Why ideal corresponding to a algebraic curve is principal ideal

Question

Let $f(x,y)=p_1(x,y)^{e_1} \cdots p_n(x,y)^{e_n}$ be the factorization of a polynomial into irreducible factors over a algebraically closed field $k$. Why the ideal $I(Z(f))$ is generated by $p_1 \cdots p_2$?

Answer 1

That is because , by Hilbert's Nullstellensatz, $$I(Z(f))=\sqrt{(f)},$$ and the radical of an ideal is the intersection of the prime ideals which contain this ideal, and also the intersection of the prime ideals minimal above the ideal.

Now the minimal prime ideals above $(f)$ are $p_1,\dots,p_n$.