Is there a short way to prove that for each irreducible polynomial $f$ in $k[x_1,…,x_n]$ the principal ideal $(f)$ is radical without using unique factorization of polynomials? A short proof of this statement (contained, for example, in the "Primer on CA of Milne") uses the fact that polynomial ring is an UFD, but is it possible to give a reasonable proof without using this fact?
[Math] If a polynomial f is irreducible then (f) is radical, without unique factorization
ac.commutative-algebrapolynomials
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Best Answer
Here are some thoughts on why such a proof might be hard to find (and interesting). They are not totally rigorous, but I think they give a different perspective, so might be worth something.
I believe the existence of an irreducible element whose ideal is not radical might be related to non-trivial torsions in the class group (I will assume the ring is normal, one can avoid it by using Chow group of codimension one instead). Indeed, just from definition, you need such an element $x$, and some elements $y,z$ such that
$$xy=z^n$$ but $z\notin (x)$. Now, if $P=(x,z)$ happens to be prime and $y \notin P$, then by computing the Weil divisor corresponding to the Cartier divisor $(x)$, one gets $n[P]=0$ in the class group. But $[P]$ is not principal: if it is, it would have to be generated by $x$ because $x$ is irreducible, but $z\notin (x)$ by assumption.
This is precisely what happened in the examples by Qiaochu ($\mathbb Z[-\sqrt{5}]$) and Gerry ($k[x,y,z]/(xy-z^2)$) (both have class group $\mathbb Z/(2)$).
So it seems to me the proof you want would rule out certain torsions in the class group but without showing that the group is trivial (which means our ring is a UFD). Unfortunately, understanding torsions in $\text{Cl}(R)$, especially over arbitrary fields, is a harder problem (think about elliptic curves!)
I hope this makes some sense.