3
$\begingroup$

Let $k'/k$ be a finite Galois extension, and $X$ an affine $k'$-scheme. Consider its Weil restriction $\mathrm{Res}_{k'/k} X$, an affine $k$-scheme.

It is well-known (for instance Weil, Adeles and algebraic groups I.3) that \begin{equation} ( \mathrm{Res}_{k'/k} X ) \times_k k' \cong \prod_{\sigma \in \mathrm{Gal} (k'/k)} X^{\sigma} \end{equation} where $X^{\sigma}$ is the base change of $X$ along the $k$-map $\sigma: k' \longrightarrow k'$.

This question Two definitions of the Weil restriction. as well as the source they mention (Weil) explains more or less how the map is defined: it is the product of the maps $p^{\sigma}$ where $p: ( \mathrm{Res}_{k'/k} X ) \times_k k' \longrightarrow X$ corresponds (I am guessing) to the identity map on $\mathrm{Res}_{k'/k} X$ under the adjunction \begin{equation} \mathrm{Hom} \left( ( \mathrm{Res}_{k'/k} X ) \times_k k', X \right) \cong \mathrm{Hom} \left( \mathrm{Res}_{k'/k} X , \mathrm{Res}_{k'/k} X \right) \end{equation} and $p^{\sigma}$ is obtained by $p$ via base change along $\sigma: k' \longrightarrow k'$ as before.

I would like to understand how $\mathrm{Gal} (k'/k)$-action on the right hand side works - on the left hand side it is simply the Galois action on the $k'$-factor. I would guess that on the right hand side the action of $\gamma$ is simply given by `permuting' the factors, via $X^{\sigma} \stackrel{\gamma}{\longrightarrow} X^{\sigma \gamma}$ - is that the case?

Of course, I tried to make the Galois action go through the isomorphism mentioned above, but I am a bit confused about whether $p^{\sigma}$ is defined as "$p$, followed by base change by $\sigma^{-1}$" or if one really wants to base change under $\sigma$ the map $p: \mathrm{Res}_{k'/k} X \times_k k' \longrightarrow X$, and then there is some canonical isomorphism between the domain and its $\sigma$-base change.

I am aware that the Galois action can be used to define the descent datum on the right hand side and ultimately prove the existence of the Weil restriction, but I am not sure how this could help me.

$\endgroup$
1
$\begingroup$

Let's suppose that $X=\text{Spec}(k'[x_i]/(f_j))$ and so $X^\sigma=(k'[x_i]/(f_j^\sigma))$ where if $\displaystyle f_j(x_1,\ldots,x_n)=\sum_I a_I x^I$ then $\displaystyle f_j^\sigma(x_1,\ldots,x_n)=\sum_I \sigma(a_I)x^I$. For every $\gamma$ we can define a map $f_\gamma:X^{\gamma \sigma}\to X^{\sigma}$ as follows. Namely, it's the one associated to the ring map

$$k'[x_i]/(f_j^{\sigma})\to k'[x_i]/(f_j^{\gamma\sigma})$$

that's trivial on the $x_i$ by acts as $\gamma$ on $k'$. Together these maps form a map, also called $f_\gamma$, from $\displaystyle \prod_\sigma X^\sigma\to\prod_\sigma X^\sigma$. This is the action of $\text{Gal}(k'/k)$ on this product.

Here is, perhaps, a slightly more compact way to write it. If $X=\text{Spec}(A)$ then we can think of $X^\sigma$ as $\text{Spec}(A\otimes_{k',\sigma}k')$. So then, we can think of $\displaystyle \prod_{\sigma\in \text{Gal}(k'/k)}X^\sigma$ as being

$$\prod_{\sigma\in\text{Gal}(k'/k)}X^\sigma=\text{Spec}\left(\bigotimes_{\sigma}(A\otimes_{k',\sigma}k')\right)$$

For every $\tau\in\text{Gal}(k'/k)$ to give an automorphism of $\displaystyle \prod_{\sigma\in\text{Gal}(k'/k)}X^\sigma$ is to give a ring automorphism of $\displaystyle \bigotimes_{\sigma\in\text{Gal}(k'/k)}(A\otimes_{k',\sigma}k')$. But, this is the ring map that sends $(x_\sigma\otimes a_\sigma)_\sigma\mapsto (x'_\sigma\otimes a'_\sigma)_\sigma$ with $x'_{\tau\sigma}\otimes a'_{\tau\sigma}:= x_\sigma\otimes \tau(a_\sigma)$.

Of course, the Galois descent then just gives that $\text{Res}_{k'/k}X$ is the spectrum of the $\text{Gal}(k'/k)$-invariants of $\displaystyle \bigotimes_\sigma (A\otimes_{k',\sigma}k')$.

Maybe a concrete example will help. As you're probably well aware, thinking of $\mathbb{C}^\times$ acting on $\mathbb{C}\cong\mathbb{R}^2$ gives a realization of $R:=\text{Res}_{\mathbb{C}/\mathbb{R}}\mathbf{G}_{m,\mathbb{C}}$ as a subgroup of $\text{GL}_{2,\mathbb{R}}$. Namely

$$\text{Res}_{\mathbb{C}/\mathbb{R}}\mathbf{G}_{m,\mathbb{C}}=\left\{\begin{pmatrix}a & -b\\ b & a\end{pmatrix}\right\}\subseteq\text{GL}_{2,\mathbb{R}}$$

Now, on points there is a natural isomorphism $R_\mathbb{C}\to\times\mathbf{G}_{m,\mathbb{C}}$ given on the level of rings as

$$\mathbb{C}[t,t^{-1},s,s^{-1}]\to \mathbb{C}[a,b]\left[\frac{1}{a^2+b^2}\right]:t\mapsto a+bi, s\mapsto a-bi$$

if we transport the $G_\mathbb{R}$ action back along this isomorphism we get the following action of conjugation $c$ on $\mathbf{G}_{m,\mathbb{C}}^2$: $c(s)=t$ and $c(\alpha)=\overline{\alpha}$ for all $\alpha\in\mathbb{C}$. This is precisely the action we described above.

$\endgroup$
  • $\begingroup$ Thanks, that's helpful. Can something be said if $k'/k$ is separable but not Galois? I imagine one wants to replace $\mathrm{Gal} (k'/k)$ by the $k$-embeddings of $k'$ into an algebraic closure $\overline k$, but those are not a group... $\endgroup$ – user94041 Oct 4 '17 at 19:43
  • $\begingroup$ Oh I think I know the answer to my last question. The statement replaces $\mathrm{Gal} (k'/k)$ by the set of embeddings I mentioned in my previous comment, and $\mathrm{Gal} (k'/k)$ (the Galois group of a normal closure of $k'$ in $\overline k$) acts on this set by post-composition, with the formulas being the same as in your answer. Right? $\endgroup$ – user94041 Oct 4 '17 at 19:49
  • $\begingroup$ @user94041 Yep! $\endgroup$ – Alex Youcis Oct 5 '17 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.