About morphism from elliptic curve to projective space and pullback divisor I'm sorry if the following question is trivial or even if it doesn't make any sense (I'm still learning about divisors).
Suppose that $E/k$ is an elliptic curve, then the divisor $nO$ on $E$ is supposed to define a map to a projective space $\alpha:E\rightarrow{\mathbb{P}^{n-1}}$ using the linear system $|nO|$. Now if I take a hyperplane $H$ in $\mathbb{P}^{n-1}$ and I regard it as a Weil divisor, then why the pullback divisor $\alpha^{*}H$ is linearly equivalent to $nO$?
I was trying to apply some results from Hartshorne, for example I think that $H$ corresponds to an invertible sheaf $\mathcal{O}(1)$, and then I considered the invertible sheaf $\alpha^{*}\mathcal{O}(1)$ and my plan was to show that this invertible sheaf is isomorphic to $\mathcal{L}(nO)$. I think that it's possible to identify $\alpha^{*}\mathcal{O}(1)$ with an invertible sheaf $\mathcal{L}(D)$ where $D$ is a Weil divisor on $E$, and if $\mathcal{L}(nO)$ is isomorphic to $\alpha^{*}\mathcal{O}(1)$ then I think that $D$ and $nO$ are linearly equivalent, but the problem is that I can't easily see what is $\alpha^{*}\mathcal{O}(1)$. I know how to define $\alpha$  but I don't know how to write $\alpha^{*}\mathcal{O}(1)$ of the form $\mathcal{L}(D)$.  
Any help is appreciated.
Edit 1: About the map $\alpha:E\rightarrow{\mathbb{P}^{n-1}}$, now I've just realized that I don't understand how it is defined. I thought that first we had to take the invertible sheaf $\mathcal{L}(nO)$ and by Riemann-Roch this should have dimension $n$, so we take a basis $s_{0},\dots,s_{n-1}$ which are supposed to be global sections which generate the invertible sheaf $\mathcal{L}(nO)$ (does it follow because it is a basis of $\Gamma(E,\mathcal{L}(nO))$?) and then theorem 7.1 from Hartshorne gives a unique $k$-morphism $\alpha:E\rightarrow{\mathbb{P}^{n-1}}$ such that $\mathcal{L}(nO)\cong{\alpha^{*}\mathcal{O}(1)}$, and I thought that this $\alpha$ is the one that is induced by $|nO|$. 
I will keep reading about this, as you can see I'm quite newbie...
 A: I am also in the process of digesting this stuff, but maybe this will be helpful for you anyway. (I probably provide too much detail in places that you are not having issues with - this is because I am explaining these things to myself at the same time. Also, since I am a novice algebraic geometer it is likely that I am made a mistake somewhere. Although I don't think so. But if something is confusing, there is a decent chance that it is wrong. Please call me out on it.)
Please let me know if this is unreadable... I can try to organize it better tomorrow.

Moral summary of what we do: We describe the pullback as pulling back an appropriate section of the bundle $O(1)$, then computing its divisor. We can find a section of $L(nO)$ which has divisor $nO$, and we know that any two global sections of a line bundle give linearly equivalent divisors. 
I spell out way too many of the details along the way, that is what makes this so long.

I think it would be helpful to unravel what you mean by the map with the linear system $|nO|$.
(Comment: If you haven't done this already - it is worthwhile to work a similar example out in the case of a line bundle on $P^1$, since in that case you can write down explicitly all of the equations, and draw some doodles that are at least metaphorically accurate.)
We are choosing $s_0, \ldots, s_n$, which are global sections of the line bundle $L(nO)$.
Our assumption that we have enough global sections to define a map means that there is at least one global section, and in fact enough so that there are no basepoints for the linear system.

We can choose $s_0$ so that the divisor it defines is $nO$.
Why? A model for $L = L(nO)$ is the sheaf assigning to a set $U$ the nonzero rational functions $f$ so that $div f|U + nO \geq 0$ - here div means the divisor that $f$ defines as a rational function. Consider the rational function "1" (what other non-zero rational functions can we write down? Basically none.)
Since the divisor we started with was already effective, $div 1 |U + nO \geq 0$ for any $U$ - so this tells us that $1$ is a global section of $L(nO)$.
We can associate to the pair $(L,1)$ an effective divisor, by working on charts where we can trivialize the line bundle $L = L(nO)$ via some $\phi : L \to O_E$, and associating to (the rational function) $\phi(1)$ a divisor. On $O^c$, $L$ already gives the regular functions, and the regular, nowhere vanishing function $1$ has empty divisor. On a neighborhood of $O$, we can find a regular function $g$ that vanishes at $O$ to order $n$ (how? the local ring is a DVR, so we know that in some neighborhood there is a uniformizer that is a regular function. Then take a power of this uniformizer.)
Then multiplication by $g$ defines a trivialization - here you have to use some tricky commutative algebra - basically coming down to the statement that a rational function with no poles is regular on normal schemes. But what you should imagine is that the condition defining $L(nO)$ only allows in functions that have poles up to order n at $O$, and which must be regular everywhere else. So multiplication by $g$ makes them regular because they have no poles (and dividing any regular function by $g$ produces a section of $L(nO)$. Under this trivialization, 1 goes to $g$, and has divisor $nO$.
So the divisor of the global section $1$ of $L(nO)$ is $nO$.

