free module is a direct sum of copies of $R$, and tensor products commute with direct sums, so every free module is flat.

As an $R$-module,is free. Therefore the polynomial algebra $R[X]$ is a flat $R$-algebra.

An element $x\in B$ is integral over $A$ when it satisfies a monic polynomial with coefficients in $A$. Equivalently, $A[x]$ is a finite $A$-module.

If $x,y$ are integral over $A$, then $A[x,y]$ is finite over $A$: it is generated by finitely many monomials $x^i{y}^j$. Multiplication by $x+y$, $x-y$, or $xy$ is an endomorphism of this finite module, so the determinant trick gives a monic annihilating polynomial. Hence the integral elements form a subring $C$ containing $A$.

The integral closure of $A$ in $B$ is this subring $C$. The ring $A$ is integrally closed in $B$ when $C=A$, and $B$ is integral over $A$ when $C=B$.

For $x\in A$, considerThis is monic, and every $\tau \in G$ permutes its factors, so all coefficients lie in $A^G$. Since $f_x(x)=0$, every $x\in A$ is integral over $A^G$. Thus

The Going-up theorem states: if $B$ is integral over $A$, ${\mathfrak{p}}_1\subseteq {\mathfrak{p}}_2$ are primes of $A$, and ${\mathfrak{q}}_1$ lies over ${\mathfrak{p}}_1$, then some prime ${\mathfrak{q}}_2\supseteq {\mathfrak{q}}_1$ lies over ${\mathfrak{p}}_2$.

Pass to $A/{\mathfrak{p}}_1\subseteq B/{\mathfrak{q}}_1$, which remains integral, and localize at the complement of ${\mathfrak{p}}_2/{\mathfrak{p}}_1$. The lying-over theorem supplies a prime of the localized upper ring over the maximal ideal of the localized lower ring. Contracting it to $B/{\mathfrak{q}}_1$, and then pulling it back to $B$, gives the required ${\mathfrak{q}}_2$.

If $x\in A$ is a unit in $B$, then $x^{-1}$ is integral over $A$:Multiplication by $x^{n-1}$ expresses $x^{-1}$ as an element of $A$, so $x$ is a unit in $A$.

Use the characterization $a\in J(A)$ exactly when $1-ra$ is a unit for every $r\in A$. If $a\in J(B)\cap A$, then $1-ra$ is a unit in $B$ and hence in $A$, proving $a\in J(A)$. Conversely, if $a\in J(A)$, every maximal ideal $\mathfrak{n}$ of $B$ contracts under the integral extension to a maximal ideal of $A$, which contains $a$. Thus every $\mathfrak{n}$ contains $a$, so $a\in J(B)$. Therefore

Put $S=A\setminus \mathfrak{p}$. Primes of $B_{\mathfrak{p}}=S^{-1}B$ correspond to primes $\mathfrak{q}$ of $B$ whose contractions are contained in $\mathfrak{p}$. By going up, each such $\mathfrak{q}$ is contained in a prime lying over $\mathfrak{p}$, and this prime is uniquely ${\mathfrak{p}}^{\prime }$. Hence ${\mathfrak{p}}^{\prime }{B}_{\mathfrak{p}}$ is the unique maximal ideal of $B_{\mathfrak{p}}$.

Localizing this already local ring at its unique maximal ideal changes nothing, soFinally, localization preserves integral extensions; therefore $B_{{\mathfrak{p}}^{\prime }}=B_{\mathfrak{p}}$ is integral over $A_{\mathfrak{p}}$.

The positive-degree part $A_{+}={⨁}_{n>0}{A}_n$ is an ideal and $A/A_{+}\simeq A_0$, so $A_0$ is Noetherian. Since $A$ is Noetherian, $A_{+}$ has finitely many homogeneous generators $x_1,\dots ,x_s$. Induction on degree shows that every positive-degree homogeneous element is a polynomial in the $x_i$ over $A_0$. Thus

For an additive length function $\lambda$ finite on the graded pieces, define the Poincare series of a graded moduleThe Hilbert-Serre theorem states thatfor a Laurent polynomial $f(t)\in \mathbb{Z}[t,t^{-1}]$.

