Galois descent in motivic cohomology Let $X_N$ denote the Fermat curve defined over $\mathbb{Q}$ by the equation $x^N+y^N-z^N=0$ and let $X_{N,\mathbb{Q}(\mu_N)}$ be the base change. Let $G$ be the Galois group of $\mathbb{Q}(\mu_N)/\mathbb{Q}$, hence $G \cong (\mathbb{Z}/N\mathbb{Z})^\times$. Consider the motivic cohomology $H^2_\mathcal{M}(X_N,\mathbb{Q}(2))$. I would like to know whether
$$ H^2_\mathcal{M}(X_N,\mathbb{Q}(2)) \cong H^2_\mathcal{M}(X_{N,\mathbb{Q}(\mu_N)},\mathbb{Q}(2))^G.$$
Motivation:
Let $W:=(\mathbb{Z}/N\mathbb{Z})^2$ act on $X_{N,\mathbb{Q}(\mu_N)}$ as
$$(r,s)(x:y:z) := (\zeta^rx:\zeta^sy:z),$$
where $\zeta$ is a primitive $N-$th root of unity. I am reading On the regulator of Fermat motives and generalized hypergeometric functions by N. Otsubo, where he proves some results of surjectivity of regulators using the splitting of the motive $h^1(X_N)$ into motives $X_N^{[a,b]}=(X_N,p_N^{[a,b]})$, where $p_N^{[a,b]}$ is in $\mathbb{Q}[W]^G$. I am working on something related, but I am not much acquainted with motivic cohomology. My definitions (which are quite ad hoc for this situation) come from Milnor $K-$theory:
$$H^2_\mathcal{M}(X_N,\mathbb{Q}(2)) \cong K_2(X_N)_\mathbb{Q} := \ker\left( K_2^M(k(X_N))\otimes \mathbb{Q} \overset{T\otimes\mathbb{Q}}{\longrightarrow} \bigoplus_{x \in X_N(\overline{\mathbb{Q}})} \overline{\mathbb{Q}}^\times \otimes \mathbb{Q} \right),$$
where $T$ denotes the Tame symbol (see for instance page 27 of the above article).  My original problem is that I want to show that if $e_N$ is in $K_2(X_N)_\mathbb{Q}$, then the elements $e_N^{[a,b]}:=p_N^{[a,b]}(e_N)$ defined by Otsubo are in $K_2(X_N)_\mathbb{Q}$ as well. But, in principle, I only know that they are in $K_2(X_{N,\mathbb{Q}(\mu_N)})_\mathbb{Q}^G$. So I would like to know whether the two vector spaces are equal. The problem is that, starting from my definition, I think that we have
$$k(X_N)^\times \otimes k(X_N)^\times \subsetneqq (k(X_{N,\mathbb{Q}(\mu_N)})^\times \otimes k(X_{N,\mathbb{Q}(\mu_N)})^\times)^G.$$
So if things work out in the end it must be thanks to the Steinberg relation $a\otimes(1-a)$ that still has to be quotiented out in order to define $K_2(k(X_N))$, or thanks to the Tame symbol. But this seems to be difficult to check, hence I was wondering if there is some general argument for the equality
$$K_2(X_N)_\mathbb{Q}=K_2(X_{N,\mathbb{Q}(\mu_N)})_\mathbb{Q}^G$$
for example coming from motivic cohomology.
 A: K-theory with rational coefficients agrees with étale K-theory with rational coefficients and consequently has étale descent (see e.g. Elden Elmanto's answer to this MO-question. In the situation at hand, this should imply the required identification $K_2(X_N)_{\mathbb{Q}}=K_2(X_{N,\mathbb{Q}}(\mu_N))^G_{\mathbb{Q}}$. Motivic cohomology with rational coefficients also has étale descent and consequently the required formula for motivic cohomology would also be true. (For étale descent properties of motivic cohomology with rational coefficients, you could look into the paper of Cisinski and Déglise.
A: What you need is the existence of the transfer map $N : K_2(L) \to K_2(K)$ for any finite field extension $L/K$, which is due to Bass and Tate, see Introduction to algebraic $K$-theory by Milnor. I don't know of any definition of the transfer map using Matsumoto's decription of $K_2$, one should rather use Milnor's definition of $K_2$. In any case, if $L/K$ is Galois and $j : K_2(K) \to K_2(L)$ is the canonical map, then $N \circ j = |G| \cdot \mathrm{id}$ and $j \circ N = \sum_{\sigma \in G} \sigma^*$. We deduce in particular that $j$ induces an isomorphism $K_2(K)_{\mathbb{Q}} \cong (K_2(L)_{\mathbb{Q}})^G$.
Granted these facts and returning to your situation, we have $K_2(\mathbb{Q}(X_N))_{\mathbb{Q}} \cong (K_2(\mathbb{Q}(\mu_N)(X_N))_{\mathbb{Q}})^G$ for the function fields, and you should get the isomorphism you want by taking the kernel of tame symbols.
I should add that transfer maps exist in much greater generality for motivic cohomology with rational coefficients, and for finite Galois covers the obvious formulas for $N \circ j$ and $j \circ N$ are still valid. See e.g. Deninger--Scholl, The Beilinson conjectures, (1.3) for a nice summary of the properties of motivic cohomology.
