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](https://arxiv.org/abs/0909.3002) 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.



Galois descent in motivic cohomology
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.


