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How do we prove the following exercise of Hartshorne?

Let $A$ be a subring of an integral domain $B$. Suppose $B$ is a finitely generated $A$-algebra. Let $b$ be a non-zero element of $B$. Then there exists a non-zero element $a$ of $A$ with the following property. If $\psi\colon A \rightarrow \Omega$ is any homomorphism of $A$ to an algebraically closed field $\Omega$ such that $\psi(a) \neq 0$, then $\psi$ extends to a homomorphism $\phi\colon B \rightarrow \Omega$ such that $\phi(b) \neq 0$.

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Check Proposition $5.23$ in Atiyah-Macdonald.

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We are supposed to prove it by ourselves. – Makoto Kato Dec 15 '12 at 7:38
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Dear @Makoto: Just refer to the proof there. I am not sure there is any point in reproducing the proof here. It's not altogether trivial. – Rankeya Dec 15 '12 at 7:42
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Thanks for the reference. I expect someone would post his original proof, though. You know there are usually several different proofs of a proposition. – Makoto Kato Dec 15 '12 at 7:59
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@MakotoKato: So you don't want to copy from book, but you do want to copy from somebody doing your work for you here? What's better about the latter? And didn't you miss a [homework] tag on the question, then? – Henning Makholm Dec 15 '12 at 23:40
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If you want a hint instead of a complete solution, then say so in your question. Otherwise this is a perfectly valid answer. – Zhen Lin Dec 18 '12 at 1:31
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