Consider the following statement:
If $K\subseteq L$ is a Galois extension of fields with Galois group $G$ and $x \in \mathbb{P}^n(L)$ is such that $\sigma(x)=x$ for all $\sigma\in G$, then $x \in \mathbb{P}^n(K)$.
Here are two proofs of it (taking for granted the fundamental fact of Galois theory that if $z\in L$ satisfies $\sigma(z)=z$ for all $\sigma\in G$ then $z\in K$):
Proof S (for “Sophisticated”): let $X \in L^{n+1}$ lift $x$ (i.e., a system of homogeneous coordinates for $x$). Then $\sigma(x)=x$ translates to $\sigma(X) = c_\sigma\, X$ for some (uniquely defined) $c_\sigma \in L^\times$, which satisfies $c_{\tau\sigma} = \tau(c_\sigma)\,c_\tau$, meaning $c \colon G \to L^\times$ is a $1$-cocycle. So by Noether's generalization of Hilbert's theorem 90 (NH90), it is a coboundary, meaning there exists $b\in L^\times$ such that $c_\sigma = \sigma(b)/b$ for all $\sigma$. But then $\sigma(X/b) = X/b$ for all $\sigma$, so $X/b \in K^{n+1}$, which represents the same $x$, so $x \in \mathbb{P}^n(K)$ as claimed.
Proof T (for “Trivial”): write $x = (x_0:\cdots:x_n)$ and let $i$ be the smallest index such that $x_i\neq 0$. Then dividing by $x_i$ we can write $x = (0:\cdots:0:1:y_{i+1}:\cdots:y_n)$ where $y_j = x_j/x_i$. The fact that $\sigma(x)=x$ gives $\sigma(y_j)=y_j$ for all $j$, so $y_j \in K$, proving $x\in\mathbb{P}^n(K)$ as claimed. (Equivalently: the decomposition $\mathbb{P}^n = \mathbb{A}^n \uplus \mathbb{A}^{n-1} \uplus \cdots \uplus \mathbb{A}^0$, where $\mathbb{A}^{n-i}$ corresponds to the vanishing of the first $i$ coordinates and the nonvanishing of the next, works both for $L$ and $K$, and is stable under $G$ so we are reduced to the statement for $\mathbb{A}^{n-i}$, which is clear.)
I do hope both proofs are correct (which is why I took the trouble to write them in such excruciating details), but here's the thing: for a long time I was so blinded by the fact that proof S seemed to be “The Right Thing” to do that I completely missed proof T (at some point I even told some students that “the same fact for $\mathbb{P}^n$ as for $\mathbb{A}^n$ depends on a fairly deep result sometimes known as ‘Hilbert's theorem 90’”, which is really embarrassing), and realizing its existence threw me into disarray. I mean, this statement really looks like NH90 was meant to prove it, and it's not supposed to be a trivial fact! So how can we have such a trivial proof? What am I missing?
To make my question a little less vague and perhaps a bit more productive, let me ask:
Question: Does proof S have added value over T? Does it perhaps tell us more or does it work in a more general context¹? Or conversely, can we explain what makes the T shortcut away from NH90 possible in this context and not in others?
- Other than trivially reproducing NH90 by saying something like “any $\mathbb{G}_{\mathrm{m}}$-torsor is trivial” (however, it might help answer the question to explain why the torsor which appears in the statement at the top of this question is “trivially trivial”, if I dare call it that).
