Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 7 (Commutative Algebra)

Set $S=A\setminus \mathfrak{m}$, a multiplicative subset because $\mathfrak{m}$ is a prime ideal. The quotient module $B=A/\mathfrak{a}$ is a cyclic module generated by $\bar{1}$, so its localization of a module $S^{-1}B$ is zero precisely when $\bar{1}/1=0$. By the vanishing criterion in a module localization, this is equivalent to the existence of $s\in S$ with $s\bar{1}=0$ in $B$. The latter condition says $s\in \mathfrak{a}$. ThereforeExplicitly, if $s\in \mathfrak{a}\setminus \mathfrak{m}$, multiplication by $s$ annihilates $B$ but becomes invertible after localization, forcing every localized element to be zero.

The forward implication is immediate: the localization of a module that is zero remains zero. Conversely, suppose $M\ne 0$ and choose $0\ne x\in M$. Its annihilatoris a proper ideal, since $1\notin J$. Choose a maximal ideal $\mathfrak{m}$ containing $J$. Then $x/1$ cannot be zero in $M_{\mathfrak{m}}$: otherwise the vanishing criterion in a module localization would give $sx=0$ for some $s\notin \mathfrak{m}$, contradicting $s\in J\subseteq \mathfrak{m}$. Thus some maximal localization is nonzero, provingIn the suggested cyclic description, $Ax\cong A/J$ and part (a) gives $(Ax{)}_{\mathfrak{m}}\ne 0$; the same fraction argument embeds it in $M_{\mathfrak{m}}$. This is localization detects zero elements, applied to an arbitrary module; finite generation is unnecessary.

Suppose $f$ is injective. If $f_{\mathfrak{m}}(x/s)=0$, some $t\notin \mathfrak{m}$ satisfies $tf(x)=f(tx)=0$. Injectivity gives $tx=0$, and the vanishing criterion in a module localization gives $x/s=0$. Hence every localized module homomorphism is injective.

Conversely, suppose all $f_{\mathfrak{m}}$ are injective. For $x\in \ker f$, its image $x/1$ in $M_{\mathfrak{m}}$ maps to zero and therefore is zero for every maximal ideal $\mathfrak{m}$. The annihilator argument in part (b) implies $x=0$. Thus $\ker f=0$ and $f$ is injective. We have proved that injectivity of a module map is detected at maximal localizations:Equivalently, exactness of localization gives $(\ker f{)}_{\mathfrak{m}}\cong \ker (f_{\mathfrak{m}})$, and part (b) detects whether this kernel is zero.

Use the defining criterion that a flat module $E$ makes $u\otimes 1_E$ injective for every injective module homomorphism $u$. By right exactness of the tensor product, preserving injections is equivalent to preserving every short exact sequence.

First assume $E$ is flat over $A$, and fix a maximal ideal $\mathfrak{m}$. Let $u:U\hookrightarrow V$ be an injection of $A_{\mathfrak{m}}$-modules. Regard it as an injection of $A$-modules. There are natural module isomorphismsFor example, $u_0\otimes (e/s)$ maps to $s^{-1}{u}_0\otimes e$, with inverse $u_0\otimes e\mapsto u_0\otimes (e/1)$. These are well-defined because every $s\notin \mathfrak{m}$ already acts invertibly on $U$. Under these isomorphisms the tensor map is $u{\otimes }_A{1}_E$, which is injective by flatness. Thus $E_{\mathfrak{m}}$ is flat over $A_{\mathfrak{m}}$.

For the converse, assume every $E_{\mathfrak{m}}$ is flat over $A_{\mathfrak{m}}$ and take any injection $u:U\hookrightarrow V$ of $A$-modules. By part (c), each $u_{\mathfrak{m}}$ is injective. Since localization commutes with tensor products, the localization of $u{\otimes }_A{1}_E$ identifies withIt is injective by the assumed local flatness. Apply part (c) again, now to $u{\otimes }_A{1}_E$, to conclude that it is injective globally. Since $u$ was arbitrary, $E$ is flat. ThereforeThis proves flatness is local without any finite-generation hypothesis.

Take $B=\mathbb{Z}$, $N=\mathbb{Z}/2\mathbb{Z}$, and the short exact sequenceMultiplication by two is injective, and the reduction map has image all of $\mathbb{Z}/2\mathbb{Z}$ and kernel $2\mathbb{Z}$, proving exactness. Any module homomorphism from $\mathbb{Z}/2\mathbb{Z}$ to $\mathbb{Z}$ is zero: its generator must map to an integer killed by two. On the other hand, a homomorphism from $\mathbb{Z}/2\mathbb{Z}$ to itself is determined arbitrarily by the image of its generator, so its Hom group is $\mathbb{Z}/2\mathbb{Z}$. Applying the Hom functor therefore givesThis sequence is not exact at $\mathbb{Z}/2\mathbb{Z}$: the incoming image is zero while the outgoing kernel is the whole nonzero group. In particular, the identity of $N$ cannot lift through the reduction map to a map $N\to \mathbb{Z}$.

