2
$\begingroup$

Is it true "If $X \subset Y$ then $\Bbb I(Y) \subset \Bbb I(X)$; here I am using proper inclusion. Couldn't prove it though. Trying for long time please help. Actually I saw here "https://people.maths.bris.ac.uk/~mp12500/teaching/Lectures5-7.pdf" in proposition 17 $\Rightarrow$ direction.

$\endgroup$
1
$\begingroup$

Looks like it's actually Prop 11 (part 2) in the notes you link that shows inclusion.

To see proper inclusion, suppose $I(Y) = I(X)$. Then $V(I(Y)) = V(I(X))$. But as $X$ and $Y$ are algebraic varieties, $V(I(X)) = X$ and $V(I(Y)) = Y$. So you must have had $X = Y$.

$\endgroup$
  • $\begingroup$ Would like to vote my question? $\endgroup$ – Gimgim Feb 11 at 21:56

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.