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.


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


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.

  • $\begingroup$ 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

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