TITLE: Does Hilbert's theorem 90 hold for local rings?
QUESTION [6 upvotes]: Let $R$ be a local ring (commutative with 1). Let $G$ be a finite subgroup of $\text{Aut}(R)$ preserving the maximal ideal. Then it seems to me that we also have:
$$H^1(G,R^\times) = 0$$
Is this correct? (The classical Hilbert theorem 90 states this when $R$ is a field).
Here's the argument:
First, you need the Lemma: If $g_1,\ldots,g_n$ are distinct automorphisms of $R$, then if for $c_i\in R$, $\sum_{i=1}^n c_ig_i = 0$ (as a function $R\rightarrow R$), then each $c_i = 0$. Indeed, one may assume that there is a minimal such relation, where $c_1g_1 + \cdots + c_rg_r = 0$ with $c_1,\ldots,c_r$ all nonzero. In this case we must have $r > 1$ since $c_1g_1 = 0$ means $c_1g_1(1) = c_1\cdot 1 = 0$ so $c_1 = 0$. Now since $g_1\ne g_r$, let $a\in R$ be such that $g_1(a)\ne g_r(a)$. Let $x\in R$ be arbitrary. We have:
$$\sum_{i=1}^r c_ig_i(ax) = \sum_{i=1}^r c_ig_i(a)g_i(x) = \sum_{i=1}^rc_ig_i(x) = 0$$
Multiplying the last sum by $g_r(a)$ and subtracting from the second sum, we get:
$$\sum_{i=1}^rc_i(g_i(a)-g_r(a))g_i(x) = \sum_{i=1}^{r-1}c_i(g_i(a)-g_r(a))g_i(x) = 0$$
Since this holds for all $x\in R$, this gives a shorter relation and by our choice of $a$, the coefficients are not all zero, since $c_1(g_1(a) - g_r(a))\ne 0$.
To prove the theorem, we now proceed as usual: Let $\alpha : G\rightarrow R^\times$ be a 1-cocycle, so $\alpha$ satisfies $\alpha(gh) = \alpha(g)\cdot (g.\alpha(h))$. In particular, we have $g.\alpha(h) = \alpha(g)^{-1}\alpha(gh)$.
Applying the above result to the residue field $R/\mathfrak{m}$, the linear combination
$$\sum_{g\in G}\alpha(g)\cdot g$$
is nonzero in the residue field. Thus, there exists a $\theta\in R$ such that
$$\beta := \sum_{g\in G}\alpha(g)\cdot g(\theta) \in R^\times$$
Then, for each $h\in G$, we have
$$h(\beta) = h\left(\sum_{g\in G}\alpha(g)\cdot g(\theta)\right) = \sum_{g\in G}h(\alpha(g))\cdot (hg)(\theta) = \sum_{g\in G}(\alpha(h)^{-1}\alpha(hg))\cdot (hg)(\theta) = \alpha(h)^{-1}\sum_{g\in G}\alpha(hg)\cdot (hg)(\theta) = \alpha(h)^{-1}\beta$$
Thus, $\alpha(h) = \frac{\beta}{h(\beta)} = \frac{h(\beta^{-1})}{\beta^{-1}}$, so $\alpha$ is a coboundary.
Does this seem right? I just want to record this here since I find it strange that I've never seen this simple generalization, which doesn't require any real additional technology to state or prove.

REPLY [7 votes]: This is false. Let $R$ be $\mathbb{Z}_2[i]$ (where $i$ denotes a choice of square root of negative one). It has an automorphism $\sigma$ exchanging $i$ and $-i$, and (writing $G = \{\text{id}, \sigma\}$) one has
$$
H^1(G, R^\times) = \frac{\text{ker} R^\times \stackrel{N}{\to} R^\times}{\text{im} R^\times \stackrel{\sigma - 1}{\longrightarrow} R^\times}
$$
Now the element $i$ of $R$ has norm 1. But it is not of the form $\alpha/\alpha^\sigma$ for any $\alpha \in R^\times$. Indeed, if
$$
\frac{a + bi}{a-bi} = i
$$
then cross-multipyling gives
$$
a + bi = ai + b.
$$
so a=b. But any element $a(1+i)$ of $\mathbb{Q}_2(i)$ with $a \in \mathbb{Z}_2$ has positive valuation, so is not a unit in $\mathbb{Z}_2[i]$.