For any injection $u:U\hookrightarrow V$, the direct sum decomposition $F=P\oplus Q$ gives natural module isomorphismsUnder these, $u\otimes 1_F$ is the block-diagonal map $(u\otimes 1_P)\oplus (u\otimes 1_Q)$. It is injective because $F$ is a flat module. If an element lies in the kernel of $u\otimes 1_P$, insert zero in the other component; injectivity of the whole map forces that element to be zero. Hence $u\otimes 1_P$ is injective for every $u$, which proves $\boxed{P\text{ is flat}}$. This establishes that direct summands of flat modules are flat, without a finite-generation assumption.

Here the Hom groups must be taken over $A$; the printed subscript $B$ in this part has no defined coefficient ring. For an arbitrary module $P$, form the free moduleand define a module homomorphism $\pi :F\to P$ by $e_p\mapsto p$. Each element of $F$ has finite support, so this prescription extends uniquely and is well-defined. It is surjective because every $p\in P$ is the image of $e_p$. Thus every module is a quotient of a free module.

Let $K=\ker \pi$. The short exact sequence $0\to K\to F\stackrel{\pi }{\to }P\to 0$ and the defining exactness condition for a projective module makesurjective. Lift $1_P$ to a map $s:P\to F$. Then $\pi s=1_P$, so $s$ is injective. Every $f\in F$ has the decompositionwhose first term lies in $K$ and second lies in $s(P)$. If $s(p)\in K$, applying $\pi$ gives $p=0$, so the intersection is zero. ConsequentlyThis is a split short exact sequence and proves that projective modules are direct summands of free modules, including modules that are not finitely generated.

First show that free modules are flat. If $F={⨁}_{j\in J}A{e}_j$, distributivity of the tensor product of modules givesExplicitly, $u\otimes {\sum }_j{a}_j{e}_j$ maps to $(a_{j}u{)}_j$, and the inverse sends a finitely supported tuple $(u_j{)}_j$ to ${\sum }_j{u}_j\otimes e_j$. For an injection $u:U\hookrightarrow V$, the tensor map is therefore the coordinatewise injection ${⨁}_{J}U\hookrightarrow {⨁}_{J}V$. This proves flatness for any free rank.

By part (c), a projective module $P$ is a direct summand of a free module $F$. The latter is flat by the preceding argument, and part (b) proves that its summand $P$ is flat. Therefore $\boxed{\text{every projective }A\text{-module is flat}}$. This proves projective modules are flat directly from the exactness and splitting definitions.

Choose $a_1,\dots ,a_d\in \mathfrak{m}$ whose classes form a basis of the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$, and put $J=(a_1,\dots ,a_d)$. Spanning gives $\mathfrak{m}=J+{\mathfrak{m}}^2$. Thus the quotient module $N=\mathfrak{m}/J$ satisfies $N=\mathfrak{m}N$. It is a finitely generated module, since $A$ is a Noetherian ring.

Here is the needed form of the Nakayama lemma, with proof. Choose generators $n_1,\dots ,n_r$ of $N$ and write $n_i={\sum }_j{c}_{ij}{n}_j$ with $c_{ij}\in \mathfrak{m}$. For the column vector $n$, $(I-C)n=0$. Multiply by the adjugate to obtain $\det (I-C)n=0$. Since $\det (I-C)\equiv 1(\mathrm{mod}\mathfrak{m})$, this determinant is a unit in the local ring $A$, so every $n_i=0$. Therefore $N=0$, proving $\mathfrak{m}=J$. Conversely, any generators of $\mathfrak{m}$ must span its cotangent space, so at least $d$ are needed. HenceHere $\mu (M)$ denotes the minimal number of generators of a module $M$.

For the additional minimum, the maximal ideal $\mathfrak{m}$ is itself $\mathfrak{m}$-primary, so it supplies an eligible ideal generated by $d$ elements. ConsequentlyThe minimum here is over primary ideals. A bound $\mu (Q)\le d$ for every $\mathfrak{m}$-primary ideal would be false. Indeed, for $A=k[[u,v]]$ and $\mathfrak{m}=(u,v)$, $d=2$ but $Q={\mathfrak{m}}^2=(u^2,uv,v^2)$ is primary and needs three generators. To verify primaryness, if $ab\in Q$ and $a\notin Q$, then $b$ cannot be a unit, so $b\in \mathfrak{m}$ and $b^2\in Q$. The three displayed monomials form a basis of $Q/\mathfrak{m}Q={\mathfrak{m}}^2/{\mathfrak{m}}^3$, proving the lower bound of three, while they already generate $Q$. This illustrates that primary ideals can need more generators than the maximal ideal.

