Past exam of the commutative algebra course of the University of Oxford

Source: released examination PDF, bundle pages 249–251; paper pages 1–3.

An ideal $\mathfrak{q}$ is $\mathfrak{p}$-primary if it is a proper primary ideal whose radical of an ideal is $\mathfrak{p}$. Explicitly,Equivalently, $ab\in \mathfrak{q}$ and $b\notin \mathfrak{p}$ imply $a\in \mathfrak{q}$: multiplication by every element outside $\mathfrak{p}$ is injective on the quotient ring $R/\mathfrak{q}$.

A minimal primary decomposition is a finite expressionin proper primary ideals such that the prime ideals ${\mathfrak{p}}_j=\sqrt{{\mathfrak{q}}_j}$ are pairwise distinct and no primary ideal can be omitted. More explicitly, for every $j$,Thus minimality means both irredundancy and distinct radicals of ideals; it does not assert that the primary ideals themselves are uniquely determined. For $I=R$, one may use the empty intersection convention.

In the terminology of a minimal primary decomposition, an associated prime of a primary decomposition of $I$ is a prime ideal occurring as $\sqrt{{\mathfrak{q}}_j}$ in such a decomposition. The first uniqueness theorem for primary decomposition says that this collection is independent of the chosen minimal primary decomposition. Among these prime ideals, an isolated prime of a primary decomposition is minimal under inclusion, equivalently a minimal prime over an ideal $I$; an embedded prime ideal strictly contains another member of the collection. ThusWhen $R$ is a Noetherian ring, these are precisely the associated primes of a module $R/I$. In that case the equivalent definition iswhere $(I:x)$ is a colon ideal. The decomposition terminology applies when a minimal primary decomposition exists; a general ring need not provide one for every ideal.

Suppose first that $\mathfrak{a}$ is $\mathfrak{m}$-primary. Since $R$ is a Noetherian ring, choose a finite generating set of an ideal $\mathfrak{m}=(x_1,\dots ,x_d)$. The equality of radicals of ideals $\sqrt{\mathfrak{a}}=\mathfrak{m}$ supplies positive integers $e_i$ with $x_i^{e_i}\in \mathfrak{a}$. SetEvery product of $n$ generators has some $x_i$ appearing at least $e_i$ times. Such products generate the product of ideals ${\mathfrak{m}}^n$, so ${\mathfrak{m}}^n\subseteq \mathfrak{a}$. Also $\mathfrak{a}\subseteq \sqrt{\mathfrak{a}}=\mathfrak{m}$. If $\mathfrak{m}=0$, take $n=1$ directly.

Conversely, suppose ${\mathfrak{m}}^n\subseteq \mathfrak{a}\subseteq \mathfrak{m}$. Taking radicals of ideals givesHence $\sqrt{\mathfrak{a}}=\mathfrak{m}$. The only maximal ideal containing $\mathfrak{a}$ is $\mathfrak{m}$: any such maximal ideal contains $\sqrt{\mathfrak{a}}$ and therefore equals $\mathfrak{m}$. By the correspondence theorem for ideals, the quotient ring $R/\mathfrak{a}$ is a local ring with maximal ideal $\mathfrak{m}/\mathfrak{a}$. Every element outside this maximal ideal is a unit.

If $ab\in \mathfrak{a}$ and $a\notin \mathfrak{a}$, then $b$ cannot be outside $\mathfrak{m}$, since multiplying by its inverse in the quotient ring would give $a\in \mathfrak{a}$. Thus $b\in \mathfrak{m}$ and $b^n\in \mathfrak{a}$, proving the required primary ideal condition. We have proved the maximal-primary ideal criterion

Put $A=\mathbb{C}[X,Y]$, $P=(X)$ and $Q=(X^2,Y)$. The quotient ring $A/P\cong \mathbb{C}[Y]$ is an integral domain, so $P$ is a prime ideal and hence a primary ideal. Put $\mathfrak{m}=(X,Y)$. ThenThe polynomial ring $A$ is a Noetherian ring by the Hilbert basis theorem, and $\mathfrak{m}$ is a maximal ideal since $A/\mathfrak{m}\cong \mathbb{C}$. Therefore the maximal-primary ideal criterion from c makes $Q$ $\mathfrak{m}$-primary. In particular its radical of an ideal is $\mathfrak{m}$, distinct from $P$.

