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Let $X$ be an algebraic variety, $D$ a principal divisor of $X$ defined over $K$, i.e. the points of $D$ are in $X(K)$ and there is a function in $\overline{K}(X)$ whose divisor is $D$. Is $D$ necessarily the divisor of a function on $X$ defined over $K$, i.e. an element of $K(X)$?

I believe that I can answer this affirmatively using Galois cohomology: any two functions with the same divisor differ by an element of $\overline{K}^*$, so we can define a cocycle of $H^1(G_K, \overline{K}^*)$ by sending an element of the Galois group to the corresponding multiplicative factor on $f$. By Hilbert's Theorem 90, this is of the form $\sigma \mapsto \frac{\sigma(\lambda)}{\lambda}$ for some $\lambda$, so then $\lambda^{-1} f$ fits the bill.

Is this right / is there a more elementary way of seeing this?

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This argument is correct (with $K^{sep}$ in place of $\overline{K}$ if $K$ is not perfect), and the argument with Hilbert's Thm. 90 is standard. If there's a more elementary argument (not that HT90 is all that sophisticated) I don't know it. (There are lots of similar contexts involving "descent of the ground field" where HT90 is used in the same way; once you've seen it used a few times like this, it starts to seem less out of place.)