2
$\begingroup$

Let $f: X \rightarrow Y$ be a morphism of schemes: $X$, $Y$ are regular schemes, let $Z_1, Z_2$ be two closed regular subschemes of $Y$, let $x \in X \times_Y Z_2$ such that $y = f(x) \in Z_1 \cap Z_2$. Then I would like to know how I can deduce that the map $$ T_xX \rightarrow (T_yY/T_yZ_2) \otimes_{k(y)} k(x) $$ is surjective knowing that

1) the image of $T_{x} (X \times_Y Z_1)$ generates $T_yY/T_yZ_2$

2) $T_x(X \times_Y Z_1) \rightarrow T_y Z_1 \otimes_{k(y)} k(x)$ is surjective.

It is supposed to follow from these things but not seeing how... I would appreciate any explanation on this. Thank you!

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ $f(x)\in Y$, but $Z_1\cap Z_2\subset X$, so did you mean something else? $\endgroup$
    – Mohan
    Commented Aug 17, 2019 at 18:20
  • $\begingroup$ @Mohan Typo fixed. Thank you. $\endgroup$
    – Johnny T.
    Commented Aug 18, 2019 at 13:52
  • $\begingroup$ Is there supposed to be a $\otimes_{k(y)} k(x)$ in 1)? Why is it missing there when it is included in the other two? $\endgroup$
    – Will Sawin
    Commented Aug 18, 2019 at 14:03
  • $\begingroup$ @WillSawin It is not there because I just wrote it as it is written in the paper I am looking at... $\endgroup$
    – Johnny T.
    Commented Aug 18, 2019 at 14:25
  • 1
    $\begingroup$ It seems crucial to the issue, because $T_x( X times_Y Z_1)$ is a subspace of $T_x X$, hence 1) should be sufficient to prove the statement unless something very strange is going on with the tensor products. $\endgroup$
    – Will Sawin
    Commented Aug 18, 2019 at 14:29

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .