0
$\begingroup$

I am actually reading the course from Hartshorne about deformation theory. (https://math.berkeley.edu/~robin/math274root.pdf ) After having defined the notion of flatness about a module, the author then defines what the deformation of a closed subscheme of a given one is, and begins with the affine case.

In this case, let $B$ be a $k$-algebra. And let's define $B' := B[t]/t^2$ and $D:=k[t]/t^2$

Let $I$ be an ideal of $B$, and $I' $ be one of $B'$ such that the image of $I' \subset B'$ onto $B'/tB'=B$ is exactly $I$.

My question is the following : why do we have in this case : $$B'/I' \otimes_D k = B/I$$ I tried to figure out on simple exemples but this fact seems weird to me. Thank your for your help.

$\endgroup$
0
$\begingroup$

As a module over $D$, $k = D/(t)$ so tensoring with it means killing $t$. This is exactly the assumption the previous sentence.

In more detail, you can apply the tensor with $k$ to the sequence $I’ \to B’ \to B’/I’$ and use right exactness and $B’\otimes k = B$ and $I’\otimes k = I$.

$\endgroup$
  • 1
    $\begingroup$ thank you for this answer it's clearer now $\endgroup$ – Axel S Jan 11 at 11:52

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.