Let $K$ be a field and $G$ an algebraic group defined over $K$. If $M\supseteq L$ are two finite Galois extensions of $K$, then the groups $\text{Gal}(M/K)$ and $\text{Gal}(L/K)$ act, respectively, on the $M$-points $G_M$ and the $L$-points $G_L$ of $G$, and one may then define $$\rho_L^M:H^1(\text{Gal}(M/K), G_M) \to H^1(\text{Gal}(L/K), G_L)$$ Well, it is the last definition that I don't get. How is $\rho_L^M$ defined? If I have a representative $\phi$ of a class in the first cohomology set, this is a function $\phi:\text{Gal}(M/K)\to G_M$, how do I get a function from $\text{Gal}(L/K)$ to $G_L$? From my point of view, it should be the other way, since $G_L$ is a subset of $G_M$ and elements in $\text{Gal}(M/K)$ map to $\text{Gal}(L/K)$ via restriction.
2025-01-13 05:27:09.1736746029
Problem defining morphism in Galois cohomology of algebraic group
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Indeed, it should go the other way, and is the inflation map : $Gal(L/K)$ is the quotient of $G=Gal(M/K)$ by $H=Gal(M/L)$ so for any $G$-group $A$ (here $A=G_M$) you have the inflation map $H^1(G/H,A^H)\to H^1(G,A)$ (and here $A^H = M_L$).