$\sqrt{I}=(f) \Rightarrow I=(f^n)$? (in a polynomial ring over a field)

Question

EDIT: imtrying46's answer in the comments is simpler than the one in the accepted answer (but both are good, and quite similar).

This is a proof-verification question. If I'm correct, I still wonder if there are even more elementary proofs (see below), and whether I really never used assumption (4) as it seems to me.

Assumptions:

1. $K$ is a field.
2. $I$ is an ideal of the polynomial ring $K[x_1,\dotsc,x_n]$.
3. The radical $\sqrt{I}$ of $I$ is generated by a single polynomial $f$.
4. $f$ is irreducible over the algebraic closure of $K$.

My question:

Is $I=(f^n)$ for some $n\geq 1$?

My attempt:

I think the answer is Yes.

Take the maximal $n$ such that $I\subset(f^n)$ ($n\geq 1$ since $I\subset\sqrt{I}=(f)$). I'll prove that $I=(f^n)$. Write $I=(f^nh_1,\dotsc,f^nh_r)$. Then (using NSZ),

$(f)=\sqrt{I}=I(V(f^nh_1,\dotsc,f^nh_r))=I(V(f^n)\cup V(h_1,\dotsc,h_r))= I(V(f^n))\cap I(V(h_1,\dotsc,h_r))=(f)\cap\sqrt{(h_1,\dotsc,h_r)}$.

In particular $f^\ell\in (h_1,\dotsc,h_r)$ for some $\ell\geq 0$. The maximality of $n$ implies that $\ell=0$.

So $1\in (h_1,\dotsc,h_r)$ and thus $f^n \in (f^nh_1,\dotsc,f^nh_r)=I$ as needed.

I think I'm probably correct, but I wonder if there's a simpler proof. In particular, can we avoid the use of the Nullstellensatz? Maybe even avoid Hilbert's Basis Theorem? Also, I think I never used the irreducibility of $f$, right?

Answer 1

This is not true. Take $((x-y)^2x,(x+y)x)\subset k[x,y]$, for instance. This cannot be generated by a single polynomial (if it were, it would have to divide $x$, but neither $1$ nor $x$ are in the ideal). Then $$\sqrt{((x-y)^2x,(x+y)x)}=\sqrt{((x-y)x,(x+y)x)} = \sqrt{(x^2-xy,x^2+xy)} = \sqrt{(x^2,xy)}=(x).$$