Does Hartshorne assume curves are irreducible when proving that pullback along a finite morphism of nonsingular curves preserves linear equivalence?

Question

The question is how it can be proved the statement in the title:

Let $f: X\to Y$ be a finite morphism of nonsingular curves.

Then the pullback morphism $f^*:\operatorname{Div} Y \to \operatorname{Div} X$ preserves linear equivalence.

I have found this very same topic answered here:

Let $f: X \to Y$ be a finite morphism of nonsingular curves. Then $f^*: \operatorname{Div} Y \to \operatorname{Div} X$ preserves linear equivalence.

In my opinion the answer linked above (and consequently the proof) is correct as long as one assumes that at least the curve $Y$ is irreducibile. So my doubt is: is it so obvious that we are dealing with irreducibile curves at the point that the book doesn't even specify it? In such case why? Or am I wrong about the fact that the proof assumes the irreducibility of ? In such case I am not able to justify this part of the proof:

"[...] On the other hand, for each $Q_i$ we can find an open set of on which $g=u\cdot g_i^{v_{Q_i(g)}}$ [...] "

since I cannot understand why it must exist an open neighborhood in which both $g$ and $g_i$ are regular.

I hope my doubt is clear, thanks in advance.

Answer 1

Yes, we are dealing with irreducible curves at that point in the book. From just before remark II.4.10.1 (page 105 in my copy):

Definition. An abstract variety is an integral separated scheme of finite type over an algebraically closed field $k$. If it is proper over $k$, we will say it is complete.

Remark 4.10.1. From now on we will use the word "variety" to mean "abstract variety" in the sense just defined. We will identify the varieties of chapter I with their associated schemes, and refer to them as quasi-projective varieties. We will use the words "curve", "surface", "three-fold", etc., to mean an abstract variety of dimension 1, 2, 3, etc.