Ideal of finite intersection of algebraic sets

Question

In general if $X_1$ and $X_2$ are two algebraic sets on $k^n$ with $k$ a field of characteristic zero, we have that $I( X_1 \cap X_2 ) = \sqrt{ I(X_1) + I(X_2) }.$ Is posible in general compute $I(X_1 \cap X_2 \cap \dots \cap X_n)$ in terms of $I(X_1), \dots , I(x_n)$?.

Thanks in advance.

Why do you assume characteristic 0? This is not necessary. – Martin Brandenburg May 22 '13 at 22:49 — Martin Brandenburg, May 22 '13 at 22:49

score 7 · Answer 1 · answered May 22 '13 at 22:51

7

In general we have $\sqrt{I+\sqrt{J}}=\sqrt{I+J}$, since both ideals are the smallest radical ideals containing $I$ and $J$. It follows by induction that $I(X_1 \cap \dotsc \cap X_n)=\sqrt{I(X_1) + \dotsc + I(X_n)}$.

answered May 22 '13 at 22:51

Martin Brandenburg

163,620

score 6 · Accepted Answer · answered May 22 '13 at 22:50

Remember that for any ring $R$ and ideal $J\subseteq R$, $$\sqrt{J}=\bigcap_{\large\substack{\text{prime ideals}\\ P\supseteq J}} \!\!\!\!\!\!\!P.$$ Also, remember that for ideals $J_1,J_2,J_3\subseteq R$, we have $$J_1\supseteq J_2+J_3\iff J_1\supseteq J_2\quad\text{and}\quad J_1\supseteq J_3.$$ Use this to show that for any ideals $J_1,J_2$ of $R$, $$\sqrt{J_1+\sqrt{J_2}}=\sqrt{J_1+J_2}$$ so that you can use induction to prove what you want.

Using prime ideals is a total overkill :). – Martin Brandenburg May 22 '13 at 22:50 — Martin Brandenburg, May 22 '13 at 22:50

Ideal of finite intersection of algebraic sets

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