A formal power series $f={\sum }_{n\ge 0}{a}_n{x}^n$ is a unit exactly when its constant coefficient is a unit in $A$. If $fg=1$, constant coefficients give $a_0{b}_0=1$. Conversely, for invertible $a_0$, setThen every positive-degree coefficient of $fg$ is zero and its constant coefficient is one, so $g=\sum b_n{x}^n$ is an inverse. This is a formal coefficient calculation and requires no convergence.

Since $A$ is a local ring, its nonunits are exactly $\mathfrak{m}$. The nonunits of $B$ are consequently the series with $a_0\in \mathfrak{m}$, which are exactly $I=\mathfrak{m}B+xB$: such a series is $a_0+xg$, and conversely an element of $I$ has constant coefficient in $\mathfrak{m}$. The map $B\to k$ taking the constant coefficient modulo $\mathfrak{m}$ is surjective with kernel $I$, so $I$ is a maximal ideal. Every proper ideal consists of nonunits and hence lies in $I$, proving that $I$ is the unique maximal ideal. ThusThis proves the description of formal power series over a local ring.

To compute the cotangent space of a formal power series ring, expandDefine the $k$-linear mapChanging a lift $a$ by ${\mathfrak{m}}^2$ or a lift $b$ by $\mathfrak{m}$ changes the image by $I^2$, so the map is well-defined. Every series in $I$ is congruent modulo $x^{2}B\subseteq I^2$ to its first two coefficients, proving surjectivity. If $a+bx\in I^2$, its constant coefficient lies in ${\mathfrak{m}}^2$ and its linear coefficient lies in $\mathfrak{m}$, so both input classes are zero. This proves injectivity. ThereforeFinally, part (a) gives generators $a_1,\dots ,a_d$ of $\mathfrak{m}$, so $I=(a_1,\dots ,a_d,x)$. Use the Krull height theorem in its lecture form: in a Noetherian ring, a prime minimal over an ideal generated by $r$ elements has height at most $r$. The problem allows Noetherianity of $B$ to be assumed. Apply the theorem to $I$ itself, whose height is the Krull dimension of the local ring $B$. Hence $\boxed{\dim B=\mathrm{ht}(I)\le d+1}$.

Since $A$ is a regular local ring, $\dim A=d$ is finite. Choose a chain of prime ideals of length $d$,Such a chain exists because the supremum defining this finite integer Krull dimension is attained; a maximal-length chain in a local ring ends at its maximal ideal.

For each $i$, the ideal ${\mathfrak{p}}_i$ is finitely generated because $A$ is Noetherian. This ensures that its extension ${\mathfrak{p}}_{i}B$ consists precisely of series all of whose coefficients lie in ${\mathfrak{p}}_i$: if ${\mathfrak{p}}_i=(b_1,\dots ,b_r)$, write each coefficient as ${\sum }_j{b}_j{c}_{jn}$ and gather them into the finitely many series ${\sum }_n{c}_{jn}{x}^n$. Thus coefficient reduction givesThe coefficient ring $A/{\mathfrak{p}}_i$ is an integral domain, and so is its formal power series ring: the first nonzero coefficient of a product is the product of the first nonzero coefficients of its factors. Hence each ${\mathfrak{p}}_{i}B$ is prime. Also ${\mathfrak{p}}_{i}B\cap A={\mathfrak{p}}_i$, so the strict inclusions persist upon extension. Finally, $x\notin \mathfrak{m}B$, since its image in $B/\mathfrak{m}B\cong k[[x]]$ is nonzero. We obtain the prime chainof length $d+1$. Therefore $\dim B\ge d+1$. Together with part (b),Its embedding dimension is also $d+1$, by part (b), so $B$ is itself a regular local ring. This proves that formal power series preserve regular local rings.

Let $S=A\setminus \mathfrak{p}$, let $S$ act on $B$ through $\phi$, and write $\mathfrak{p}B$ for the extension of an ideal generated by $\phi (\mathfrak{p})$. Since $\kappa (\mathfrak{p})=S^{-1}(A/\mathfrak{p})$, the fiber ring has the canonical presentationsThe first isomorphism sends $b\otimes (\bar{a}/s)$ to $\phi (a)b/s$ modulo $\mathfrak{p}B$. Its inverse sends the class of $b/s$ to $b\otimes (\bar{1}/s)$. The tensor balancing relation, reduction modulo $\mathfrak{p}$ and the fraction relations show that these maps are well-defined and inverse. Thus the description is canonical and does not depend on chosen representatives.