How is the map to projective space defined? It is given in coordinates by $\alpha = [s_0: \ldots: s_n]$., where the rest of $s_i$ just complete $s_0$ to a basis of the global sections of $L(nO)$. (Technically, what we are doing here is using the map on schemes defined by the ring morphism $k[x_j / x_i] \to \Gamma(O_E, D(s_i)) $ that sends $x_j / x_i$ to $s_j / s_i$. But this intuition about coordinates is good, and makes formal sense via the functor of points.)
Intuitively, $s_0$ is the pullback of the coordinate function $x_0$: the "values" of $s_0 / s_i$ on $D(s_i)$ are the values of $x_0 / x_i$ on $D(x_i)$. This is not just intuition, since $\alpha^* O(1) \cong L(nO)$ and $\alpha^* x_0 = s_0$ (equality after applying this isomorphism between the line bundles).
(About proving this equality: You may want to describe $x_0$ under trivializations on the charts $D(x_i)$, then see that this pulls back over each of the trivializations to the description of $s_0$ under trivializations on the charts $D(s_i)$. Maybe there is a slicker way to do this? Probably. But anyway, it is built into the construction of this morphism.)
So this suggests that the divisor of $s_0$, which is $nO$, is the pullback of the divisor of $x_0$, which is in the same class as $H$.
So let's try to understand what pullback is.

Disclaimer: I am not completely sure that this definition is correct - I did a little googling and couldn't find it. I think I can prove that it is equivalent to the definition in terms of pulling back local equations of a Cartier divisor (when it makes sense to do that). I do this in the next block.
If $\alpha : Y \to X$ is a morphism of smooth varieties, we can (I think!!!!) define the pullback as follows. If $D$ on $X$ is an effective divisor, then $L(D)$ is a line bundle with a global section $1$ (we need locally factorial in order for this to be a line bundle). The divisor associated to $(1)$ is $D$. We know how to pullback line bundles and their sections, so $\alpha^*(1)$ gives a section of a line bundle on $Y$, and we can define this to be the pullback of the divisor $D$.
Now we extend this linearly to a map between the divisor groups.
I also claim that this definition of pullback preserves linear equivalence - if two effective divisors are linearly equivalent, then they can be written as $(s) = div (s)$ and $div(t)$ , where $s,t$ are two sections of the same line bundle $L$. Then $s/t$ is a rational section of $L^* \otimes L \cong O_{X}$ defined on $D(t)$ (otherwise known as a rational function). But then $s/t$ pulls back to a rational function on $Y$, which gives a linear equivalence between the pullbacks of $div(s)$ and $div(t)$.
If $E - D$ and $E' - D'$ are differences of effective divisors which are linearly equivalent, then we have shown that $E + D'$ and $E' + D$ pullback to linearly equivalent divisors. So it follows from linearity that $E - D$ and $E' - D'$ pullback to linearly equivalent divisors.
Thus $\alpha^*$ induces a map on class groups.

Let me try to give this definition some credibility (after I tried hard to take away its credibility...) by proving that it is the same as the definition in terms of Cartier classes. (You can skip to the next block to see how it solves your question.)
Let $D$ be a Cartier divisor on $X$, that is, a global section of the sheaf $K^* / O^*$. ( This means that for each open set in some cover of $X$, we have a chosen rational function, and the ratio of these rational functions on the overlaps is regular. )
We can encode this data as $(U_i, f_i)$, for the said rational functions on $U_i$. Then, under the condition that $\alpha(Y)$ does not land in the support of $D$, we can define a Cartier divisor $\alpha^*$ on $Y$ by the open sets $\alpha^{-1} ( U_i)$ and rational functions $\alpha^* (f_i)$. (our condition about the support is to ensure that it makes sense to pullback a rational function.)
Suppose that $D$ is that Cartier divisor, and $W$ is a Weil divisor, so that they are equivalent under the Weil-Cartier divisor correspondence on a smooth variety (really, an integral, separated, Noetherian locally factorial scheme) (given a Cartier divisor, you can work in the patches where it is defined as a rational function, and compute the divisor of that rational function... to go the other way, the section 1 of the line bundle $L(W)$ is a rational section, after trivializations (any two of which differ by an invertible regular function) it describes a bunch of rational functions whose pairwise ratios are regular on the overlaps). 
Claim: Suppose that $\alpha^* (D)$ is defined, i.e. the support of $D$ doesn't contain the image of $\alpha$. Then $\alpha^* D$ and $\alpha^* W$ are again equivalent under the Weil-Cartier divisor correspondence.
Lazy sketch of proof: For this, it suffices to work with an effective Weil divisor, since all the maps are linear. Then we can turn it into a Cartier divisor and pullback, or alternatively pull it back and then turn it into an effective Cartier divisor. We must check that these processes agree. In $L(W)$ 1 is a regular section of the corresponding line bundle, and since trivializations pull back, and since pulling back sections "commutes" with trivializations, we have defined the same Cartier divisor in two ways. (Symbolically, I mean that if $\phi : L \to O_X$ is a trivialization, then $\alpha^*( \phi(s)) = \alpha^*(\phi) (\alpha^*(s))$ - which can be checked on affines using properties of the tensor product. Or alternatively by describing a section as a map $s : O_X \to L$.)

Now we apply this definition to $x_0$, thought of as a global section of $O(1)$. As we discussed, $x_0$ pulls back to $s_0$. Moreover, the class of $(x_0)$ is the class of $H$, and we checked in the previous box that this pullback is well defined on divisor classes.
So if this definition of pullback agrees with yours, then I think this is a more or less complete argument (though there are still lots of details that can be filled in, depending on how pedantic you are feeling).
Do you buy it?
