# Ideals of subsets in $k[x_1,..,x_n]$

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 $$p \in X_1 \cap X_2$$.
• 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? – Manoel Apr 22 at 21:15