Vtyjkyuk

I am trying to prove that $Z(bigcap_lambdainLambdaI_lambda)subset bigcup_lambdainLambda Z(I_lambda)$ if $Lambda$ is a finite index set.

$I_lambda$ is a set of ideals of polynomials over $k[x_1,cdots,x_n]$. $Z(I)=ainmathbbA^n: f(a)=0 text for all f in I$ .
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 $Z(bigcap_lambdainLambdaI_lambda)$ is a closed set, while $bigcup_lambdainLambda Z(I_lambda)$ is not necessarily closed.
Can anyone help me?
Thanks.

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

If I understand correctly your question, you are trying to show that, if a prime $p$ contains $displaystylebigcap_lambdainLambdaI_lambda$, then it contains one of $I_lambda$.

Suppose $p$ does not contain $I_lambda$ for any $lambdainLambda$ except for $lambda_0$, then we show that $p$ contains $I_lambda_0$.

For $lambdanelambda_0$, by our assumption there exists $x_lambdain I_lambdasetminus p$.
Then for any $xin I_lambda_0$, $$x_lambda_0cdotprod_lambdanelambda_0x_lambdainprod_lambdainLambdaI_lambdasubseteqbigcap_lambdainLambdaI_lambdasubseteq p,$$ but $prod_lambdanelambda_0x_lambdanotin p$. Thus $x_lambda_0in p$ as desired.

Notice that $prod_lambdanelambda_0x_lambdanotin p$ because $Lambda$ is finite, so this proof does not apply to infinite $Lambda$, as expected.

Hope this helps.