Say $f : \mathbb{A}^2 \rightarrow \mathbb{A}$ is projection on the first coordinate. Given a sheaf $\mathcal{N}$ of $\mathcal{O}_Y$modules defined by a finitely generated $\mathbb{C}[x]$module, what is $f^*\mathcal{N}$? Does it come from a $\mathbb{C}[x, y]$module?
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Let $A \to B$ be a ring morphism, and $f: X=Spec B \to Spec A=Y$ be the associated map of affine schemes. For an $A$module $N,$ the pullback $f^{\star}(\mathcal{N}) \cong (N \otimes_A B)^{\sim}$ where $\mathcal{N}=(N)^{\sim}$ is the associated $\mathcal{O}_Y$module. So in your case, $f^{\star}(\mathcal{N}) \cong (N \otimes_{\mathbb{C}[x]}\mathbb{C}[x,y])^{\sim}\cong \mathcal{N}\otimes_{\mathcal{O}_Y}\mathcal{O}_X.$ Finitely generated assumption is of course redundant. – Ehsan M. Kermani Oct 09 '13 at 01:39

Neglect the last statement! – Ehsan M. Kermani Oct 09 '13 at 03:21
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I assume by $Y$ you mean $\mathbb A^1 = Spec(\mathbb C [x])$. Then the answer to your last question is yes, which is the content of Proposition 5.8, Chapter II in Hartshorne's book.
(just as Ehsan Kermani says in his comment)
Since you are dealing with affine schemes, you can identify $\mathcal N$ with its set of global sections $N=\Gamma(\mathbb A^1, \mathcal N)$. Then, again as Ehsan says, to compute the pullback, you just tensor with $k[x,y]$, where now $k[x,y]$ acts on $N$ as follows: $x$ acts as before, but $y \cdot n = 0$ for all $n$.
Fredrik Meyer
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