Induct on $s$. For the last generator $x_s$ of degree $k_s$, multiplication gives an exact sequence whose kernel is the $x_s$-torsion and whose cokernel is $M/x_{s}M$. Additivity of $\lambda$ yieldsBoth modules on the right are finite graded modules over the algebra generated by $x_1,\dots ,x_{s-1}$. The induction hypothesis gives the asserted denominator. The case $s=0$ is a finite Laurent polynomial because $M$ is finitely generated over $A_0$.

When every $k_i=1$, cancel common factors to writewhere $d$ is the pole order at $t=1$. Sincethe coefficient of $t^n$ in $P(M,t)$ is, for all sufficiently large $n$, a fixed linear combination of shifted binomial polynomials. It is therefore a polynomial in $n$ of degree exactly $d-1$ unless $P=0$, in which case its degree is $-1$.

The Krull dimension $\dim R$ is the supremum of lengths $n$ of strict chainsof prime ideals. The transcendence degree ${\mathrm{trdeg}}_{k}L$ is the cardinality of a transcendence basis of $L/k$.

By Noether normalization lemma, there are algebraically independent $y_1,\dots ,y_d\in A$ such that $A$ is finite, hence integral, over $k[y_1,\dots ,y_d]$. Their fraction field has transcendence degree $d$, and $L$ is algebraic over it, so ${\mathrm{trdeg}}_{k}L=d$. Going up and incomparability show that an integral extension preserves Krull dimension, while a polynomial ring in $d$ variables over a field has dimension $d$. Hence

A chain of length $s$ in $R/\mathfrak{p}$ lifts to a chainin $R$. Since $R$ is a domain and $\mathfrak{p}\ne 0$, prepending $(0)⊊\mathfrak{p}$ gives a chain of length $s+1$. Therefore $s+1\le r$ and

Going up lifts every prime chain in $A$ to one in $B$, so $\dim B\ge \dim A$. Conversely, contracting a strict chain of primes of $B$ gives a chain in $A$, and the incomparability theorem for integral extensions ensures that no strict inclusion contracts to equality. Thus $\dim A\ge \dim B$, and

The height $\mathrm{ht}\mathfrak{p}$ is the supremum of lengths of strict chains of primes ending at $\mathfrak{p}$. The Krull principal ideal theorem says that in a Noetherian ring every prime minimal over a principal proper ideal has height at most one.

Induct on $n$. If $\mathfrak{p}$ is minimal over $I=(x_1,\dots ,x_n)$, choose a prime $\mathfrak{q}\subseteq \mathfrak{p}$ minimal over $(x_1,\dots ,x_{n-1})$ and localize appropriately. In $R/\mathfrak{q}$, the prime $\mathfrak{p}/\mathfrak{q}$ is minimal over the principal ideal generated by $x_n$, so its relative height is at most one. Induction gives $\mathrm{ht}\mathfrak{q}\le n-1$, hence

Write $\mathfrak{p}=(x)$. The principal ideal theorem and the hypothesis give $\mathrm{ht}\mathfrak{p}=1$. We use the standard principal prime in a Noetherian local ring lemma: a principal prime of positive height in a Noetherian local ring is generated by a nonzerodivisor and is the unique minimal prime above zero. The lemma follows by applying the associated-prime description of zero divisors and Nakayama's lemma to $\mathrm{ann}(x)$; if a nonzero annihilator or another minimal component existed, the principal prime would have height zero.

