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I am looking for a proof of the following statement without using full power of Chevalley's theorem 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 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 from MSE to here.

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  • $\begingroup$ Intersection of all primes is the nilradical, there cannot be nonzero nilpotents in a domain. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 9, 2023 at 4:23
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    $\begingroup$ @მამუკაჯიბლაძე Note that I wrote "non-zero primes" of $A$, not "all primes" of $A$. $\endgroup$
    – William Sun
    Commented Apr 9, 2023 at 4:25
  • $\begingroup$ You are right, sorry $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 9, 2023 at 4:27

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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] 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.

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

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  • $\begingroup$ "so there is a nonzero element $x\in A$ such that $A_x\hookrightarrow B_x$ is finite" I do not follow this part. If $B$ were to be replaced by a field $k$, the claim becomes "given finite type $A\hookrightarrow k$, there is a nonzero $x\in A$ such that $A_x\hookrightarrow k$ is finite". The best I could do is "finitely presented" instead of "finite". See stacks.math.columbia.edu/tag/00FG $\endgroup$
    – William Sun
    Commented Apr 10, 2023 at 14:44
  • $\begingroup$ We already know that $\operatorname{Frac} A \to B$ is finite by the weak Nullstellensatz. Choosing generators $x_1,\ldots,x_n$ of $B$ over $A$, we get monic minimal polynomials $f_i \in (\operatorname{Frac} A)[t]$ with $f_i(x_i) = 0$, and clearing denominators we get an element $x \in A$ such that all $f_i$ live in $A_x[t]$. Then $A_x \to B$ is generated by elements satisfying a monic polynomial over $A_x$, hence is a finite extension. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 10, 2023 at 14:46
  • $\begingroup$ It would be good to have a reference for this "weak Nullstellensatz", since it contains the core of the argument, and also is not a standard name $\endgroup$
    – math54321
    Commented Apr 12, 2023 at 0:19
  • $\begingroup$ @math54321 See for instance Atiyah–MacDonald, Corollary 7.10. But most places that cover the Nullstellensatz cover this weak version first (even if not explicitly by that name). See for instance [Tag 00FV] or Lang's Algebra, Corollary IX.1.2. Most other results by the name "weak Nullstellensatz" are reformulations of this (often under the additional assumption that $K$ is algebraically closed). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 12, 2023 at 9:15

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