Edit: I should point out that I'm working over an algebraically closed field $k$.
Let $X_1,X_2\subset\mathbb{A}^n$ be affine algebraic sets. Show that $I(X_1\cap X_2)=\sqrt{I(X_1)+I(X_2)}$. Show by example that taking the radical here is necessary. Can you see geometrically what it means if $I(X_1\cap X_2)\neq I(X_1)+I(X_2)$?
I showed this using that for ideals $\mathfrak{a}_1,\mathfrak{a}_2$ in a ring, $\sqrt{\mathfrak{a}_1+\mathfrak{a}_2}=\sqrt{\sqrt{\mathfrak{a}_1}+\sqrt{\mathfrak{a}_2}}$. As for an example, I took $X_1=V(x)$, $X_2=V(x+y^2)$ in $\mathbb{A}^2$. Then $I(X_1)+I(X_2)=\langle x\rangle+\langle x+y^2\rangle=\langle x,y^2\rangle,$ which is not radical. But I don't know exactly how to interpret that geometrically.
The intersection of those two varieties is just the origin. I would have supposed that it has something to do with the origin occurring "more than once" (read from the ideal $\langle x,y^2\rangle$), but I'm not sure why that is the case. Is it because the variety $X_1$ is a tangent to $X_2$ at the origin, and thus it counts as a "double" intersection? Then I don't understand it, because the point $0$ occurs only once in $X_1$, so why twice in the intersection? And what is the difference - geometrically - between $\langle x,y^2\rangle$ and $\langle x^2,y\rangle$?
Thank you very much for helping me understand these things!