Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\phi(b) \neq 0$.

share|improve this question

2 Answers 2

Check Proposition $5.23$ in Atiyah-Macdonald.

share|improve this answer
We are supposed to prove it by ourselves. –  Makoto Kato Dec 15 '12 at 7:38
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
@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
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
@ZhenLin I don't think it's a perfectly valid answer in that not everybody can see the book. –  Makoto Kato Dec 19 '12 at 20:22

I've just come up with the following proof(since this is an exercise, I have been trying to solve it by myself).

We use induction on the number of generators of $B$ over $A$. It suffices to to prove the assertion when $B = A[x]$. Let $K$ be the field of fractions of $A$. Suppose $x$ is not algebraic over $K$. Let $b = a_0x^n + \cdots + a_1x + a_n$, where $a_i \in A$ for $i = 0,1,\dots,n$ and $a_0 \neq 0$. Let $\psi\colon A \rightarrow \Omega$ be a homomorphism such that $\psi(a_0) \neq 0$. Let $\alpha$ be an element of $\Omega$ which is not a root of the polynomial $\psi(a_0)X^n + \cdots + \psi(a_1)X + \psi(a_n)$. There exists a unique homomorphism $\phi\colon B \rightarrow \Omega$ extending $\psi$ such that $\phi(x) = \alpha$. Then $\phi(b) \neq 0$.

It remains to prove the asssertion when $x$ is algebraic over $K$. Then $b$ is also algebraic over $K$. Suppose $a_0x^n + \cdots + a_1x + a_n = 0$, where $a_i \in A$ for $i = 0,1,\dots,n$ and $a_0 \neq 0$. Suppose $c_0b^m + \cdots + c_1b + c_m = 0$, where $c_i \in A$ for $i = 0,1,\dots,m$ and $c_0 \neq 0$ We may assume that $c_m \neq 0$. Let $a = a_0c_m$. Then $B_a = A_a[x]$ is integral over $A_a$, where $A_a$ is the localization of $A$ with respect to $\{1, a, a^2,\dots\}$. Let $\psi\colon A \rightarrow \Omega$ be a homomorphism such that $\psi(a) \neq 0$. $\psi$ is uniquely extended to a homomorphism $\psi'\colon A_a \rightarrow \Omega$. Hence by this question, $\psi'$ can be extended to a homomorphism $\phi\colon B_a \rightarrow \Omega$. We claim $\phi(b) \neq 0$. Suppose $\phi(b) = 0$. Then $\psi(c_m) = 0$. Hence $\psi(a) = 0$. This is a contradiction.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.