Here is the needed argument directly. For every $n\ge 1$,Indeed, if $y{x}^n=0$ with $y\notin \mathfrak{p}$, then $y$ is a unit in $R_{\mathfrak{p}}$, so $x^n=0$ there. The maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}=(x)$ would then be nilpotent, making $R_{\mathfrak{p}}$ zero-dimensional, contrary to $\mathrm{ht}\mathfrak{p}\ge 1$.

Now suppose $xy=0$. The displayed containment gives $y=x{y}_1$, then $x^2{y}_1=0$ gives $y_1=x{y}_2$, and inductivelyBecause $R$ is Noetherian, the ascending chain $\mathrm{ann}(x)\subseteq \mathrm{ann}(x^2)\subseteq \cdots$ stabilizes, say at $n=N$. Then $y_N\in \mathrm{ann}(x^{N+1})=\mathrm{ann}(x^N)$, and hence $y=x^N{y}_N=0$. Thus $x$ is a nonzerodivisor.

Let $N={\bigcap }_{n\ge 0}{x}^{n}R$. It is a finitely generated ideal. If $y=xz\in N$, then $xz\in x^{n+1}R$ for every $n$; cancellation of the nonzerodivisor $x$ gives $z\in x^{n}R$ for every $n$. Hence $N=xN$, and Nakayama lemma gives $N=0$ because $x$ lies in the maximal ideal.

Every nonzero element consequently has a finite $x$-adic order. If nonzero $a,b$ satisfied $ab=0$, write $a=x^r{a}^{\prime }$ and $b=x^s{b}^{\prime }$ with $a^{\prime },b^{\prime }\notin (x)$. Cancelling $x^{r+s}$ gives $a^{\prime }{b}^{\prime }=0$, contradicting primality of $(x)$. Equivalently, $(0)$ is prime, so

The formal power series ring $P=k[[X]]$ is Noetherian, so the finite product $R=P\times P$ is Noetherian. Its maximal ideals areso there are exactly two.

The idealis principal and prime because $R/\mathfrak{p}\simeq k$. The prime chainshows that it has height one, and no longer chain exists because $\dim P=1$. Yetwith both factors nonzero, so $R$ is not a domain. This shows why locality is essential in part i.

A stable $I$-filtration of $M$ is a descending sequence $M=M_0\supseteq M_1\supseteq \cdots$ with $I{M}_n\subseteq M_{n+1}$ for all $n$ and equality for all sufficiently large $n$. The Rees ring and associated Rees module areThe filtration condition makes multiplication by $I^r{t}^r$ send $M_n{t}^n$ into $M_{n+r}{t}^{n+r}$, so $M^{\ast }$ is a graded $R^{\ast }$-module.

If the filtration is stable from degree $\ell$, then $M^{\ast }$ is generated over $R^{\ast }$ by finite generating sets for $M_0,\dots ,M_{\ell }$. Conversely, let homogeneous elements of degrees at most $\ell$ generate $M^{\ast }$. In every degree $n>\ell$, each expression for an element of $M_n$ uses a positive-degree coefficient from $R^{\ast }$, so $M_n=I{M}_{n-1}$. Hence finite generation is equivalent to stability.

For $M^{\prime }\subseteq M$, takeThen $M^{\ast }$ is a graded submodule of the finite Rees module $⨁I^{n}M$. Since $R$ is Noetherian and $I$ is finitely generated, $R^{\ast }$ is Noetherian; hence $M^{\ast }$ is finite and the filtration is stable. Therefore, for some $\ell$ and all $n\ge \ell$,This is the Artin-Rees lemma.

Let $N={\bigcap }_{n\ge 0}{I}^{n}M$. Apply Artin--Rees to $N\subseteq M$. Since $N\subseteq I^{n}M$ for every $n$, stability givesfor all sufficiently large $n$, and in particular $N=IN$. The module $N$ is finitely generated because $M$ is Noetherian. The determinant trick applied to a finite generating set of $N=IN$ produces $a\in I$ withTaking $x=-a\in I$ yieldswhich is Krull intersection theorem.