Integral extension implies surjective map on spectra: Difference between revisions

Revision as of 20:01, 2 February 2008

Statement

Suppose $S$ is an integral extension of a ring $R$ , in other words $f : R \to S$ is an injective homomorphism of commutative unital rings with the property that every element of $S$ is integral over the image of $R$ . Then, the map:

$f^{*} : S p e c (S) \to S p e c (R)$

from the spectrum of $S$ to that of $R$ , that sends a prime ideal of $S$ to its contraction in $R$ , is surjective. In other words, every prime ideal of $R$ occurs as the contraction of a prime ideal of $S$ .

Proof

The goal is to prove that starting with a prime ideal $P$ of $R$ , we can find a prime ideal $Q$ of $S$ such that $f^{- 1} (Q) = P$ .

We localize $R$ at $P$ , and localize $S$ at the image of $U = R ∖ P$ to get $S^{'}$ . Then $R_{P}$ is a local ring with unique maximal ideal $P^{'} = P R_{P}$ , and $f$ induces a map $R_{P} \to S^{'} = S [U^{- 1}]$ .

We thus have an inclusion $f : R_{P} \to S^{'}$ . Consider the image $P^{'} S^{'}$ . This is an ideal of $S^{'}$ . If $P^{'} S^{'}$ is a proper ideal, it is contained in some maximal ideal $M$ , and the contraction of that maximal ideal to $R_{P}$ is precisely $P^{'}$ . Contracting back along the localization, we find a prime ideal of $S^{'}$ , whose contraction is exactly $P$ . (we are using the fact that contracting a maximal ideal of $S^{'}$ yields a prime, though not necessarily maximal, ideal of $S$ ).

Thus, the main goal is to show that $P^{'} S^{'} \neq S^{'}$ (this is where we need to use integrality). The idea is to construct a $R$ -subalgebra of $S^{'}$ , called $S^{″}$ , that is finite over $R_{P}$ , and use Nakayama's lemma to derive a contradiction. Here are the steps:

Since $S$ is integral over $R$ , $S^{'}$ is integral over $R_{P}$
If $P^{'} S^{'} = S^{'}$ , then the element $1 \in S^{'}$ can be written as a $P^{'}$ -linear combination of finitely many elements from $S^{'}$
Let $S^{″}$ be the $R_{P}$ -subalgebra generated by these finitely many elements. Then $S^{″}$ is finitely generated and integral over $R_{P}$ , hence it is finitely generated as a module over $R_{P}$ . For full proof, refer: finitely generated and integral implies finite
We thus have $P^{'} S^{″} = S^{″}$ (since $1 \in P^{'} S^{″}$ ). Since $P^{'}$ is the Jacobson radical of $R_{P}$ , Nakayama's lemma tells us that $S^{″} = 0$ , yielding a contradiction.

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 ==Proof==
-{{fillin}}
+The goal is to prove that starting with a prime ideal <math>P</math> of <math>R</math>, we can find a prime ideal <math>Q</math> of <math>S</math> such that <math>f^{-1}(Q) = P</math>.
+We localize <math>R</math> at <math>P</math>, and localize <math>S</math> at the image of <math>U = R \setminus P</math> to get <math>S'</math>. Then <math>R_P</math> is a local ring with unique maximal ideal <math>P' = PR_P</math>, and <math>f</math> induces a map <math>R_P \to S' = S[U^{-1}]</math>.
+We thus have an inclusion <math>f:R_P \to S'</math>. Consider the image <math>P'S'</math>. This is an ideal of <math>S'</math>. If <math>P'S'</math> is a proper ideal, it is contained in some maximal ideal <math>M</math>, and the contraction of that maximal ideal to <math>R_P</math> is precisely <math>P'</math>. Contracting back along the localization, we find a prime ideal of <math>S'</math>, whose contraction is exactly <math>P</math>. (we are using the fact that contracting a maximal ideal of <math>S'</math> yields a prime, though not necessarily maximal, ideal of <math>S</math>).
+Thus, the main goal is to show that <math>P'S' \ne S'</math> (this is where we need to use integrality). The idea is to construct a <math>R</math>-subalgebra of <math>S'</math>, called <math>S''</math>, that is ''finite'' over <math>R_P</math>, and use [[Nakayama's lemma]] to derive a contradiction. Here are the steps:
+* Since <math>S</math> is integral over <math>R</math>, <math>S'</math> is integral over <math>R_P</math>
+* If <math>P'S' = S'</math>, then the element <math>1 \in S'</math> can be written as a <math>P'</math>-linear combination of finitely many elements from <math>S'</math>
+* Let <math>S''</math> be the <math>R_P</math>-subalgebra generated by these finitely many elements. Then <math>S''</math> is finitely generated and integral over <math>R_P</math>, hence it is finitely generated as a module over <math>R_P</math>. {{proofat|[[finitely generated and integral implies finite]]}}
+* We thus have <math>P'S'' = S''</math> (since <math>1 \in P'S''</math>). Since <math>P'</math> is the [[Jacobson radical]] of <math>R_P</math>, Nakayama's lemma tells us that <math>S'' = 0</math>, yielding a contradiction.