I am looking for a proof of the following statement without using full power of [Chevalley's theorem](https://stacks.math.columbia.edu/tag/00FE) on constructible sets. We say a domain $A$ is $0$*-open* if $\{(0)\}$ is open in $\operatorname{Spec}(A)$. Equivalently, there is an element $x\in A$ such that $A\_x$ is a field. And equivalently, either the only prime is $(0)$, or there is a non-zero element $x\in A$ contained in all non-zero primes $\mathfrak p\subset A$.



> 
> For domains $A$ and $B$, let $A\hookrightarrow B$ be injective and of
> finite type. Then $A$ is $0$-open if $B$ is $0$-open.
> 
> 
> 


The above statement could be proved using Chevalley's theorem as below:


If $B$ is $0$*-open*, then let $B\_y$ be a field for some $y\in B$ and the composition $A\hookrightarrow B\hookrightarrow B\_y$ would be injective and finite type. By [this](https://stacks.math.columbia.edu/tag/00FG) elementary lemma, there is a nonzero element $x\in A$ such that $A\_x\hookrightarrow B\_y$ is injective and of finite presentation. Chevalley's theorem would give us that $A\_x$ is $0$-open.


If $A\_x$ is 0-open, then there is a non-zero element $x'\in A$ contained in all non-zero primes $\mathfrak p$ not containing $x$. The element $xx'\neq 0$ is then contained in all non-zero primes of $A$. Thus $A$ is $0$-open, as desired.


I am cross posting the [same question](https://math.stackexchange.com/questions/4672282/localization-at-a-point-is-a-field) from MSE to here.



Finite type injective ring map between domains preserves the open point $(0)$
Here's a more down to earth argument that uses the weak Nullstellensatz¹ instead of Chevalley's theorem:


**Lemma.** *Let $\phi \colon A \hookrightarrow B$ be an injective ring homomorphism of finite type between integral domains, and assume there exists a nonzero element $y \in B$ such that $B\_y$ is a field. Then there exists a nonzero element $x \in A$ such that $A\_x$ is a field.*


*Proof.* If $A \hookrightarrow B$ is of finite type, then so is $A \hookrightarrow B\_y$. Replacing $B$ by $B\_y$, we may assume that $B$ is a field. Then $A \hookrightarrow B$ is a finite type ring homomorphism, hence so is $\operatorname{Frac} A \hookrightarrow B$. The weak Nullstellensatz then says that $\operatorname{Frac} A \hookrightarrow B$ is a finite extension, so there exists a nonzero element $x \in A$ such that the ring homomorphism $A\_x \hookrightarrow B\_x = B$ is finite.


Now if $\mathfrak p \subseteq A\_x$ is a nonzero prime ideal, then 'going up' [Tag [00GU](https://stacks.math.columbia.edu/tag/00GU)] for the inclusion $(0) \subseteq \mathfrak p$ shows that there exists a nonzero prime ideal $\mathfrak q \subseteq B$ with $\mathfrak q \cap A\_x = \mathfrak p$. This is impossible since $B$ is a field, so we conclude that $A\_x$ is a field. $\square$


**Remark.** You could also try to use generic flatness instead of generic finiteness (this is possibly more natural for this question). But the lemma that a finite type flat map is open usually depends on Chevalley's theorem, or at least gets very close to proving Chevalley. By contrast, the proof above only uses 'going up' without discussing closedness or openness of morphisms of schemes.




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¹ The weak Nullstellensatz says that if $K \to L$ is a finite type ring homomorphism between fields, then it is actually finite.


