So certainly every line bundle can be written this way, and injectivity follows from the properties of $\mathcal O_{\mathbb P(\mathcal E)}$ and the projection formula for locally free sheaves.

, the line bundle $\mathcal N$ is on $Y$, $\mathcal M\cong \mathcal O_{\mathbb P(\mathcal E)}(m)\otimes \pi^* \mathcal N$

I understand your question now. In my answer I just showed why if we take the line bundle $M$ on $\mathbb P(\mathcal E)$ it is isomorphic on each fiber to the same $\mathcal O_{\mathbb P^n}(m)$ and that this $m$ doesn't change. Once we have that, we in the situation of Hartshorne exercise III,12.4 with a flat projective morphism $\pi:\mathbb P(\mathcal E)\rightarrow Y$ and two invertible sheaves $\mathcal M$ and $\mathcal O_{\mathbb P(\mathcal E)}(m)$ which are isormorphis on each fiber. By that exercise (which is essentially Grauert's theorem), we get a line bundle $\mathcal N$ with

Thanks for the suggestion about the Euler characteristic. I'm only aware of, but comfortable yet with, the generalized Grothendieck-Riemann-Roch theorem so I came up with a more elementary argument just using semicontinuity and you comment about the Euler char.

I don't yet know too much above the Chow ring, so I didn't understand your answer so much. Either way, I am in fact trying to prove this for a general integral scheme, and it seems to be true there, so I'm not assuming anything about the singularities of my scheme.

Fair enough. I just haven't seen the degree defined that way before so I'm not yet comfortable with it. Care to provide a good reference? Couldn't one argue something similar, namely that since the fibres are isomorphic the induced invertible sheaves must be isomorphic? I'm not being precise here, I admit, since that doesn't mean that the induced invertible sheaf by some random isomorphism will be the invertible sheaf induced by restriction to the other fiber, but I believe this idea should lead to a precise answer, no?

Doesn't effectiveness then follow from the degree being 0? Is there a way of seeing this wihtout the notion of degree? Thanks for your help.

How do you see that it's effective and of degree 0 on every fibre?

Dear Matt, as often seems to happen, I realized my error like ten minutes after I asked that question. I always get confused which sign to put inside $\mathcal O(D)$ to get what I want. Thanks for the hint how to prove the residue theorem, I think I have it.

Dear Matt, I am indeed unfamiliar with that statement, and only know of the result I quoted from Hartshorne's brief discussion in III.7 of his book. Do know of a reference for the stronger statement? Also, in you answer above, isn't $\Omega^1_{\tilde{C}}(x+y)$ the differentials with zeroes at x and y? Wouldn't $\Omega^1_{\tilde{C}}(-x-y)$ be the sheaf of differentials with poles there? This may just be a stupid question, but I want to make sure I understand. Thanks for all the help.

How does the residue theorem imply the existence of differentials with non-zero residues? I thought it just implied that the sum of the residues of any meromorphic differential was zero.

How does Serre duality imply this? In order to prove Serre Duality for a singular curve one first needs the dualizing sheaf, no?