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?