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Apr 12, 2023 at 9:15 comment added R. van Dobben de Bruyn @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).
Apr 12, 2023 at 0:19 comment added math54321 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
Apr 10, 2023 at 14:51 vote accept William Sun
Apr 10, 2023 at 14:46 comment added R. van Dobben de Bruyn 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.
Apr 10, 2023 at 14:44 comment added William Sun "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
Apr 10, 2023 at 14:40 history edited R. van Dobben de Bruyn CC BY-SA 4.0
Simplified the proof and replaced scheme language by ring language.
Apr 10, 2023 at 0:58 history edited R. van Dobben de Bruyn CC BY-SA 4.0
Added a comment about yet another method of proof.
Apr 9, 2023 at 22:13 history answered R. van Dobben de Bruyn CC BY-SA 4.0