Lemma 10.107.4. Let $R$ be a ring. The rule $I \mapsto V(I)$ determines a bijection

**Proof.**
Let $I$ be a pure ideal. Then since $R \to R/I$ is flat, by going up generalizations lift along the map $\mathop{\mathrm{Spec}}(R/I) \to \mathop{\mathrm{Spec}}(R)$. Hence $V(I)$ is closed under generalizations. This shows that the map is well defined. By Lemma 10.107.3 the map is injective. Suppose that $Z \subset \mathop{\mathrm{Spec}}(R)$ is closed and closed under generalizations. Let $J \subset R$ be the radical ideal such that $Z = V(J)$. Let $I = \{ x \in R : x \in xJ\} $. Note that $I$ is an ideal: if $x, y \in I$ then there exist $f, g \in J$ such that $x = xf$ and $y = yg$. Then

Verification left to the reader. We claim that $I$ is pure and that $V(I) = V(J)$. If the claim is true then the map of the lemma is surjective and the lemma holds.

Note that $I \subset J$, so that $V(J) \subset V(I)$. Let $I \subset \mathfrak p$ be a prime. Consider the multiplicative subset $S = (R \setminus \mathfrak p)(1 + J)$. By definition of $I$ and $I \subset \mathfrak p$ we see that $0 \not\in S$. Hence we can find a prime $\mathfrak q$ of $R$ which is disjoint from $S$, see Lemmas 10.9.4 and 10.16.5. Hence $\mathfrak q \subset \mathfrak p$ and $\mathfrak q \cap (1 + J) = \emptyset $. This implies that $\mathfrak q + J$ is a proper ideal of $R$. Let $\mathfrak m$ be a maximal ideal containing $\mathfrak q + J$. Then we get $\mathfrak m \in V(J)$ and hence $\mathfrak q \in V(J) = Z$ as $Z$ was assumed to be closed under generalization. This in turn implies $\mathfrak p \in V(J)$ as $\mathfrak q \subset \mathfrak p$. Thus we see that $V(I) = V(J)$.

Finally, since $V(I) = V(J)$ (and $J$ radical) we see that $J = \sqrt{I}$. Pick $x \in I$, so that $x = xy$ for some $y \in J$ by definition. Then $x = xy = xy^2 = \ldots = xy^ n$. Since $y^ n \in I$ for some $n > 0$ we conclude that property (5) of Lemma 10.107.2 holds and we see that $I$ is indeed pure. $\square$

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