Suppose $\mathfrak{q}$ is a prime ideal of $B$ with ${\phi }^{-1}(\mathfrak{q})=\mathfrak{p}$. Then $\mathfrak{p}B\subseteq \mathfrak{q}$ and $\mathfrak{q}$ is disjoint from $\phi (S)$. Its localization $S^{-1}\mathfrak{q}$ is a proper prime, since the quotient is the localization of the integral domain $B/\mathfrak{q}$. It contains $\mathfrak{p}{S}^{-1}B$, and therefore gives a primeConversely, for a prime $\mathfrak{Q}\subset C$, take its contraction $\mathfrak{q}$ under the ring homomorphism $\psi :B\to C$, $b\mapsto b\otimes 1$. Contraction preserves primality. Every $\phi (a)$ with $a\in \mathfrak{p}$ maps to zero, while every $\phi (s)$ with $s\in S$ maps to a unit and hence cannot belong to a proper prime. Thus ${\phi }^{-1}(\mathfrak{q})=\mathfrak{p}$.

These constructions are inverse. For a prime disjoint from $\phi (S)$, the fraction $b/1$ belongs to its localized extension exactly when some $s\in S$ has $\phi (s)b\in \mathfrak{q}$; primality then implies $b\in \mathfrak{q}$. In the other direction, every element of $C$ has a representative $b/s$, and membership in $\mathfrak{Q}$ is equivalent to membership of $b/1$ because the denominator is a unit. Hence $\mathfrak{Q}$ is recovered by extending its contraction and quotienting. We have proved the canonical fiber of the map on spectra correspondence:If a denominator maps to zero, the fiber ring is the zero ring and both sets are empty; the same argument includes this case.

Let $\mathfrak{q}$ be a prime ideal of $C$. Its quotient ring $D=C/\mathfrak{q}$ is a nonzero integral domain and a finite-dimensional $K$-vector space. For $0\ne a\in D$, multiplication by $a$ is an injective $K$-linear map $D\to D$, because $D$ has no zero divisors. An injective endomorphism of a finite-dimensional vector space is surjective. In particular $ab=1$ for some $b\in D$. Every nonzero element is therefore invertible, so $D$ is a field and $\mathfrak{q}$ is a maximal ideal.

To prove finiteness, select any $r$ distinct maximal ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$. Distinct maximal ideals are comaximal: ${\mathfrak{m}}_i+{\mathfrak{m}}_j$ strictly contains ${\mathfrak{m}}_i$, so must equal $C$. The Chinese remainder theorem gives a surjectionHere is the needed surjectivity explicitly. For each $j\ne i$, choose $a_{ij}\in {\mathfrak{m}}_j$ with $a_{ij}\equiv 1(\mathrm{mod}{\mathfrak{m}}_i)$, using comaximality. Then $e_i={\prod }_{j\ne i}{a}_{ij}$ is one modulo ${\mathfrak{m}}_i$ and zero modulo the other selected ideals. Given any residue tuple represented by $b_i$, the element ${\sum }_i{b}_i{e}_i$ maps to that tuple. Thus the displayed map really is surjective.

Each nonzero field quotient has $K$-dimension at least one, soThere cannot be more than ${\dim }_{K}C$ maximal ideals, because otherwise one could select a finite list exceeding this bound. ConsequentlyThis proves the description of the spectrum of a finite-dimensional commutative algebra. If $C=0$, its spectrum is empty and the same conclusion holds.

Choose algebra generators $b_1,\dots ,b_n$ for $B$ over $A$. Because the extension is integral, each $b_i$ satisfies a monic equationRepeatedly reduce any exponent at least $e_i$ using this equation. Since the generators commute, every polynomial in them becomes an $A$-linear combination of the finitely many monomialsThus $B$ is a finitely generated module over $A$, with at most $D={\prod }_i{e}_i$ generators. This proves the relevant subalgebra generated by finitely many integral elements assertion rather than merely invoking it.

For a fixed prime ideal $\mathfrak{p}\subset A$, form the fiber ring $C=B{\otimes }_A\kappa (\mathfrak{p})$. Tensoring the finite generating list shows ${\dim }_{\kappa (\mathfrak{p})}C\le D$: its pure tensors with one span the fiber over the residue field $\kappa (\mathfrak{p})$. Part (b) therefore gives only finitely many primes in $C$. By the canonical bijection in part (a), these are precisely the primes of $B$ contracting to $\mathfrak{p}$. HenceNo Noetherian hypothesis on $A$ is needed. For an empty algebra-generating list, $B=A$ and the empty product $D=1$ gives the same conclusion.