I am stuck on a claim in the proof of Theorem IV.4.1 in Hartshorne.
Let $X$ be a elliptic curve (a nonsingular projective variety of dimension 1 and genus 1 over an algebraically closed field), $T$ be a scheme of finite type (over the same field), and $p: X \times T \to T$ be the projection. In paragraph 4 of the proof, we have a invertible sheaf $\mathcal{M'}$ such that the restriction of $\mathcal{M'}$ to each fibre of $p$ has degree 1, and such that $p_*(\mathcal{M'})=\mathcal{O}_T$.
Now the section $1 \in \Gamma(T, \mathcal{O}_T)$ gives a section $s \in \Gamma(X \times T, \mathcal{M'})$ which defines an effective divisor $Z \subset X \times T$. Hartshorne claims that the restricted morphism $Z \to T$ is an isomorphism, but I am having trouble verifying this.
Since the restriction of $\mathcal{M'}$ to each fibre of $p$ has degree 1, I can see that $Z$ intersects each fibre of $p$ in exactly one point (without multiplicity). So the morphism $Z \to T$ is a bijection of sets. In fact, by basic topology, it is even a homeomorphism.
However I am having difficulty seeing what happens at the level of sheafs. Pick a local affine nbd $Spec B$ of $T$, then $s \in \mathcal{M'}(X \times Spec B)$ corresponds to $1 \in B$. Also $\mathcal{M'}(X \times Spec B)$ restricted to, say $Spec A \times Spec B \subset X \times T$ is nothing but $A \otimes B$, since $\mathcal{M'}$ is invertible. So we have $s\in A \otimes B$, and the structure sheaf of $Z$ is $A\otimes B/<s>$ in this local nbd. So I am supposed to prove $A\otimes B/<s>=B$? Something doesn't seem too right to me...
I apologize if there are any stupid mistakes above, I am still new to the subject. Any help is appreciated!
Use $s$ to get an exact sequence, $0\to \mathcal{O}_{X\times T}\to \mathcal{M}'\to \mathcal{M}'|Z\to 0$. Apply $p_*$ to get $0\to \mathcal{O}_T\to\mathcal{O}_T\to p_*(\mathcal{M}'|Z)\to R^1p_*\mathcal{O}_{X\times T}=\mathcal{O}_T\to 0$. So, we get $p_*(\mathcal{M}'|Z)=\mathcal{O}_T$. Since $p:Z\to T$ is a finite map (bijective) and $\mathcal{M}'|Z$ is a line bundle on $Z$, easy to check that this implies $p_*\mathcal{O}_Z=\mathcal{O}_T$ and then $p:Z\to T$ is an isomorphism.