Past exam of the commutative algebra course of the University of Oxford / 2017 (A12191W1)

Source: released examination PDF, bundle pages 398–399.

A commutative ring $R$ is a Noetherian ring if every ideal has a finite generating set of an ideal. Equivalently, it satisfies the ascending chain condition on ideals. Indeed, the union of an ascending chain is an ideal, and finitely many generators all belong to one term, after which the chain stabilizes. Conversely, an ideal without a finite generating set of an ideal would allow us to choose successively $a_{n+1}\notin (a_1,\dots ,a_n)$, producing a strictly ascending chain of ideals.

Let $\varphi :R\to S$ be a surjective ring homomorphism, where $R$ is a Noetherian ring. For any ideal $J$ of $S$, its inverse image is an ideal of $R$, say ${\varphi }^{-1}(J)=(a_1,\dots ,a_r)$. Taking images givesThus homomorphic images of Noetherian rings are Noetherian. This also applies to the image of an arbitrary ring homomorphism, by taking its image as the codomain.

The Hilbert basis theorem states that if $R$ is a Noetherian ring, then the polynomial ring $R[t]$ is a Noetherian ring. Repeating this gives the result for finitely many variables.

To prove it, fix an ideal $K\subseteq R[t]$. For $n\ge 0$, define the coefficient ideals of a polynomial idealHere the zero polynomial is allowed. Each $A_n$ is an ideal: addition and multiplication by constants preserve the required degree bound. Multiplying a representing polynomial by $t$ gives $A_n\subseteq A_{n+1}$. The ascending chain condition therefore supplies $N$ such that $A_n=A_N$ for every $n\ge N$.

For each $0\le n\le N$, choose a finite generating set of an ideal $a_{n1},\dots ,a_{n{r}_n}$ for $A_n$, omitting zero generators, and choose $f_{nj}\in K$ of polynomial degree $n$ with leading coefficient $a_{nj}$. If $A_n=0$, choose no such polynomials. We claim that the finitely many $f_{nj}$ generate $K$.

Take a nonzero $f\in K$ of polynomial degree $d$. Put $n=\min (d,N)$. Its leading coefficient belongs to $A_d=A_n$, so write it as ${\sum }_j{b}_j{a}_{nj}$. The polynomialbelongs to $K$ and is either zero or has polynomial degree strictly below $d$. Induction on polynomial degree now expresses $f$ as a combination of the chosen $f_{nj}$; when $d=0$, the displayed difference is zero. Hence $K$ is finitely generated. Since $K$ was arbitrary, $R[t]$ is a Noetherian ring.

For $x\in X$, let $e_x$ be the indicator function of $\{x\}$. In the ring of functions $R={\mathbb{R}}^X$, multiplication is pointwise, soSuppose first that $X$ is finite, and let $J$ be an ideal of $R$. Define $S=\{x\in X:f(x)\ne 0\text{ for some }f\in J\}$. If $x\in S$, choose such an $f$; multiplying it by $e_x/f(x)$ shows that $e_x\in J$. Every $f\in J$ satisfieswhere the real scalars are constant functions. Thus $J=(e_x:x\in S)$ has a finite generating set of an ideal, so $R$ is a Noetherian ring. If $X$ is empty, $R$ is the zero ring, and the conclusion still holds.

If $X$ is infinite, consider the idealIt is an ideal because a sum has support contained in the union of the two supports, and multiplication by any function cannot enlarge support. If $f_1,\dots ,f_r$ generated $F$, their supports would have a finite union $T$. Every element of $(f_1,\dots ,f_r)$ would vanish outside $T$, whereas $e_x\in F$ for any $x\in X\setminus T$. This contradiction proves

No. Write $x_i=t_i+I$ in the quotient ring $R=L/I$. Its defining relations areConsequently $x_i^{2^{i-1}}=x_1\ne 0$, using the permitted assumption, while $x_i^{2^i}=0$. Thus every $x_i$ is a nonzero nilpotent element.

