ideals and variety

Question

I am trying to prove that $Z(\bigcap_{\lambda\in\Lambda}I_\lambda)\subset \bigcup_{\lambda\in\Lambda} Z(I_\lambda)$ if $\Lambda$ is a finite index set.

$\{I_{\lambda}\}$ is a set of ideals of polynomials over $k[x_1,\cdots,x_n]$. $Z(I)=\{a\in\mathbb{A}^n: f(a)=0 \text{ for all } f \in I\}$ . I know that if $\Lambda$ is infinite, this statement is false, because it is very easy to show the other direction "$\supseteq$" even in infinite case, and $Z(\bigcap_{\lambda\in\Lambda}I_\lambda)$ is a closed set, while $\bigcup_{\lambda\in\Lambda} Z(I_\lambda)$ is not necessarily closed. Can anyone help me? Thanks.

Answer 1

By induction, we only have to prove $Z(I\cap J)\subset Z(I)\cup Z(J)$. If p is not contained in the right, then there exist $f\in I, g\in J$ such that $f(p)\neq 0, g(p)\neq 0$. $fg\in I\cap J$, but $fg$ does not vanish at p, which means $p\notin Z(I\cap J)$.

Answer 2

If I understand correctly your question, you are trying to show that, if a prime $p$ contains $\displaystyle\bigcap_{\lambda\in\Lambda}I_\lambda$, then it contains one of $I_\lambda$.

Suppose $p$ does not contain $I_\lambda$ for any $\lambda\in\Lambda$ except for $\lambda_0$, then we show that $p$ contains $I_{\lambda_0}$.

For $\lambda\ne\lambda_0$, by our assumption there exists $x_\lambda\in I_\lambda\setminus p$. Then for any $x\in I_{\lambda_0}$, $$x_{\lambda_0}\cdot\prod_{\lambda\ne\lambda_0}x_\lambda\in\prod_{\lambda\in\Lambda}I_\lambda\subseteq\bigcap_{\lambda\in\Lambda}I_\lambda\subseteq p,$$ but $\prod_{\lambda\ne\lambda_0}x_\lambda\not\in p$. Thus $x_{\lambda_0}\in p$ as desired.

Notice that $\prod_{\lambda\ne\lambda_0}x_\lambda\not\in p$ because $\Lambda$ is finite, so this proof does not apply to infinite $\Lambda$, as expected.

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Hope this helps.