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?
$\begingroup$
$\endgroup$
8
-
3$\begingroup$ Why are you so intent on avoiding the fact that polynomial rings are UFDs? $\endgroup$– Qiaochu YuanCommented Mar 8, 2011 at 9:40
-
4$\begingroup$ If that was your motivation, you should have said something to that effect in your question. In any case, I feel like undergraduates ought to know that polynomial rings are UFDs anyway. $\endgroup$– Qiaochu YuanCommented Mar 8, 2011 at 10:29
-
8$\begingroup$ Dear agleaner, I strongly recommend proving the UFD property for polynomial rings. Remember that your undergraduates will have been factoring polynomials since they were 11 or 12 years old, so their intuition behind unique factorization should be pretty solid. Also, later on, if you want to discuss examples like hypersurfaces and so on, you will be rather hamstrung if you don't have the UFD property to help you. Regards, Matthew $\endgroup$– EmertonCommented Mar 8, 2011 at 12:59
-
12$\begingroup$ I am growing annoyed by the tendency to answer questions like "how to prove X with Y" with "why don't you like Y?". There can be lots of reasons for this, beginning with "the classical proof of Y is non-constructive", "the proof of Y does not generalize as far as I want to generalize X", "I hate the author of Y" or simply "I was not able to come up with a proof of Y on my own and now I have sworn never to use Y". This is all irrelevant. Mathematics has always profited from having various proofs for one and the same result. $\endgroup$– darij grinbergCommented Mar 8, 2011 at 18:59
-
4$\begingroup$ @darij: yes, but I would strongly prefer that the OP give at least one such reason. In this case I see no benefit in seeking an alternate proof and IMO the burden of proof is on the OP to suggest a reason to believe that there is one. $\endgroup$– Qiaochu YuanCommented Mar 8, 2011 at 19:18
|
Show 3 more comments
3 Answers
$\begingroup$
$\endgroup$
1
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.
answered Mar 9, 2011 at 9:39
-
$\begingroup$ Thanks a lot, I have to play a bit with what you say to understand better, but it is interesting! $\endgroup$ Commented Mar 9, 2011 at 13:51
$\begingroup$
$\endgroup$
1
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.
In $\S 15.5$ of my commutative algebra notes 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.
answered Mar 8, 2011 at 18:02
-
$\begingroup$ Hasse sketched a proof a la Zermelo in his Vorlesungen über Zahlentheorie, which appeared in 1949 but was written in the late 1930s. This is not the number theory book that was translated into English. $\endgroup$ Commented Mar 9, 2011 at 10:47
$\begingroup$
$\endgroup$
11
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 $3 \not \in (2 + \sqrt{-5})$.
(The lesson here is that irreducibility is not a useful idea in non-UFDs.)
answered Mar 8, 2011 at 9:51
-
12$\begingroup$ Is the goal an example with polynomials? In the ring $k[x,y,z]/(xy-z^2)$, $y$ is irreducible, $z^2$ is in $(y)$, but $z$ is not in $(y)$. $\endgroup$ Commented Mar 8, 2011 at 11:53
-
10$\begingroup$ Re irreducibility is not a useful idea in non-UFDs: UFD is a much stronger condition than necessary. The statement follows if every irreducible element is prime; this holds for example in every GCD domain, and more generally, in every pre-Schreier domain. $\endgroup$ Commented Mar 8, 2011 at 12:52
-
5$\begingroup$ Irreducible elements in arbitrary domains are a huge topic of contemporary research among commutative algebraists. Because they behave in complicated ways, they are interesting to study (at least for some people). Just look up the notion of elasticity of a domain to get an idea of how much has been done here. $\endgroup$ Commented Mar 8, 2011 at 17:51
-
2$\begingroup$ Sorry to insist on this, but I am one of the 'some people' Pete L. Clark mentioned. Actually, one could even say that the notion 'irreducible' is only useful in non-UFDs, since in UFDs it coincides with the notion 'prime' and thus it is redundant. Teaching at least the differene between the two notions is I think already useful, and can be done in a short way in an elementary number theory context; for example, by considering all positive integers congruent $1$ modulo $4$ with multiplication (sometimes called Hilbert semigroup, as he is said to have used it as instructional example). $\endgroup$– user9072Commented Mar 9, 2011 at 11:43
-
4$\begingroup$ @Qiaochu: by the way, your first example $k[x^2,xy,y^2]$ works, just take $(x^2)$. In fact, it is the same as Gerry's example. $\endgroup$ Commented Mar 9, 2011 at 11:49