Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 (Commutative Algebra)

An $R$-module $M$ is Noetherian when every submodule is finitely generated, equivalently when every ascending chain of submodules stabilizes.

A free module is an $R$-module with a basis: every element has a unique expression as a finite linear combination of basis elements.

A flat module $M$ is one for which the tensor functor $-{\otimes }_{R}M$ is exact. Since tensor products are always right exact, it is equivalent to require that tensoring with $M$ preserve injections.

A projective module $P$ has the lifting property: for every surjection $E\twoheadrightarrow F$ and every map $P\to F$, there is a map $P\to E$ making the resulting triangle commute. Equivalently, $P$ is a direct summand of a free module.

The statement is true. If $M$ is generated by $m_1,\dots ,m_r$, there is a surjection $R^r\twoheadrightarrow M$. The right exactness of the tensor product of modules gives a surjectionA finite direct sum of Noetherian modules is Noetherian, and a quotient of a Noetherian module is Noetherian. Hence $M{\otimes }_{R}N$ is Noetherian.

The statement is true. Choose a maximal ideal $\mathfrak{m}\subset B$ and write $L=B/\mathfrak{m}$, a field extension of $k$. Taking the quotient by $A{\otimes }_k\mathfrak{m}$ givesso $A{\otimes }_{k}L$ is a finitely generated algebra over $k$, and therefore over $L$.

By finite generation descends along a field extension, $A$ is finitely generated over $k$. Indeed, collect the finitely many coefficients from $A$ occurring in a finite set of $L$-algebra generators of $A{\otimes }_{k}L$. If $A_0$ is the $k$-subalgebra generated by those coefficients, then $(A/A_0){\otimes }_{k}L=0$; because a field extension is a faithfully flat module, $A=A_0$. Interchanging $A$ and $B$ proves that $B$ is also finitely generated.

The statement is true. Fix an isomorphism $\varphi :M{\otimes }_{R}N\to R^n$, and write the inverse image of the first standard basis vector asDefineSince $\psi ((n_j{)}_j)=1$, this is a split surjection. Tensor its splitting with $M$. The resulting split surjection has the formThus $M$ is a direct summand of a finite free module, so it is a projective module. Symmetry gives the same conclusion for $N$. This is projectivity of factors of a nonzero finite free tensor product.

The statement is false. Give $\mathbb{C}[X,Y]$ its $\mathbb{C}[T]$-algebra structure byand use the quotient mapsThe induced maps send $T$ to $X$ and $Y$, respectively, so both polynomial rings are free of rank one, hence flat modules, over $\mathbb{C}[T]$.

Their tensor product over the middle ring isHere $T$ acts by zero, so this is the torsion module $\mathbb{C}[T]/(T)$. It is not flat: tensoring the injection $\mathbb{C}[T]\stackrel{T}{\to }\mathbb{C}[T]$ with it produces the zero map on a nonzero module.

The maximum isThe product of two fields $K\times L$ attains it: its only nonzero proper ideals are $K\times 0$ and $0\times L$, and both are maximal ideals.

For the upper bound, suppose ${\mathfrak{m}}_1$ and ${\mathfrak{m}}_2$ are distinct maximal ideals of $R$. Their intersection cannot be nonzero, since a nonzero proper ideal is maximal and cannot be properly contained in either of two distinct maximal ideals. Hence ${\mathfrak{m}}_1\cap {\mathfrak{m}}_2=0$. Also ${\mathfrak{m}}_1+{\mathfrak{m}}_2=R$, so the Chinese remainder theorem givesBoth factors are fields, and this product has exactly two maximal ideals. Thus a third maximal ideal is impossible.

For a commutative ring $R$, the Jacobson radical isLet $A\subseteq B$ be integral. If $\mathfrak{n}$ is maximal in $B$, then its contraction $\mathfrak{n}\cap A$ is maximal in $A$. Therefore every $a\in J(A)$ belongs to every $\mathfrak{n}$, andConversely, the Lying-over theorem puts a maximal ideal $\mathfrak{n}$ of $B$ above every maximal ideal $\mathfrak{m}$ of $A$. Hence an element of $J(B)\cap A$ lies in every $\mathfrak{m}$, provingThis is the Jacobson radical under an integral extension formula.

