Timeline for Field of fractions of etale stalk of Dedekind domain (Example from Milne's LEC)
Current License: CC BY-SA 4.0
12 events
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| Sep 19, 2023 at 22:57 | comment | added | user267839 | Theorem 41.2 in Nagata's Local Rings provides - if I'm not missing something - an argument without flatness assumption. This should give the full answer to my concern in "one nitpick" comment above | |
| Aug 17, 2023 at 9:25 | comment | added | R. van Dobben de Bruyn | By degree counting, I mean the statement that $\Gamma(f^{-1}(x),\mathcal O_{f^{-1}(x)})$ is a $\kappa(x)$-algebra of dimension $[L:K]$ if $f \colon Y \to X$ is a finite extension of normal schemes with generic fibre $K \to L$ (and $x \in X$ is any point). This holds in the flat case (e.g. for Dedekind schemes), but not in general. (This approach is useful because Milne's Thm. 3.21 shows that the étale locus agrees with the unramified locus, and checking that something is unramified can be done fibrewise.) | |
| Aug 17, 2023 at 7:49 | comment | added | user267839 | Just to clarify: What do you mean in this context by the "degree counting argument" ( which seemingly only work if the considered map is flat)? A guess: Do you refer by this to the additivity of the field extension degrees with resp the composition of field extensions, which in turn carry information about ramification behavior in this sense: en.m.wikipedia.org/wiki/… . Or mean by "argue by degree counting" another issue? | |
| Aug 16, 2023 at 21:00 | comment | added | R. van Dobben de Bruyn | I did see your other (previously bountied) question, but the answer is I don't really know! What makes it difficult (to me) is that the finite map $X' \to X$ is no longer flat, so you can't argue by degree counting. For instance, I haven't thought about classical questions like whether Galois acts transitively on the fibre (in some scheme-theoretic way?), although I assume this is known. | |
| Aug 16, 2023 at 13:01 | comment | added | user267839 | In other words, how much of the Galois ramification theory applied in above setting still survive in higher dimension? | |
| Aug 16, 2023 at 12:55 | comment | added | user267839 | Now the natural question is if the result above that the field of fractions of $\mathcal{O}_{X,\overline{x}}^{\text{et}}$ equals to the fixed field $(K^{\text{sep}})^{I(\widetilde{\mathfrak{p}})}$ still could work, if we weaken our "Dedekind assumption" to "normality"only, ie in higher dimensional setting? | |
| Aug 16, 2023 at 12:53 | comment | added | user267839 | one nitpick: the assumption was that $X$ is Dedekind scheme, in of dimension one $1$ and therefore the tools from classical algebraic theory for number fields fit here. Nevertheless, if we start with arbitrary local normal ring $A$ with field of fractions $K$, any separable extension field extension $L$ and intergral closure $B$ of $A$ in $L$, then picking an maximal ideal $p$ of $B$ lying over the max ideal $m$ of $A$, allows still to define decomposition and inertia subgroups wrt $p \subset B$ inside the Galois group of $L/K$. So the terminology also makes sense for higher dimension. | |
| Aug 15, 2023 at 10:16 | vote | accept | user267839 | ||
| Aug 15, 2023 at 8:53 | comment | added | Vik78 | Perhaps that should say instead: the data of the etale neighborhood determines $\sigma$ as a map $L \to K^{sep}$ (up to composition with elements of $I(\tilde{p}))$,and therefore induces a natural map $L \to (K^{sep})^{I(\tilde{p})}$. Taking the limit we get our desired isomorphism | |
| Aug 15, 2023 at 7:06 | comment | added | Vik78 | so we get a natural embedding $L \to (K^{sep})^{I(\tilde{\mathfrak{p}})}$. So a choice of $\tilde{\mathfrak{p}}$ gives a canonical isomorphism to this field. which is not encoded by a geometric point over $\mathfrak{p}$ alone. | |
| Aug 15, 2023 at 7:03 | comment | added | Vik78 | I think there’s a subtlety here: you are implicitly thinking of finite separable field extensions $K \to L$ as sitting inside a fixed $K^{sep}$. This is how the geometric point $\overline{x}$ over $\mathfrak{p}$ remembers the prime ideal $\tilde{\mathfrak{p}}$ which induces it: if we have a finite separable $K \to L$ and an unramified prime $\mathfrak{q}$ over $\mathfrak{p}$ in $B$, we can always conjugate by an element $\sigma$ of Gal($K^{sep}/K)$ so that $\sigma{\mathfrak{q}} = \tilde{\mathfrak{p}} \cap B$. The coset $\sigma I(\tilde{\mathfrak{p}})$ is determined by $\mathfrak{q}$, | |
| Aug 14, 2023 at 23:13 | history | answered | R. van Dobben de Bruyn | CC BY-SA 4.0 |