# Silverman *Arithmetic of Elliptic Curves* Problem 1.12 (a)

I am trying to understand Problem 1.12 (a) in Silverman's Arithmetic of Elliptic Curves. Here is the problem

Let $$K$$ be a perfect field, $$V/K$$ be an affine variety, and let $$G_K = \operatorname{Gal}(\overline{K}/K)$$. Prove that $$K[V] = \{f\in\overline{K}[V]:f^\sigma=f\;\;\forall\sigma\in G_K\}$$.

Here is the hint: the $$\subset$$ direction is clear. Conversely, if $$F\in\overline{K}[X]$$ is a representative of $$f$$, the map $$G_K\to\mathcal{I}(V)$$ via $$\sigma\mapsto F^\sigma-F$$ is a 1-cocycle (indeed, if $$f^\sigma=f$$ in $$\overline{K}[V]$$, then $$F^\sigma-F\in\mathcal{I}(V)$$ and $$F^{\sigma\tau}-F = (F^{\sigma}-F)^\tau+ (F^\tau-F)$$). It's the next point that I do not understand. Using $$H^1(G_K,\overline{K}^+) = 0$$, I'm supposed to deduce that there is some $$G\in\mathcal{I}(V)$$ such that $$F+G\in K[X]$$. I see that this would solve the problem, since in that case $$f = (F+G)+\mathcal{I}(V)\in K[V]$$.

Here's what I don't understand. Since $$H^1(G_K,\overline{K}^+) = \frac{Z^1(G_K,\overline{K}^+)}{B^1(G_K,\overline{K}^+)}$$, so if $$H^1 = 0$$, this means $$Z^1(G_K,\overline{K}^+) = B^1(G_K,\overline{K}^+)$$, so the function $$\sigma\mapsto F^\sigma-F$$ out to come from a boundary.

Question 1: It seems that $$\sigma\mapsto F^\sigma-F$$ is already in $$B^1(G_K,\overline{K}^+)$$ by definition, so how am I supposed to use the condition that $$H^1 = 0$$?

Question 2: In addition, it seems to me that $$H^1(G_K,\overline{K}^+)$$ is also the wrong group to be looking at. We're treating $$\overline{K}[X]$$ as an additive $$G_K$$-module, not $$\overline{K}^+$$, so shouldn't I be looking at $$H^1(G_K,\overline{K}[X])$$ or $$H^1(G_K,\overline{K}[V])$$?

Any elucidation would be much appreciated.

• @reuns $\overline{K}[V] := \overline{K}[X]/\mathcal{I}(V)$ and similarly for $K[V]$. Indeed, $V$ is assumed to be geometrically irreducible. I'm afraid that trace argument would only work in the case that $L/K$ is a finite Galois extension.
– Nico
Commented Dec 16, 2020 at 4:51
• I was saying that in characteristic 0 it is obvious that for $f\in L[V]$ fixed by all $\sigma \in G_K$ then $f = \frac1{[L:K]} Tr_{L[X]/K[X]}(f)\in K[V]$ Commented Dec 16, 2020 at 4:52
• In characteristic $p$ it is not true: $t$ is fixed by all the $\sigma$ but it is not in $K=\Bbb{F}_p(t^p)$. What is true is that $f^{p^m}\in K[V]$ for some $m$. Commented Dec 16, 2020 at 5:13
• Correct, I forgot to mention that $K$ is perfect.
– Nico
Commented Dec 16, 2020 at 5:15

$$f\in \overline{K}[X]$$

$$c(\sigma)=f^\sigma-f$$ is in $$Z^1(G_K,\overline{K}[X])$$

We are told that $$f^\sigma \equiv f\in \overline{K}[V]$$ ie. $$c\in Z^1(G_K,I(V)\overline{K}[X])$$.

We are told that $$H^1(G_K,\overline{K})=0$$.

This implies that $$H^1(G_K,I(V)\overline{K}[X])=0$$. This is because a $$K$$-vector space basis of $$I(V)$$ gives a $$G_K$$-invariant $$\overline{K}$$-vector space basis of $$I(V)\overline{K}[X]$$.

This gives that $$c(\sigma)=g^\sigma-g$$ with $$g\in I(V)\overline{K}[X]$$

so that $$(f-g)^\sigma-(f-g)=0\in \overline{K}[X]$$ ie. $$f-g\in K[X]$$ and $$f\equiv f-g\in K[V]$$

• Wonderful! Could you explain the point about $H^1(G_K,I(V)\overline{K}[X]) = 0$ in a bit more detail?
– Nico
Commented Dec 16, 2020 at 6:00
• $c(\sigma)=\sum_j c_j(\sigma) b_j$ where $(b_j)$ is a $K$-basis of $I(V)$ and $c_j\in Z^1(G_K,\overline{K})$ Commented Dec 16, 2020 at 6:24