Take $R=\mathbb{Z}$, $M=\mathbb{Q}/\mathbb{Z}$, and $k=2$. The group $M$ is nonzero and divisible. Since every element of $M$ has finite order, $M$ is the filtered union of finite cyclic groups. For every $n\ge 1$,Tensor products commute with filtered colimits, so

Now suppose a nonzero finitely generated $R$-module $M$ satisfied $M^{\otimes k}=0$. Choose a maximal ideal $\mathfrak{m}$ in the support of $M$. The localized module $M_{\mathfrak{m}}$ is nonzero and finitely generated. By Nakayama lemma,This is a nonzero vector space over the residue field $\kappa (\mathfrak{m})$, so its $k$-fold tensor power is nonzero. But it is the reduction modulo $\mathfrak{m}$ of $(M^{\otimes k}{)}_{\mathfrak{m}}$, a contradiction. Thus a nonzero tensor-nilpotent module cannot be finitely generated.

Yes. Let $\mathfrak{m}$ be maximal and setThen $L$ is a field generated as a $\mathbb{C}$-algebra by countably many elements. Since the polynomial ring in countably many variables has a countable monomial basis, $L$ has at most countable dimension as a $\mathbb{C}$-vector space.

Suppose $t\in L$ were transcendental over $\mathbb{C}$. The familywould be linearly independent over $\mathbb{C}$. Indeed, after multiplying a finite relation by ${\prod }_j(t-x_j)$, evaluation at $t=x_i$ forces the $i$th coefficient to vanish. This would be an uncountable linearly independent subset of the countable-dimensional vector space $L$, a contradiction.

Thus $L/\mathbb{C}$ is algebraic. Since $\mathbb{C}$ is an algebraically closed field, $L=\mathbb{C}$. If $x_i$ is the image of $T_i$, the quotient map is evaluation at $(x_1,x_2,\dots )$ and

Take $V=U={\mathbb{R}}^3$ and identify $V{\otimes }_{\mathbb{R}}U$ with the space of $3\times 3$ real matrices, where pure tensors correspond to matrices of rank at most one. Letand let $f$ be the quotient map. It is not injective because its kernel is the nonzero line $\mathbb{R}{I}_3$.

If $f(v_1\otimes u_1)=f(v_2\otimes u_2)$, thenfor some $\lambda \in \mathbb{R}$. The left side has matrix rank at most two. If $\lambda \ne 0$, the right side has rank three, which is impossible. Thus $\lambda =0$ and the two pure tensors were equal. Hence $f$ is injective on the set of pure tensors while failing to be injective linearly.

The assertion is false. LetThe element $1/y$ does not belong to $A$: assigning degrees $\mathrm{deg}x=\mathrm{deg}y=1$ and $\mathrm{deg}(x/y)=0$ shows that every element of $A$ has nonnegative total degree, whereas $1/y$ has degree $-1$.

If $A=T^{-1}R$ for some multiplicative subset $T$, the membership $x/y\in T^{-1}R$ would give $x/y=a/t$ for some $a\in R$ and $t\in T$. Since $R$ is a unique factorization domain and $x,y$ are coprime, the equation $xt=ay$ implies $y\mid t$. Write $t=yu$. As $t$ is invertible in $T^{-1}R$, so iscontrary to $1/y\notin A$. Thus a subring of a localization of a ring need not itself be a localization of the original ring.

Choose $a\in K\setminus F$. The multiplication homomorphismhasThis element is nonzero. Since $1,a$ are linearly independent over $F$, there is an $F$-linear functional $\lambda :K\to F$ with $\lambda (1)=0$ and $\lambda (a)=1$. Applying $\lambda \otimes {\mathrm{id}}_K$ sends the displayed tensor to $1$.

Thus the unital ring homomorphism $\mu$ has a nonzero kernel. A unital homomorphism from a field is injective, so $K{\otimes }_{F}K$ is not a field.

