Timeline for Origin of the theorem on the existence of the smallest field of definition of an affine variety
Current License: CC BY-SA 3.0
9 events
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| Nov 3, 2012 at 17:57 | comment | added | user27056 | @Makoto: If $W$ descends to a subspace $W_0$ of $V_{K_0}$ then the classes $v_j \bmod W_0$ in $V_{K_0}/W_0$ for $j \in J$ are a $K_0$-basis since the classes $v_j \bmod W$ in $V_K/W$ are a $K$-basis. Expanding each $v_{i'} \bmod W_0$ ($i'\not\in J$) in this basis yields coefficients in $K_0$ that must be the coefficients over $K$, so these are exactly the $a_{i'j}$'s, and so each $a_{i'j}$ lies in $K_0$. Conversely, since the $v_{i'} - \sum_j a_{i'j} v_j$ span $W$ over $K$, their span over $F(a_{i'j})$ is a descent to that minimal possibility. QED Is that Weil's "somewhat involved" proof? | |
| Oct 29, 2012 at 23:17 | comment | added | Makoto Kato | The point of the proof is to prove the subextension you constructed is indeed the desired one. That is what Weil and Bourbaki did. | |
| Oct 29, 2012 at 4:37 | comment | added | user27056 | 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. | |
| Oct 29, 2012 at 4:36 | history | edited | user27056 | CC BY-SA 3.0 |
added 141 characters in body
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| Oct 29, 2012 at 2:50 | comment | added | Joël | 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? | |
| Oct 29, 2012 at 2:24 | comment | added | user27056 | @Makoto: Ah, then I'm glad I never looked up the Bourbaki proof to which EGA punted. :) | |
| Oct 29, 2012 at 2:23 | comment | added | Makoto Kato | 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. | |
| Oct 29, 2012 at 1:34 | comment | added | user27056 | 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. | |
| Oct 29, 2012 at 1:32 | history | answered | user27056 | CC BY-SA 3.0 |