# Beware of the Two Galois Actions

Posted by Jason Polak on 22. December 2015 · Write a comment · Categories: algebraic-geometry, commutative-algebra · Tags: ,

Let $F$ be a field and $E/F$ be a nontrivial Galois extensions with Galois group $\Gamma$. If $V$ is an $F$-scheme then the points $V(E)$ carry a natural action of $\Gamma$ via the action on $\mathrm{Spec}(E)$. Sometimes, however, $V$ might have two Galois actions. How does this arise?

Perhaps the most natural setting is when $V$ is defined using the restriction of scalars functor. For example, if $X$ is an $E$-scheme, then $V = \mathrm{Res}_{E/F}(X)$ is by definition the scheme whose points in an $F$-algebra $R$ are given by
$$V(R) = X(R\otimes_F E).$$
Then, for any such $R$, the set $V(R)$ has a natural action of $\Gamma$ acting on $E$ in the tensor product $R\otimes_F E$. On the other hand, if $R$ is an $E$-algebra, then $V(R)$ will also have a $\Gamma$-action, via the action of $\Gamma$ on $R$. And, these two actions won’t be the same!

To see how this works, we’ll take the simplest example where $E/F$ is quadratic. The general case is exactly the same, but the notation is not as nice for exposition. Let’s see how the group $\Gamma$ acts on $E\otimes_F E$. In actual computations, like when you’re working with algebraic groups, it’s convenient to choose an isomorphism $E\otimes_F E\to E\oplus E$. How does this work? We can write the second $E$ in the tensor product as $F[x]/f$ where $f$ is an irreducible, monic, quadratic polynomial. Then the tensor product becomes
$$E\otimes_F F[x]/f \cong E[x]/f \cong E[x]/(x-a)(x-b)$$
where $a,b\in E$ are the roots of $f$ in $E$. Once we do this, we use the isomorphism
$$E[x]/(x-a)(x-b) \xrightarrow{\sim} E\oplus E\\f \mapsto (f(a),f(b))$$
to get an isomorphism $E\otimes_FE\cong E\oplus E$.

Now, we can describe the action of the nontrivial element of $\mathrm{Gal}(E/F)$. If we act on first factor of $E\otimes_FE$, then we the action is on the coefficients of the polynomials in $E[x]/f$. Hence, in this quadratic case we see that it corresponds in $E\oplus E$ to $(a,b)\mapsto (\overline{b},\overline{a})$, where the bar corresponds to applying the nontrivial element of $\mathrm{Gal}(E/F)$. On the other hand, if we act on the second factor of $E\otimes_FE$, then this corresponds to switching $a$ and $b$ in $E[x]/(x-a)(x-b)$, and so in $E\oplus E$ the action is $(a,b)\mapsto (b,a)$!

Therefore, in computations with schemes defined by restriction of scalars, beware of the two Galois actions!