Lemma 10.30.8. Let $A \subset B$ be an inclusion of domains induces an algebraic extension of fraction fields. If $J \subset B$ is a nonzero ideal, then $A \cap J$ is nonzero too. Thus the image of a proper closed subset of $\mathop{\mathrm{Spec}}(B)$ is not dense in $\mathop{\mathrm{Spec}}(A)$.
Proof. Let $x \in J$ be a nonzero element. Since $x$ is algebraic over the fraction field of $A$, there exists a $d \geq 1$ and $a_0, \ldots , a_ d \in A$ with $a_0, a_ d \not= 0$ such that $a_ d x^ d + a_{d - 1} x^{d - 1} + \ldots + a_0 = 0$ in $B$. Then $a_0 \in I$. $\square$
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Comment #10980 by Junyan Xu on