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Weil proved the following theorem in his book Foundations of Algebraic Geometry, p.19. The proof is somewhat involved. I wonder if the theorem is his original.

Theorem Let $K[X_1,\dots, X_n]$ be the polynomial ring over a field $K$. Let $I$ be an ideal of $K[X_1,\dots, X_n]$. There exists a smallest subfield $k$ of $K$ such that $I$ is generated by polynomials in $k[X_1,\dots,X_n]$.

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    $\begingroup$ I am sorry, I don't know the answer to your question but I just realized that you can prove it using Gröbner basis. Let $E$ and $F$ be subfields of $K$ such that $I$ is generated by polynomials with coefficients in $E$ and in $F$, respectively. Then choose reduced Gröbner bases $G$ and $H$ of $I$ with respect to the same term ordering having all coefficients in $E$ and in $F$, respectively. Now both $G$ and $H$ are reduced Gröbner bases of $I$ also over $K$. Because of the unicity of the reduced Gröbner basis, we have $G=H$. Hence $I$ is generated by polynomials with coefficients in $E\cap F$. $\endgroup$ Commented Oct 28, 2012 at 21:51

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As far as I can see, Weil was indeed the main source for this viewpoint on fields of definition in algebraic geometry. However, it may be hard to pin down the specific result quoted here in his 1935 paper. This paper is probably most readily found in the first volume of Weil's papers published by Springer, but the later book presents his notion of variety and the related field theory (with generic points) in far more detail.

What I'd like to add is a reference to Dieudonne's book History of Algebraic Geometry (especially VII.4). This was first published in French in 1974 and then in English translation in 1985. Dieudonne took a strong interest in this kind of history and assembled a lot of material about older origins of ideas while emphasizing the key role played by Weil. Naturally names like van der Waerden, E. Noether, and Siegel are part of that history as well.

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I suspect that this theorem is indeed due to Weil.

"Foundations of Algebraic Geometry" by Weil was published in 1946, but the 1944 paper "Some Properties of Ideals in Rings of Power Series" by Claude Chevalley (Transactions of the American Mathematical Society, Vol. 55, No. 1 (Jan., 1944), pp. 68-84) attributes to Weil the development of the theory around "ideals in polynomial rings" over a decade earlier in "Arithmetique et geometrie sur les varietes algebriques" in 1935 (see footnote on p. 83).

Reading the AMS review it seems the only other possible originators would have been Siegel, or perhaps Noether or van der Waeden. I don't have a copy of Weil's 1935 work, but you might track it down and (if you can read enough French) check for this particular result.


Edit: For remarks which are perhaps related/interesting (in terms of Weil's background and his familiarity with Kronecker's work) read from the last paragraph of page 12 here and the referenced ICM address by Weil in 1950.

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This has a very easy proof if one generalizes it to (infinite-dimensional) linear algebra and forgets about commutative algebra. Let $K/F$ be an extension of fields (e.g., could take $F$ to be a prime field), $V$ an $F$-vector space (such as a polynomial ring over a prime field), and $W$ a $K$-subspace of $V_K := K \otimes_F V$.

Among all subfields $K_0$ of $K$ over $F$ such that $W = K \otimes_{K_0} W_0$ for a (visibly unique) $K_0$-subspace $W_0$ of $V_{K_0}$, we claim that the intersection of these fields works too.

(In case $V$ is an $F$-algebra and $W$ is an ideal of $V_K$, obviously $W_0$ is an ideal of $V_{K_0}$, so this really does imply Weil's result. In fact, it gives a more general result: no need to assume the algebras are finitely generated.)

Proof: Choose an $F$-basis $\{v_i\}_{i \in I}$ of $V$, so there is a subset $J$ of $I$ such that $\{v_j \bmod W\}_{j \in J}$ is a $K$-basis of $V/W$. For $i' \in I - J$, expand $v_{i'} \bmod W \in V/W$ in this basis:
$$v_{i'} \equiv \sum_{j \in J} a_{i'j} v_j \bmod W$$ with $a_{i'j} \in K$. The necessary and sufficient condition on $K_0$ for $W_0$ to exist is that $K_0$ contains every $a_{i'j}$ (for $j \in J$ and $i' \in I - J$). So the subfield $F(a_{i'j})_{i', j}$ is the desired minimal subextension of $K$ over $F$. QED

There is a very elegant modern discussion of the theme of field of definition (for closed subschemes, morphisms, etc.) without any finiteness hypotheses in EGA IV$_2$, 4.8.

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  • $\begingroup$ I know that what I wrote above doesn't answer the question, but I wanted to communicate that the proof needn't be "somewhat involved" if it is generalized in a suitable way. (I have never read Weil's proof, so I have no idea what he does.) And that part of EGA provides a modern reference if one is desired. $\endgroup$
    – user27056
    Commented Oct 29, 2012 at 1:34
  • $\begingroup$ EGA IV_2, Corollary 4.8.7 is essentilly the same as the above claim of yours. It refers Bourbaki's Algebra Ch. II for its proof. However the Bourbaki's proof is quite different from yours. It is not so involved, but not so simple as yours. $\endgroup$ Commented Oct 29, 2012 at 2:23
  • $\begingroup$ @Makoto: Ah, then I'm glad I never looked up the Bourbaki proof to which EGA punted. :) $\endgroup$
    – user27056
    Commented Oct 29, 2012 at 2:24
  • $\begingroup$ I don't get it. As soon as $W$ is not $0$, the subfield of $K$ that your argument constructs is $K$ itself. For if you take $w = \sum a_i(w) v_i$ a non-zero element in $W$, one of the $a_i(w)$ is non-zero, and for every $\lambda$ in $K$, $a_i(\lambda w)=\lambda a_i(w)$ so $a_i(\lambda w)$ may be any element you want in $K$. Am I wrong? $\endgroup$
    – Joël
    Commented Oct 29, 2012 at 2:50
  • $\begingroup$ Dear Joel: Sorry, I garbled the argument. I have replaced it with what I meant to say (equally short, but now correct). Thanks for catching it. If you delete your comment (assuming you consider it now to be moot) then I will delete this one. $\endgroup$
    – user27056
    Commented Oct 29, 2012 at 4:37

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