Moving a divisor on a (reducible, non-reduced) curve I am trying to understand the first sentence of the proof of 9.1/5 in "Neron models." There we have a proper curve $X$ over a field $K$ and a line bundle $\mathscr{L}$ on $X$. Our ultimate goal is to express the degree of $\mathscr{L}$ in terms of the degrees of $\mathscr{L}|_{X_i}$, where the $X_i$ are the irreducible components of $X$. We represent $\mathscr{L}$ as $\mathscr{O}(D)$ for some Cartier divisor $D$ on $X$ (to find such a $D$ one could use, for instance, [EGA IV$_4$, 21.3.4 a)]). Then it seems to be implicitly claimed in the proof of 9.1/5 that we may even choose $D$ so that its support does not contain any intersection points $X_i \cap X_j$ for $i \neq j$. My question is: how to find such a $D$?
 A: First of all, given where you need this, let's start by passing to the algebraic closure of $K$. That does not effect the degree of divisors and makes our lives easier.
I started writing this up and realized that while I think what I was saying could be carried out, doing it precisely is a bit (i.e., a lot) of work. Doing it with one intersection point is a piece of cake: the line bundle is trivial in a neighbourhood of that point and the generator gives a meromorphic function whose associated divisor agrees with the one corresponding to the line bundle, so we have our principal divisor we needed. The trouble is that if we have several intersection points then fixing it at one point may screw it up at another... So we'd need some sort of a Riemann-Roch (or Mittag-Leffler) argument to say that if we have locally given principal divisors we can find a global one that agrees with those at those points. I think this can be done, but it seems too much work when there is a shorter way.
Ultimately, what you want is equivalent to the following statement:

Claim For any finite set of points on $X$ there exists an open set $U$ containing all those points in that finite set and such that $\mathscr L\left|_U\right.$ is trivial.

Proof: 
We are over and algebraically closed field now, so a proper curve is actually projective. Therefore we may write our line bundle as $\mathscr L \simeq \mathscr L_1 \otimes \mathscr L_2^{-1}$ where $\mathscr L_1$ and $\mathscr L_2$ are very ample line bundles. Now for any finite set of points  a very ample line bundle has a global section which is non-zero at any of those points, so if we take one section of each of $\mathscr L_1$ and $\mathscr L_2$, then on the complement of the zero loci of these sections all three of our line bundles are trivial. $\square$
Of course, I could have done this without making that claim, the union of the zero loci of those sections is actually the support of a divisor representing $\mathscr L$, which is really what you asked, but for some reason I like this formulation better. 
