Let $p$ be prime and $(a,b)\in \mathbb{Z}^2$. Prove $I=\{f(X,Y)| f(a,b)\equiv 0 \mod p\}$ can be generated by three explicit elements.
Hilbert's Theorem tells me that $I$ must really be finitely generated because $\mathbb{Z}$ is principal, hence Noetherian.
However, the proof of Hilbert's Theorem asks for too many polynomials in order to conclude this, so I believe we must not simply copy Hilbert's Theorem's proof.
In the case, $a$ and be $b$ are congruent to $0$, I was able to show $I=(p)+(X)+(Y)$.
In the case $a\equiv0$ and $b$ is invertible, I have $(p)+(x)\subseteq I$, but there are certain elements which are not spanned by these two...
Finally, for the case both of them are invertible I am clueless.
Clearly $(p)+(X-a)+(Y-b) \subseteq I.$ For the other direction, suppose $f \in I$ and write $f(X,Y) = \sum\limits c_{ij} (X-a)^i (Y-b)^j$ for some coefficients $c_{ij}.$