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For a commutative ring $R$ that is a Noetherian ring and a multiplicatively closed set $S$, a finitely generated module $N$ has a free resolution with every term finitely generated. Hom from a finite free module commutes with localization, and exactness of localization commutes with cohomology. Thus the displayed isomorphism holds for every $i\ge 0$ and arbitrary $M$. The finiteness condition is on the first argument; the second need not be finitely generated.

For a multiplicatively closed set $S$ containing one, the localization of a ring $S^{-1}R$ consists of fractions $r/s$ with $r\in R,s\in S$, under the equivalenceThe factor $u$ is needed when the ring has zero divisors. Addition and multiplication areThe canonical map sends $r$ to $r/1$ and makes every element of $S$ a unit. It is universal with this property: a ring homomorphism $\varphi :R\to T$ taking $S$ into units extends uniquely by $r/s\mapsto \varphi (r)\varphi (s{)}^{-1}$. Usually zero is excluded from $S$; if it is included, this construction is the zero ring.

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 $R$.

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 ideal $I$ in a commutative ring, every prime ideal containing $I$ contains $\sqrt{I}$, because $a^n\in P$ implies $a\in P$. Conversely, if $a\notin \sqrt{I}$, the multiplicatively closed set $T=\{1,a,a^2,\dots \}$ is disjoint from $I$. By Zorn's lemma, there is an ideal $P\supseteq I$ maximal among those disjoint from $T$.

This $P$ is a prime ideal: if $bc\in P$ but $b,c\notin P$, maximality makes both $P+(b)$ and $P+(c)$ meet $T$. Multiplication of representatives from these two intersections puts an element of $T$ in $P$, a contradiction. Thus there is a prime ideal containing $I$ but avoiding $a$, establishing the identity. For $I=R$, the intersection is empty and means $R$. It follows that an ideal is a radical ideal exactly when it is an intersection of ideals that are all prime ideals.