Neither primary ideal can be omitted: $X\in P\setminus Q$, while $Y\in Q\setminus P$. Thus the given intersection is already a minimal primary decomposition. To express the ideal by generators, write an element of $P\cap Q$ as $Xh=X^{2}u+Yv$. Reducing modulo the prime ideal $(X)$ gives $Y\bar{v}=0$ in the integral domain $\mathbb{C}[Y]$, so $v=Xw$. Cancelling $X$ in the integral domain $A$ gives $h=Xu+Yw$. HenceBy the first uniqueness theorem for primary decomposition, the associated primes of a module $A/J$ are exactly the two radicals of ideals. Their strict inclusion $P⊊\mathfrak{m}$ givesOne can also see both associated primes of a module directly through annihilators:For the first colon ideal, reducing $hY\in J$ modulo $(X)$ forces $h\in (X)$, and the reverse inclusion follows from $XY\in J$. For the second colon ideal, cancel $X$ in $hX\in X(X,Y)$. Both represented elements of the quotient module $A/J$ are nonzero.

Because $(r)$ is a proper prime ideal, each $(r^k)$ is proper and its radical of an ideal isIndeed, $(r^k)\subseteq (r)$ gives one inclusion of radicals of ideals, while $(rc{)}^k\in (r^k)$ gives the other.

We prove the saturation condition for a primary ideal: if $ab\in (r^k)$ and $a\notin \mathfrak{p}$, then $b\in (r^k)$. For $k=1$ this is the defining property of a prime ideal. For $k>1$, write $ab=r^{k}c$. Since $a\notin (r)$, the prime ideal condition gives $b=r{b}_1$. The element $r$ is a non-zero-divisor, so cancellation givesInduction gives $b_1\in (r^{k-1})$, hence $b\in (r^k)$. Together with the computed radical of an ideal, this proves

Since the radical of an ideal $\sqrt{\mathfrak{q}}$ is $(r)$, some positive power of $r$ belongs to $\mathfrak{q}$. Choose the least such exponent $k$. Then $(r^k)\subseteq \mathfrak{q}$ and $r^j\notin \mathfrak{q}$ for $0\le j<k$, with $r^0=1$ excluded because the primary ideal $\mathfrak{q}$ is proper.

Take $x\in \mathfrak{q}$. If $x\notin (r^k)$, choose the largest $j\in \{0,\dots ,k-1\}$ for which $x\in (r^j)$, and write $x=r^{j}u$. Maximality of $j$ implies $u\notin (r)=\sqrt{\mathfrak{q}}$: otherwise $x\in (r^{j+1})$, a contradiction, including when $j=k-1$. The primary ideal condition applied to $r^{j}u\in \mathfrak{q}$ therefore forces $r^j\in \mathfrak{q}$, contrary to the choice of $k$.

Consequently every $x\in \mathfrak{q}$ lies in $(r^k)$, andTogether with i, this classifies the primary ideals for a regular principal prime. The exponent is unique: if $(r^k)=(r^{k+1})$, then $r^k=r^{k+1}c$, and cancellation of the non-zero-divisor $r^k$ would give $1=rc$, contradicting that $(r)$ is proper. No Noetherian hypothesis is needed.

