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Assume that $ B \hookrightarrow A $ is an injective ring homomorphism.

Question: If $\mathfrak{p}$ is an associated prime of $B$, is there an associated prime $ \mathfrak{q}$ of $A$ such that $ \mathfrak{q} \cap B = \mathfrak{p}$?

Some comments:

  • This is true if $ \mathfrak{p}$ is a minimal prime of $B$. Just take an associated prime of $ (B \backslash \mathfrak{p})^{-1} A$. So the only problem is with embedded primes
  • The question is an algebraic translation of the following more geometric question: If $X \to Y$ is a quasicompact morphism of noetherian schemes, is every associated point of the schematic image hit by some associated point of $X$?
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2 Answers 2

up vote 6 down vote accepted

Let me explain why the statement is true. Let $i:B\to A$ be an injection of noetherian rings and $\mathfrak{p}$ be an associated prime of $B$ : it is the annihilator of a nonzero element $f\in B$.

Let us first localize at $\mathfrak{p}$. More precisely, replacing $B\to A$ by $B_{\mathfrak{p}}\to A\otimes_B B_{\mathfrak{p}}$ , it is possible to assume that $B$ is local with maximal ideal $\mathfrak{p}$.

Let $I$ be the ideal of $A$ generated by $i(f)$, viewed as an $A$-module. Since $I\neq 0$ by injectivity of $i$, it has an associated prime $\mathfrak{q}\subset A$. This means that there is $a\in A$ such that $\mathfrak{q}$ is the annihilator of ai(f). Note that $\mathfrak{q}$ is also associated to $A$, and let us check that it has the required properties. As $\mathfrak{p}$ annihilates $f$, $\mathfrak{q}\cap B$ contains $\mathfrak{p}$. Since $\mathfrak{p}$ is maximal, $\mathfrak{q}\cap B=\mathfrak{p}$.

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Put $B=k[x,y]/(xy,y^2)$ and $A=B[z]/(yz-x)$. Then $(y)$ is an associated (in fact minimal!) prime of $B$. But any prime of $A$ that contains $(y)$ must also contain $x$, and so cannot contract to just $(y)$.

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$B\to A$ does not seem to be injective here, as $x^2$ that was nonzero in $B$ becomes equal to $xyz=0z=0$ in $A$. –  Olivier Benoist Jul 4 '13 at 23:10
    
@OlivierBenoist: You are right, of course. I retract my example. –  Steven Landsburg Jul 4 '13 at 23:14

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