The Krull intersection theorem givesbecause $(R,\mathfrak{m})$ is a Noetherian local ring and $x\in \mathfrak{m}$. Consequently every nonzero $a\in R$ has a largest $x$-adic order: one can write

Suppose nonzero elements $a,b$ satisfy $ab=0$. Write $a=x^r{a}^{\prime }$ and $b=x^s{b}^{\prime }$ with $a^{\prime },b^{\prime }\notin (x)$. Since $x$ is a non-zero-divisor, cancellation of $x^{r+s}$ gives $a^{\prime }{b}^{\prime }=0$. Reducing modulo $(x)$ now gives a product of two nonzero elements equal to zero in $R/(x)$, contradicting that this quotient is an integral domain. Hence $R$ is an integral domain.

For an algebraically closed field $k$, the Weak Hilbert Nullstellensatz says that every maximal ideal of $k[x_1,\dots ,x_n]$ isfor a unique $a\in k^n$. Equivalently, every proper ideal has a common zero. The Strong Hilbert Nullstellensatz says that for every ideal $I$,

To deduce the strong form, let $f$ vanish on $V(I)$ and introduce a variable $y$. The equations in $I$ together with $1-yf$ have no common zero: a common zero would satisfy both $f=0$ and $yf=1$. The weak theorem therefore givesSubstitute $y=f^{-1}$ in the localization $k[x_1,\dots ,x_n,f^{-1}]$. The last term vanishes, and clearing a power of $f$ yields $f^N\in I$. Thus $f\in \sqrt{I}$. The reverse inclusion is immediate, completing the Rabinowitsch trick proof.

TakeThenEvery prime ideal in the image ofis disjoint from the multiplicative set $\{1,t_1,t_1^2,\dots \}$. In particular, the prime ideal $(t_1)$ is not in the image, so the contraction map is not surjective.

Let $d=\mathrm{deg}f$ and let $f_d$ be the nonzero highest homogeneous part. Choose one coordinate, after a permutation, such thatis not the zero polynomial. A polynomial of degree at most $d$ in each variable cannot vanish on the entire grid $\{0,1,\dots ,d{\}}^{n-1}$, by induction on the number of variables. Hence there are $a_i\in \{0,1,\dots ,d\}$ such that

SetThis is given, up to the initial coordinate permutation, by an integer matrix $M$ with determinant $\pm 1$ andIn the inverse coordinates $t_i=y_i+a_i{y}_n$, the coefficient of $y_n^d$ in $f$ is the nonzero real number $f_d(a_1,\dots ,a_{n-1},1)$. Dividing by it makes the defining equation monic in $y_n$. Thus $A$ is integral over $\mathbb{R}[y_1,\dots ,y_{n-1}]$ by linear Noether normalization for a hypersurface. The Lying-over theorem now makessurjective.

Pass to the integral extensionThe ring $B/I$ is an integral domain, and the ideal $J/I$ has zero contraction to the base. If $J/I$ contained a nonzero element $b$, choose an integral equation for $b$ of least degree:Its constant term is nonzero, since otherwise the domain property would let us cancel $b$ and obtain an equation of lower degree. Butis a nonzero element of $(J/I)\cap (A/(I\cap A))$, a contradiction. Therefore $J/I=0$ and $I=J$. This proves ideal contraction rigidity under an integral extension.

It is enough to prove that every prime ideal $\mathfrak{q}$ of $S$ is the intersection of the maximal ideals containing it. Replace $R\subseteq S$ byThe new extension is integral, both rings are domains, and the base remains a Jacobson ring.

Let $0\ne s\in S$. Choose an integral equation of least degreeAs above, $a_0\ne 0$. Since the zero ideal of $R$ is the intersection of its maximal ideals, choose a maximal ideal $\mathfrak{m}$ with $a_0\notin \mathfrak{m}$. By the Lying-over theorem, some maximal ideal $\mathfrak{n}$ of $S$ contracts to $\mathfrak{m}$. If $s\in \mathfrak{n}$, the integral equation would imply $a_0\in \mathfrak{n}\cap R=\mathfrak{m}$, a contradiction. Thus every nonzero $s$ is omitted by some maximal ideal, so their intersection is zero. Therefore $S$ is Jacobson, proving Integral extension of a Jacobson ring.

