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Let $X$ be a scheme over a field $k$, and let $K$ be a field extension of $k$. Let $X_K$ denote the base change. I want to know whether for any open $U\in X,\mathscr O(U_K)=\mathscr O(U)\otimes_k K.$

I believe this should be true, since for any affine open $V=specA$ inside $U$, we have $\mathscr O (V_K)=A\otimes_k K=\mathscr O(V)\otimes_k K,$ so maybe, since the sections agree locally, they should be the same. But I'm not sure how to put this rigorously. Any help would be appreciated.

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  • $\begingroup$ There is no reason to introduce $X$: just start with the $k$-scheme $U$ (as correctly realized by Minseon). $\endgroup$ Commented Sep 1, 2017 at 9:16

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This is true if $U$ is quasi-compact and quasi-separated. (A more general statement is e.g. here.)

Since $U$ is quasi-compact, there is a finite cover $U = \bigcup_{i=1}^{n} U_{i}$ where each $U_{i}$ is affine open. Thus we obtain a diagram

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} 0 & \ras{} & \mathscr{O}(U) \otimes_{k} K & \ras{} & \prod_{i=1}^{n} (\mathscr{O}(U_{i}) \otimes_{k} K) & \ras{} & \prod_{i,j=1}^{n} (\mathscr{O}(U_{i} \cap U_{j}) \otimes_{k} K) \\ & & \da{g_1} & & \da{g_2} & & \da{g_3} & \\ 0 & \ras{} & \mathscr{O}(U \times_{k} K) & \ras{} & \prod_{i=1}^{n} \mathscr{O}(U_{i} \times_{k} K) & \ras{} & \prod_{i,j=1}^{n} (\mathscr{O}((U_{i} \cap U_{j}) \times_{k} K)) \\ \end{array} $$ where both rows are exact (the first row is exact since $k \to K$ is flat). Here $g_{2}$ is an isomorphism since each $U_{i}$ is affine; thus $g_{1}$ is injective. Applying the above argument with $U$ replaced by $U_{i} \cap U_{j}$ (which is quasi-compact since $U$ is quasi-separated), we obtain that $g_{3}$ is injective. Thus $g_{1}$ is an isomorphism by a diagram chase.


Here is an example showing that the map $$ \mathcal{O}(U) \otimes_{k} K \to \mathcal{O}(U \times_{k} K) $$ may not be surjective if $U$ is not quasi-compact. (This is essentially same example given by Najib Idrissi in Tensor products over field do not commute with inverse limits?.) Let $k$ be a field and set $K := k(\{t_{n}\}_{n \in \mathbb{Z}})$; here $K/k$ is a field extension of countably infinite transcendence degree. Set $U := \amalg_{n \in \mathbb{Z}} \operatorname{Spec} k$; then $U \times_{k} K \simeq \amalg_{n \in \mathbb{Z}} \operatorname{Spec} K$ and $\mathcal{O}(U) = \prod_{n \in \mathbb{Z}} k$ and $\mathcal{O}(U \times_{k} K) = \prod_{n \in \mathbb{Z}} K$. The inclusion $k \to K$ gives a canonical inclusion $\prod_{n \in \mathbb{Z}} k \to \prod_{n \in \mathbb{Z}} K$ of $k$-algebras, where $k$ acts diagonally; then base change induces the desired morphism \begin{align*} \textstyle \varphi : (\prod_{n \in \mathbb{Z}} k) \otimes_{k} K \to \prod_{n \in \mathbb{Z}} K \end{align*} of $K$-algebras. We show that $\varphi$ is not surjective. The image of $\varphi$ consists of elements of the form $\alpha = \sum_{i=1}^{\ell} (\mathbf{a}_{i} \otimes x_{i})$ with $\mathbf{a}_{i} \in \prod_{n \in \mathbb{Z}} k$ and $x_{i} \in K$. Then $\alpha \in \prod_{n \in \mathbb{Z}} F$ where $F$ is the finitely generated subextension $F = k(x_{1},\dotsc,x_{\ell})$ of $K/k$. Thus for example the element $(t_{n})_{n \in \mathbb{Z}} \in \prod_{n \in \mathbb{Z}} K$ is not contained in the image of $\varphi$.

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  • $\begingroup$ Sorry, but what are the second maps in both horizontal sequences(the ones in between $g_2$ and $g_3$)? I only know of an equalizer diagram of that sort, but can't seem to find the homomorphisms. $\endgroup$ Commented Sep 1, 2017 at 19:12
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    $\begingroup$ It's the difference of the two morphisms in the equalizer diagram. More precisely if $U = \bigcup_{i=1}^{n} U_{i}$ is a covering, then the map $\prod_{i=1}^{n} \mathcal{O}(U_{i}) \to \prod_{i,j=1}^{n} \mathcal{O}(U_{i} \cap U_{j})$ sends the $n$-tuple $(s_{i})_{i=1,\dotsc,n}$ (where $s_{i} \in \mathcal{O}(U_{i})$) to $(s_{i}|_{U_{i} \cap U_{j}}-s_{j}|_{U_{i} \cap U_{j}})_{i,j=1,\dotsc,n}$. $\endgroup$ Commented Sep 1, 2017 at 19:33
  • $\begingroup$ Ah, thanks! I also thought of the same map, but for some reason I thought the maps should be algebra homomorphisms, which is not true. Being $k$-module is enough for the argument. My bad! $\endgroup$ Commented Sep 1, 2017 at 20:15

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