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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.

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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 $X = Y$.

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

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