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Suppose that $k$ is algebraically closed field and $X_1, X_2 \subset \mathbb{A}^n_k$. Show that $$ I(X_1 \cap X_2)=\sqrt{I(X_1)+I(X_2)}. $$

I thought the first thing to do was to use the Nullstellensatz Hilbert to vanish with that square root:

$$ \sqrt{I(X_1)+I(X_2)}= I(Z(\sqrt{I(X_1)+I(X_2)}))= I(Z(I(X_1)+I(X_2))). $$

I thought it was easier to show that $$ I(X_1 \cap X_2)=I(Z(I(X_1)+I(X_2))). $$ But I could not get out of here. Maybe it was a bad start. Does anyone suggest anything to me? A new beginning or how to get out of that point.

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It's enough to show that $ X_1 \cap X_2 = Z(I(X_1) + I(X_2)).$

  • Suppose $p \in X_1 \cap X_2$. Any element $f \in I(X_1) + I(X_2)$ takes the form $f = g + h$, where $g \in I(X_1)$ and $h \in I(X_2)$. So $f(p) = g(p) + h(p) = 0$. Hence $p \in Z(I(X_1) + I(X_2))$.

  • Suppose $p \in Z(I(X_1) + I(X_2))$. Any $g \in I(X_1)$ is also in $I(X_1) + I(X_2)$, so $g(p) = 0$ for all $g \in I(X_1)$; hence $p \in Z(I(X_1)) = X_1$. By a similar argument, $p \in X_2$. So $p \in X_1 \cap X_2$.

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  • $\begingroup$ For me, it was a great help. But I understand that you have used $X_1$ and $X_2$ are algebraic sets to conclude that $ Z(I(X_1))=X_1$ and $Z(I(X_2))=X_2$. Could we solve for $X_1$ and $X_2$ as sets of points only? $\endgroup$ – Manoel Apr 22 at 21:15
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    $\begingroup$ @Manoel Not sure off the top of my head. Unfortunately I have to go - perhaps someone else can help? $\endgroup$ – Kenny Wong Apr 22 at 21:19

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