I tried the following problem in Liu's book, 7.3.1 but I don't see where it was needed that $X$ is integral - maybe someone can help me here. Is the following true without supposing that $X$ is integral? What goes wrong otherwise?
Let $X$ be a projective curve over a field $k$ and let $k' /k$ be an extension. We let $p:X_{k'} \rightarrow X$ denote the projection and $D \in Div(X)$. Then, one is supposed to show that:
Suppose X is integral. Show that D is principal if and only if $p^*D$ is.
One direction is easy. The argument I have in mind for the other direction goes something along the lines as follows.
Let us set $\mathcal{K}_X$ to be the sheaf of regular sections on $X$ and $\mathcal{K*}_X$ those that are ''invertible'' . Let us note that by the flat base change theorem we have that $H^0(X_{k'},\mathcal{K^*}_{X_{k'}}) \cong H^0(X, \mathcal{K^*}_X) \otimes_k k'.$ Now, we have that for a cartier divisor $D$, that $D$ is principal iff $$H^0(X,\mathcal{K^*}_X \cap \mathcal{O}_X(D)(X) \neq \emptyset$$ and $\text{deg } D=0.$ We see that if $p^*D$ is principal, clearly $\text{deg } D = 0$.Now, if $$(H^0(X, \mathcal{K^*}_X) \otimes_k k') \cap (\mathcal{O}_X(D)(X) \otimes_k k') \neq \emptyset$$
my thinking was that, after clearing denominators surely this would imply that $$H^0(X,\mathcal{K^*}_X \cap \mathcal{O}_X(D)(X) \neq \emptyset$$
but maybe this is not true in general or my argument fails somewhere.
So, my questions are twofold, I suppose:
1. Is integral really needed?
2. If not, is the argument above correct? I understand if you don't have the time to look at it and would be happy with an answer to $1$ as well.
Just as a first comment: note that $H^0(X,\mathcal K_X^{\times})$ is just an abelian group (under multiplication); it is not a $k$-vector space or anything like that. So tonsuring it with $k'$ doesn't make sense, and the relationship between invertible elements in $K(X)$ and in $K(X_{k'})$ is more complicated than you seem to think.
Also, regarding "this would surely imply": it's pretty hard to see what the argument actually is here, or what "clearing denominators" means. (It might be that what you have in mind here is intertwined with your previous misconception discussed in the preceding paragraph.)
So, unfortunately, your argument is essentially completely non-existent as it stands.