For $Y_1,Y_2$ closed subsets of $\mathbb{A}_k^n$, $\mathcal{I}(Y_1 \cap Y_2)= \text{rad}(\mathcal{I}(Y_1)+\mathcal{I}(Y_2))$

Asked Mar 13 '20 at 21:55

Active Mar 13 '20 at 21:55

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Let $Y_1,Y_2$ are two (Zariski) closed subsets of $\mathbb{A}_k^n$, for any field $k$. Then how do I show that $\mathcal{I}(Y_1 \cap Y_2)= \text{rad}(\mathcal{I}(Y_1)+\mathcal{I}(Y_2))$.

Clearly $\text{rad}(\mathcal{I}(Y_1)+\mathcal{I}(Y_2)) \subseteq \mathcal{I}(Y_1 \cap Y_2)$. How can I prove the converse ? I need some help. Thanks.

asked Mar 13 '20 at 21:55

user371231

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Does this answer your question? Proving an inclusion related to algebraic sets and interpreting it – KReiser Mar 13 '20 at 22:30
Is this not true if $k$ is not algebraically closed ? The solution of above link is under the assumption that $k$ is algebraically closed. – user371231 Mar 14 '20 at 10:15

For $Y_1,Y_2$ closed subsets of $\mathbb{A}_k^n$, $\mathcal{I}(Y_1 \cap Y_2)= \text{rad}(\mathcal{I}(Y_1)+\mathcal{I}(Y_2))$

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