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Let A = $k[x_1, x_2, \ldots, x_n]$ and let $I_{\lambda}$ be an ideal of A.

Let J = $\sum_{\lambda \in \Lambda} I_{\lambda}$ be a finite sum. Show that $V(J) = \cap_{\lambda \in \Lambda} V(I_{\lambda})$.

It seems quite obvious that $V(\sum I_{\lambda}) \subseteq \cap I_{\lambda} $.

However, I am not at all sure where to start for showing the other inclusion. Ideas?

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2 Answers 2

Suppose $x\in \cap_{\lambda\in\Lambda}V(I_\lambda)$, so $x$ vanishes on every polynomial $f_\lambda\in I_\lambda$ for any $\lambda\in\Lambda$. Let $f\in\sum_{\lambda\in\Lambda I_\lambda}$, so $f=\sum_{\lambda\in\Lambda} f_\lambda$ for some collection of polynomials $f_\lambda\in I_\lambda$. Then $f(x)=\sum_{\lambda\in\Lambda} f_\lambda(x)=\sum_{\lambda\in\Lambda}0=0$ so $x\in V(\sum_{\lambda\in\Lambda} I_\lambda)$. This proves $\cap_{\lambda\in\Lambda}V(I_\lambda)\subseteq V(\sum_{\lambda\in\Lambda} I_\lambda)$.

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Let $f\in J$, then $f=\sum_\lambda f_\lambda$, where the sum is finite and $f_\lambda\in I_\lambda$. Then, if $x\in \cap_\lambda V(I_\lambda)$, we have $$ f(x) = \sum_\lambda f_\lambda(x)=0 $$ since $x\in V(I_\lambda)$ for all $\lambda$, consequently $x\in V(J)$.

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