Nullstellenzats

The other day I was trying to explain the Nullstellenzats theorem to my mother. It was somewhat hard because she had never taken a course in mathematics, and it was even worse because I'm still having problems understanding and using this theorem LOL, but anyway I managed to draw a simple picture which illustrates it: $\hspace{100mm}$

I explained to her the next nulltellensatz version: Let $k$ be an algebraically closed field, and let $I$ be an ideal of the ring of polynomials $k[x_{1},\dots,x_{n}]$. If a polynomial $f\in{k[x_{1},\dots,x_{n}]}$ is such that $f(\alpha)=0$ for all $\alpha\in{Z(I)}$, then $f\in{\sqrt{I}}$, where $Z(I)=\{\beta\in{k}\;|\;g(\beta)=0\;\forall\;g\in{I}\}$. It's simple: We draw two bags. The first bag on the left hand side represents $I$, and each point inside it represents an element $f\in{I}$. The second bag represents $Z(I)$, and each point inside it represents an element $\beta\in{Z(I)}$. An arrow that goes from a point belonging to the first bag (i.e., an element $f\in{I}$) to a point belonging to the second bag (i.e., $\beta\in{Z(I)}$) indicates that the first point "vanishes" on the second point (i.e., $f(\beta)=0$). From the hypothesis, we find that there is arrow from each point belonging to the first bag to a fixed point belonging to the second bag, and this happens for each fixed point in the second bag. We also have a point outside the first bag (i.e., the polynomial $f$) which vanishes on every point belonging to the second bag (so, there's an arrow that goes from the point $f$ to each point inside the second bag). So, Nullstellezats basically says that somehow $f$ "belongs to" the first bag, namely, choosing an appropriate positive integer $n$ such that $f^{n}\in{I}$. And finally, the white background of the picture reminds us that we can do all this only if $k$ is algebraically closed :D