The principal ideals satisfyEvery inclusion is strict. Otherwise $x_{i+1}=a{x}_i=a{x}_{i+1}^2$ for some $a\in R$, and henceThe fact that one minus a nilpotent element is a unit applies: if $z^m=0$, then $(1-z)(1+z+\cdots +z^{m-1})=1$. Here $z=a{x}_{i+1}$ is nilpotent, so multiplication by this inverse would give $x_{i+1}=0$, a contradiction. The strictly ascending chain of ideals violates the ascending chain condition. Therefore

The radical of an ideal $I$ isWe first establish the properties needed below. If $a^m,b^n\in I$, every term of the binomial theorem expansion of $(a+b{)}^{m+n-1}$ contains either $a^m$ or $b^n$, so $a+b\in \sqrt{I}$. Also $(-a{)}^m\in I$ and $(ra{)}^m=r^m{a}^m\in I$. Thus $\sqrt{I}$ is an ideal. The definition gives $I\subseteq \sqrt{I}$ and shows that $I\subseteq J$ implies $\sqrt{I}\subseteq \sqrt{J}$. Finally, if $a^n\in \sqrt{I}$, then $(a^n{)}^m\in I$ for some $m$, so $a\in \sqrt{I}$. Hence $\sqrt{\sqrt{I}}=\sqrt{I}$; in other words, $\sqrt{I}$ is a radical ideal.

We prove the stronger fact that the radical of an ideal is the intersection of the prime ideals containing it:If $a\in \sqrt{I}$ and $P\supseteq I$ is a prime ideal, then $a^n\in P$ for some $n$, and repeated application of the definition of a prime ideal gives $a\in P$. This proves one inclusion.

Conversely, fix $a\notin \sqrt{I}$. The multiplicatively closed set $T=\{1,a,a^2,\dots \}$ is disjoint from $I$. Consider the ideals containing $I$ and disjoint from $T$. This collection is nonempty, and the union of any chain in it is another such ideal. By Zorn's lemma, there is a maximal member $P$. It is proper because $1\in T$.

To prove that $P$ is a prime ideal, suppose $bc\in P$ but $b,c\notin P$. Maximality forces both $P+(b)$ and $P+(c)$ to meet $T$. Thus, for some $u,v\ge 0$,Multiplication gives $a^{u+v}\in P$, since every term lies in $P$, including $rsbc$. This contradicts disjointness from $T$. Therefore $P$ is a prime ideal containing $I$ and avoiding $a$. This proves the other inclusion, including the required prime ideal avoiding a multiplicative subset argument without using any unproved facts about the nilradical.

If $I=\sqrt{I}$, the displayed intersection represents $I$ as required. Conversely, an intersection of ideals that are all prime ideals is a radical ideal: if $a^n$ belongs to every one of them, then so does $a$. ConsequentlyThe unit ideal $I=R$ is included by interpreting the empty intersection as $R$.

If $\sqrt{I}+\sqrt{J}=R$, write $1=a+b$ with $a\in \sqrt{I}$ and $b\in \sqrt{J}$. Choose $m,n\ge 1$ such that $a^m\in I$ and $b^n\in J$. In the binomial theorem expansioneach term has either at least $m$ factors $a$, or at least $n$ factors $b$. Hence every term belongs to the sum of ideals $I+J$, andThus comaximal ideals can be recognized from their radicals of ideals.

For the general identity, $I+J\subseteq \sqrt{I}+\sqrt{J}$ givesOn the other hand, $I,J\subseteq I+J$ implies $\sqrt{I},\sqrt{J}\subseteq \sqrt{I+J}$. Since the latter is an ideal, $\sqrt{I}+\sqrt{J}\subseteq \sqrt{I+J}$. Taking radicals of ideals and using the idempotence proved in a yieldsTherefore the radical of a sum of ideals satisfies

Fix $n\ge 1$ and suppose, for a contradiction, that $I^n=I^{n+1}$. Set $M=I^n$. Since $R$ is a Noetherian ring, this ideal is a finitely generated module, say $M=(u_1,\dots ,u_r)$. It is nonzero: if $0\ne a\in I$, then $a^n\ne 0$ because $R$ is an integral domain.

