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Let $R$ be a domain, $R\subset S$ is an integral extension of rings. Suppose $0\neq s\in S$ such that for any $r\in R$ and any integer $n$, $rs^n=0$ implies $r=0$.Then there exists $0\neq x\in R$, such that for every ring homomorphism $\phi : R\to K$ with $\phi(x)\neq 0$, where $K$ is an algebraically closed field, there exists an extension $\phi':S\to K$ such that $\phi' (s)\neq 0$.

Let $p=\text{ker}\phi$. Then $p$ is a prime ideal of $R$. So there is a prime ideal $P$ of $S$ lying over $p$.The injection $R/p\to K$ extends to $\operatorname{Frac}(R/p)\to K$. Since $\operatorname{Frac}(S/P)$ is algebraic over $\operatorname{Frac}(R/p)$, $\operatorname{Frac}(R/p)\to K$ extends to $\operatorname{Frac}(S/P)\to K$. Thus $\phi':S\to S/P\to \operatorname{Frac}(S/P)\to K$ is an extension of $\phi$.

It remains to show $\phi'(s)\neq 0$. How can I deduce such $s$ isn't in the kernel of $\phi'$?

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    I don't get the first part, see $R=S=\Bbb{Z}, s=2,K=\overline{\Bbb{F}}_2$. – reuns Nov 16 '22 at 21:53
  • The correct statement is that if $S$ is integral over $R$ then any prime ideal $p\subset R$ is of the form $P\cap R$ for some prime ideal $P\subset S$ so that $Frac(S/P)$ is an algebraic extension of $Frac(R/p)$. – reuns Nov 16 '22 at 21:57
  • @reuns You are right. The last sentence of the first part should be " Then there exists $0\neq x\in R$ such that for every ring homomorphism $\phi: R\to K$ with $\phi(x)\neq 0$ where $K$ is an algebraically closed field, there exists an extension...." – Display name Nov 28 '22 at 01:28

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