The image $f/1$ is a unit in the localization at a prime ideal $R_{\mathfrak{p}}$, because $f\notin \mathfrak{p}$. The universal property of localization, applied to the ring homomorphism $R\to R_{\mathfrak{p}}$ and the multiplicative subset $\{1,f,f^2,\dots \}$, gives a unique ring homomorphism fixing the image of $R$:Fixing the image of $R$ is exactly the required compatibility with the algebra over a ring structures. Uniqueness also follows directly: such a ring homomorphism must send $1/f$ to the inverse of $f/1$, and these elements together with $R$ generate $R_f$.

Choose $f=1$, so that $R_f=R$ and $f\notin \mathfrak{p}$. For an integral domain, the canonical ring homomorphism $R\to R_{\mathfrak{p}}$ is injective: if $a/1=0$, the vanishing criterion in a module localization gives $sa=0$ for some $s\notin \mathfrak{p}$. Such an $s$ is nonzero, so the absence of zero divisors gives $a=0$. Thus

Consider the kernel of a ring homomorphismSince $R$ is a Noetherian ring, this ideal has a finite generating set of an ideal, say $K=(a_1,\dots ,a_t)$. For each $i$, choose $s_i\notin \mathfrak{p}$ with $s_i{a}_i=0$, and putThen $f\notin \mathfrak{p}$ by the prime ideal condition, and $fK=0$. If $K=0$, take the empty product $f=1$.

If $a/f^n$ maps to zero under the ring homomorphism ${\theta }_f$ from a, then $a/1=0$ in $R_{\mathfrak{p}}$ because $f/1$ is a unit there. Hence $a\in K$, so $fa=0$. The vanishing criterion in a module localization now gives $a/f^n=0$ already in $R_f$. ThereforeThis is the finite generation clears localization kernels argument: the finite generating set of an ideal permits a single element outside $\mathfrak{p}$ to annihilate the entire kernel of a ring homomorphism.

For $n\ge 1$, let $K_n=\ker ({\varphi }^n)$, where the power denotes repeated composition of the ring endomorphism. These ideals form an ascending chain:The ascending chain condition for a Noetherian ring gives $K_N=K_{N+1}$ for some $N\ge 1$. Take $a\in \ker \varphi$. Since $\varphi$ is surjective, its iterate ${\varphi }^N$ is surjective, so choose $b\in R$ with ${\varphi }^N(b)=a$. Thenso $b\in K_{N+1}=K_N$. Therefore $a={\varphi }^N(b)=0$.

Thus the kernel of a ring homomorphism $\varphi$ is zero. This proves that a surjective endomorphism of a Noetherian ring is an automorphism, and in particular

Use the direct product of ringswith coordinatewise addition and multiplication and unit $(1,1,\dots )$. Define the shiftCoordinatewise operations show that $g$ preserves addition, multiplication and the unit, so it is a ring endomorphism. It is surjective because $(0,b_0,b_1,\dots )$ maps to any prescribed $(b_0,b_1,\dots )$. Its kernel of a ring homomorphism isso it is not injective. This is the noninjective surjective shift on a product of fields.

This ring fails the Noetherian hypothesis in c. Indeed, let $e_n$ have a single $1$ in coordinate $n$. The ideals $I_n=(e_0,\dots ,e_n)$ form the strictly ascending chainsince $e_{n+1}\notin I_n$.

A ring is reduced if its only nilpotent element is zero. Equivalently,The second expression says that its nilradical vanishes. A reduced ring may have zero divisors; for example, the direct product of rings $\mathbb{R}\times \mathbb{R}$ is reduced but $(1,0)(0,1)=0$.

Write $N=\sqrt{(0)}$. Every element of the extension of an ideal $N{R}_{\mathfrak{p}}$ is a fraction $a/s$ with $a\in N$ and $s\notin \mathfrak{p}$: a finite sum of such fractions has a common denominator and numerator still in the ideal $N$. If $a^n=0$, then $(a/s{)}^n=0$, soConversely, let $a/s$ be a nilpotent element of the localization at a prime ideal $R_{\mathfrak{p}}$. For some $n\ge 1$, $a^n/s^n=0$. The vanishing criterion in a module localization supplies $t\notin \mathfrak{p}$ with $t{a}^n=0$. It follows thatso $ta\in N$. Since $ts\notin \mathfrak{p}$,This proves that the nilradical commutes with localization:The multiplication by $t$ is necessary: the numerator $a$ of a nilpotent element in a localization need not itself be nilpotent in $R$.