The equality $M=IM$ allows us to writeLet $A=(a_{ij})$ and $B={\mathrm{Id}}_r-A$. Then $B(u_1,\dots ,u_r{)}^{\mathsf{T}}=0$. The adjugate matrix identity $\mathrm{adj}(B)B=(\det B){\mathrm{Id}}_r$ implies that $\det B$ annihilates every $u_i$. In the determinant expansion of ${\mathrm{Id}}_r-A$, the constant term is $1$ and every other term contains an entry of $A$, soThis is the determinant trick. Since some $u_i\ne 0$, multiplicative cancellation in an integral domain gives $1-b=0$. But then $1=b\in I$, contradicting $I\ne R$. We have proved the strict descent of powers of a nonzero proper ideal in a Noetherian domain:All these powers of an ideal have the same radical of an ideal. Indeed, $I^n\subseteq I$ gives $\sqrt{I^n}\subseteq \sqrt{I}$. Conversely, if $x^m\in I$, then $x^{mn}\in I^n$, so $x\in \sqrt{I^n}$. Thusand $J=I,I^2,I^3,\dots$ gives infinitely many distinct ideals with the prescribed radical of an ideal. If the convention $0\in \mathbb{N}$ is used, the additional inequality $I^0=R\ne I$ follows immediately from properness.

No. A domain with a nonzero idempotent maximal ideal is a counterexample. ConsiderThis can be realized concretely as a ring of functions on $t>0$: its elements are finite sums of rational multiples of nonnegative dyadic rational powers of $t$. Each stage $\mathbb{Q}[u_n]$ is a polynomial ring, because a nonzero polynomial cannot vanish for all $u_n>0$. Any two elements lie in one stage, which is an integral domain, so their product is nonzero if both are nonzero. Thus $R$ is an integral domain.

The constant term defines a surjective ring homomorphism $ϵ:R\to \mathbb{Q}$. It is well defined because replacing $u_n$ by $u_{n+1}^2$ preserves the constant term. Its kernel of a ring homomorphismis a maximal ideal, since $R/\mathfrak{m}\cong \mathbb{Q}$ is a field, by the maximal ideal quotient criterion. Also $u_0=t\in \mathfrak{m}$ is nonzero.

If $f\in \mathfrak{m}$, choose a stage in which $f=u_{n}g(u_n)$ for a polynomial $g\in \mathbb{Q}[z]$. ThenBoth factors belong to $\mathfrak{m}$, so $f\in {\mathfrak{m}}^2$. The reverse inclusion holds for every ideal, proving that $\mathfrak{m}$ is an idempotent ideal. HenceThe Noetherian ring hypothesis in c is therefore essential even for a maximal ideal.

The Krull dimension of a commutative ring $R$ isThe length of a prime chain counts strict inclusions, rather than the number of prime ideals.

Let $R$ be a principal ideal domain that is not a field. We first show that every nonzero prime ideal is a maximal ideal. Write such a prime ideal as $P=(p)$ with $p\ne 0$, and suppose $P\subseteq (a)$. Then $p=ab$ for some $b\in R$. Since $ab\in P$, either $a\in P$, in which case $(a)=P$, or $b\in P$. In the second case $b=pc$ for some $c$, and $p=apc$ gives $1=ac$ by multiplicative cancellation in an integral domain. Thus $(a)=R$. This proves maximality.

Since $R$ is an integral domain, $(0)$ is a prime ideal, and every other prime ideal is maximal. Consequently every prime chain has length at most one. For the opposite inequality, choose a nonzero nonunit $a$, which exists because $R$ is not a field. The proper principal ideal $(a)$ is contained in a maximal ideal $\mathfrak{m}$; existence follows from Zorn's lemma applied to the proper ideals containing $(a)$. A maximal ideal is a prime ideal, since its quotient ring is a field. Therefore $(0)⊊\mathfrak{m}$ is a prime chain of length one. Thus the Krull dimension of a principal ideal domain that is not a field is

Let $P$ be a prime ideal of $R$ disjoint from the multiplicatively closed set $Y$. In the localization $S=Y^{-1}R$, its extension of an ideal isIndeed, finite sums of such fractions have a common denominator in $Y$ and numerator in $P$, and every generator $p/1$ has this form.

