If a polynomial f is irreducible then (f) is radical, without unique factorization? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T16:58:34Z http://mathoverflow.net/feeds/question/57801 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57801/if-a-polynomial-f-is-irreducible-then-f-is-radical-without-unique-factorizatio If a polynomial f is irreducible then (f) is radical, without unique factorization? aglearner 2011-03-08T09:35:47Z 2011-03-13T04:14:57Z <p>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? </p> http://mathoverflow.net/questions/57801/if-a-polynomial-f-is-irreducible-then-f-is-radical-without-unique-factorizatio/57803#57803 Answer by Qiaochu Yuan for If a polynomial f is irreducible then (f) is radical, without unique factorization? Qiaochu Yuan 2011-03-08T09:51:48Z 2011-03-08T11:19:38Z <p>No, in the sense that this statement is false in a ring without unique factorization. For example, the element $2 + \sqrt{-5}$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$, and $9 \in (2 + \sqrt{-5})$ but $3 \not \in (2 + \sqrt{-5})$.</p> <p>(The lesson here is that irreducibility is not a useful idea in non-UFDs.)</p> http://mathoverflow.net/questions/57801/if-a-polynomial-f-is-irreducible-then-f-is-radical-without-unique-factorizatio/57866#57866 Answer by Pete L. Clark for If a polynomial f is irreducible then (f) is radical, without unique factorization? Pete L. Clark 2011-03-08T18:02:03Z 2011-03-09T06:43:14Z <p>This answer is directed at the comment that a proof that $R$ a UFD $\implies$ $R[t]$ a UFD takes two pages, i.e., one lecture two give.</p> <p>In $\S 15.5$ of <a href="http://www.math.uga.edu/~pete/integral.pdf" rel="nofollow">my commutative algebra notes</a> I give a "lemmaless" proof of this fact modelled after the Lindemann-Zermelo direct proof that $\mathbb{Z}$ is a UFD. This proof shows up several times in the literature; the earliest I found was in a 1950 paper of S. Borofsky (but I wouldn't be surprised to hear that Hasse and/or Zermelo knew it, for instance). The proof takes a little over half a page, including displayed equations. It does use the decomposition of the UFD property into "ACCP" (ascending chain condition on principal ideals) plus "EL" (irreducible elements are prime), but the same can be said about the classic proof that $\mathbb{Z}$ is a UFD. All in all I think this would be a short amount of class time to cover some very important results, thus a good investment. Certainly I would rather do this than try to give some independent proof of the much weaker result that irreducible elements in polynomial rings over fields generate radical ideals. </p> http://mathoverflow.net/questions/57801/if-a-polynomial-f-is-irreducible-then-f-is-radical-without-unique-factorizatio/57935#57935 Answer by Hailong Dao for If a polynomial f is irreducible then (f) is radical, without unique factorization? Hailong Dao 2011-03-09T09:39:04Z 2011-03-13T04:14:57Z <p>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. </p> <p>I believe the existence of an irreducible element whose ideal is not radical might be related to <em>non-trivial torsions</em> 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 </p> <p>$$xy=z^n$$ 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.</p> <p>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 $\mathbb Z/(2)$).</p> <p>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!)</p> <p>I hope this makes some sense. </p>