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I am confused by this exercise in Ravi's notes:

16.3F (paraphrasing)

Suppose $i : p \to A_k^1$ is the inclusion of the origin. Consider the associated short exact sequence of quasi-coherent modules: $0 \to O(-p) \to O \to i_* (O|p) \to 0$, and use it to show that the pullback is not left exact.

This short exact sequence is the $\tilde{M}$ construction of the exact sequence of $k[t]$ modules $0 \to t k[t] \to k[t] \to k[t]/(t) \to 0$. In this case pullback is given by $\_ \otimes_{k[t]} k[t]/(t)$. But the map $0 \to t k[t] \to k[t]$ remains injective after tensoring, because $tk[t] \otimes_{k[t]} k[t]/(t) = 0$.

(I am doubly confused because I could interpret $0 \to t k[t] \to k[t]$ as $\times t: k[t] \to k[t]$, which gives a different answer. But based on what I understand, the first is the correct way to describe the quasicoherent sheaf associated to a closed subscheme, and the second is just a local trivialization.)

Help?

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Your computation of the tensor product is wrong: the base-change of a free module of rank one (which $tk[t]$ is) is again a free module of rank one (not zero, as you assert).

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  • $\begingroup$ Oh, of course. You can't move the $t$ to the other side of the tensor, since there is nothing to move it off of. Thanks. $\endgroup$ – Lorenzo Najt Jan 4 '16 at 18:43

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