We show that its contraction of an ideal recovers $P$:One inclusion is immediate. Conversely, if $r/1=p/y$ with $p\in P$ and $y\in Y$, equality of fractions in the field of fractions of the integral domain $R$ gives $yr=p\in P$. Since $y\notin P$ and $P$ is a prime ideal, $r\in P$.

If $S{P}_1=S{P}_2$, contracting would therefore give $P_1=P_2$, contrary to the hypothesis. HenceWe will also need the fact that $SP$ is a prime ideal, part of the prime ideal correspondence for localization. It is proper by the contraction identity, since otherwise $1\in P$. If $(a/u)(b/v)\in SP$, multiplication by the denominator gives $ab/1\in SP$, and contraction gives $ab\in P$. Thus $a\in P$ or $b\in P$, so one of the two fractions belongs to $SP$.

Suppose, for a contradiction, that $P_2\cap \mathbb{Z}=(0)$. Then $P_1\cap \mathbb{Z}=(0)$ too. Both prime ideals are disjoint from the multiplicatively closed set $Y=\mathbb{Z}\setminus \{0\}$. The localization iseach fraction on the left has rational coefficients, and every polynomial with rational coefficients has a common nonzero integer denominator.

By b, the extended prime ideals form a strict inclusionThe first is nonzero, because $P_1$ contains a nonzero polynomial and the map $\mathbb{Z}[t]\to \mathbb{Q}[t]$ is injective.

The polynomial ring $\mathbb{Q}[t]$ is a principal ideal domain: in any nonzero ideal, a nonzero polynomial of least polynomial degree generates the ideal, since the polynomial division algorithm gives a remainder of smaller polynomial degree, which must be zero. By a, every nonzero prime ideal of this principal ideal domain is maximal. The strict inclusion of extended prime ideals is therefore impossible. We obtain

For any prime number $p$, there is a prime chainin $\mathbb{Z}[t]$. These are prime ideals because the respective quotient rings are $\mathbb{Z}[t]$, ${\mathbb{F}}_p[t]$, and ${\mathbb{F}}_p$, all integral domains. The inclusions are strict because $p\ne 0$ and $t\notin (p)$. Hence $\dim \mathbb{Z}[t]\ge 2$.

To prove the upper bound, suppose there were a prime chain of length at least three. Since $\mathbb{Z}[t]$ is an integral domain, we could insert $(0)$ at its beginning if necessary and select a chainBy c, $P_2\cap \mathbb{Z}\ne (0)$. This intersection is a proper prime ideal of $\mathbb{Z}$: if a product of integers lies in it, primality of $P_2$ puts one factor in it. Therefore $P_2\cap \mathbb{Z}=(p)$ for a prime number $p$, and $(p)\subseteq P_2$.

We need to check that $P_2\ne (p)$. Here prime ideals below a principal prime give the necessary restriction: no nonzero prime ideal can lie strictly below $(p)$. If $0\ne P⊊(p)$, then $p\notin P$. Choose $0\ne f\in P$ and factor out the largest common power of $p$ from its integer coefficients:Such a finite $k$ exists because $f$ has at least one nonzero integer coefficient. Since $p\notin P$ and $P$ is a prime ideal, repeated primality gives $g\in P\subseteq (p)$, a contradiction. Applied to $P_1⊊P_2$, this rules out $P_2=(p)$.

It follows that $P_2/(p)$ is a nonzero prime ideal ofAs in c, the polynomial division algorithm makes ${\mathbb{F}}_p[t]$ a principal ideal domain, so its nonzero prime ideals are maximal by a. Thus $P_2/(p)$ is maximal. The ideal correspondence theorem makes $P_2$ a maximal ideal of $\mathbb{Z}[t]$, contradicting $P_2⊊P_3$. Consequently