Past exam of the commutative algebra course of the University of Oxford / 2017 / 2

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.