For a prime ideal $\mathfrak{p}$, its height is the supremum of the lengths of strict chains of prime ideals ending at $\mathfrak{p}$. For a proper ideal $I$,This is the height of an ideal.

The Krull height theorem states that if $R$ is Noetherian, $I=(a_1,\dots ,a_m)$, and $\mathfrak{p}$ is minimal over $I$, then

We prove it by induction on $m$. The case $m=1$ is the Krull principal ideal theorem. For the induction step, let ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_s$ be the finitely many minimal primes over $(a_2,\dots ,a_m)$ that lie below $\mathfrak{p}$. By induction each has height at most $m-1$; if one equals $\mathfrak{p}$, we are done.

Otherwise, suppose $\mathfrak{p}$ has finite height $d$ and choose a chainwhose first nonminimal term ${\mathfrak{p}}_1$ is contained in none of the ${\mathfrak{q}}_i$. Such a chain is obtained by prime avoidance and the principal ideal theorem: a three-term segment can be replaced by a prime minimal over ${\mathfrak{p}}_0+(b)$ for an element $b$ avoiding the finitely many unwanted primes. Choose

The prime $\mathfrak{p}$ is minimal over $(b,a_2,\dots ,a_m)$. Otherwise a prime strictly between some ${\mathfrak{q}}_i$ and $\mathfrak{p}$ would show that $\mathfrak{p}/(a_2,\dots ,a_m)$ has height at least two, although it is minimal over the principal ideal generated by $a_1$; this contradicts the principal ideal theorem. In $R/(b)$, the prime $\mathfrak{p}/(b)$ is therefore minimal over an ideal generated by $m-1$ elements, so induction bounds its height by $m-1$. The $d-1$ strict inclusions from ${\mathfrak{p}}_1/(b)$ to $\mathfrak{p}/(b)$ giveand hence $d\le m$. If the height were infinite, the same argument applied to arbitrarily long finite chains would give the same fixed bound, which is impossible. This completes the proof.

Let $S=k[x_1,\dots ,x_r]$ with positive degrees $d_i=\mathrm{deg}{x}_i$, and let $M$ be a finitely generated graded $S$-module whose graded pieces are finite-dimensional over $k$. The Hilbert-Serre theorem states thatfor some Laurent polynomial $P(t)\in \mathbb{Z}[t,t^{-1}]$. For the standard grading, the Hilbert function $n\mapsto {\dim }_k{M}_n$ consequently agrees for all sufficiently large $n$ with a polynomial in $n$.

For the proof, induct on $r$. When $r=0$, $M$ is finite-dimensional and its Hilbert series is a Laurent polynomial. For $r>0$, multiplication by $x_r$ gives an exact sequence of graded modulesBoth $K$ and $C$ are annihilated by $x_r$, so they are finitely generated graded modules over $k[x_1,\dots ,x_{r-1}]$. Additivity of the Hilbert series yieldsThe induction hypothesis supplies the required denominator for the right side and proves the rational formula. When all $d_i=1$, expanding $(1-t{)}^{-r}$ shows that its coefficients are binomial polynomials in $n$, which proves eventual polynomiality.

The cases $n=0$ and $n=d$ can indeed be finite: a Noetherian ring has finitely many minimal primes, and a semilocal ring may have finitely many maximal ideals. We prove that every intermediate height occurs infinitely often.

Because $d$ is a finite integer equal to the supremum of prime-chain lengths, there is a chainEach ${\mathfrak{p}}_i$ has height exactly $i$: its displayed lower chain gives height at least $i$, while any longer lower chain could be extended by the remaining displayed primes and would contradict $\mathrm{ht}{\mathfrak{p}}_d=d$.

Suppose $0<n<d$. The three primesfall under prime ideals between a three-prime chain, so infinitely many primes $\mathfrak{q}$ satisfyEvery such $\mathfrak{q}$ has height exactly $n$: the lower inclusion gives height at least $n$, and height at least $n+1$ would, after appending ${\mathfrak{p}}_{n+1}$, contradict its height $n+1$. Therefore there are infinitely many height-$n$ primes. The assumed finiteness forcesThis is infinitude of intermediate-height prime ideals.