If $R$ is reduced, its nilradical $N$ is zero. By b, every localization at a prime ideal has zero nilradical, so in particular $R_{\mathfrak{m}}$ is reduced for every maximal ideal $\mathfrak{m}$.

Conversely, suppose every such localization is reduced. Take a nilpotent element $x\in R$. Then $x/1$ is nilpotent in every $R_{\mathfrak{m}}$, hence zero there. To see directly that $x=0$, suppose otherwise and consider the proper annihilatorChoose a maximal ideal $\mathfrak{m}$ containing this annihilator. The equality $x/1=0$ in $R_{\mathfrak{m}}$ would imply $sx=0$ for some $s\notin \mathfrak{m}$, by the vanishing criterion in a module localization. But then $s\in {\mathrm{Ann}}_R(x)\subseteq \mathfrak{m}$, a contradiction. This is the principle that localization detects zero elements. ThereforeThe argument also includes the zero ring, which is reduced and has no maximal ideals.

Let $N$ be the nilradical of $R$. By b, the reduced locus of a ring isThe Noetherian ring hypothesis makes the ideal $N$ finitely generated; write $N=(n_1,\dots ,n_t)$. Fix $\mathfrak{p}\in \mathcal{R}$. For every $i$, the vanishing criterion in a module localization gives $s_i\notin \mathfrak{p}$ with $s_i{n}_i=0$. The product $f=s_1\cdots s_t$ lies outside the prime ideal $\mathfrak{p}$ and satisfies $fN=0$. When $N=0$, take $f=1$.

For every prime ideal $\mathfrak{q}$ in the distinguished open setthe element $f/1$ is a unit in $R_{\mathfrak{q}}$. The equality $fN=0$ therefore implies $N{R}_{\mathfrak{q}}=0$, so $R_{\mathfrak{q}}$ is reduced. Thus $\mathfrak{p}\in D(f)\subseteq \mathcal{R}$, proving that $\mathcal{R}$ is a Zariski-open subset of the spectrum of a ring.

More precisely, let $A={\mathrm{Ann}}_R(N)$, the annihilator of a module $N$. The argument shows that $N{R}_{\mathfrak{p}}=0$ if and only if some $f\in A$ lies outside $\mathfrak{p}$. HenceEquivalently, the complement is the support of a module $N$. This is vanishing of a finitely generated module near a prime applied to $N$: its finite generation lets the separately chosen denominators be combined into one element annihilating all of $N$.

Set $L=M/\mathfrak{a}M$, a quotient module. The exactness of localization gives, for every maximal ideal $\mathfrak{m}$,If $\mathfrak{m}\supseteq \mathfrak{a}$, the hypothesis $M_{\mathfrak{m}}=0$ makes this quotient module zero. If $\mathfrak{m}⊉\mathfrak{a}$, choose $s\in \mathfrak{a}\setminus \mathfrak{m}$. Its image in $R_{\mathfrak{m}}$ is a unit, so $\mathfrak{a}{R}_{\mathfrak{m}}=R_{\mathfrak{m}}$ and the displayed quotient module is again zero. ThusFor completeness, if $0\ne \ell \in L$, its annihilator is a proper ideal, and hence lies in a maximal ideal $\mathfrak{m}$. The vanishing criterion in a module localization shows $\ell /1\ne 0$ in $L_{\mathfrak{m}}$: any element annihilating $\ell$ lies inside $\mathfrak{m}$. This contradicts the vanishing just proved. Hence $L=0$, andThe use of localization detects zero elements is valid for every module; neither finite generation nor a Noetherian hypothesis is needed.

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