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.
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 $p \in X_1 \cap X_2$.