# о факториальности кольца многочленов

Эта запись будет совершенно непонятна вам, если у вас нет высшего образования по математике, извините.

Искал разные доказательства того, что R is UFD => R[x] is UFD, или русскими словами, кольцо многочленов над факториальным кольцом само факториально. Традиционное док-во (использующее лемму Гаусса и единственность разложения в многочленах над полем частных) мне очень не по душе, искал более прямые. Нашел несколько версий в lecture notes Пита Кларка и по ссылкам оттуда.

В итоге написал автору следующее письмо (по-английски, с некоторыми опущениями). Если настоящие математики хотят прокомментировать, буду рад любым замечаниям.

Having obsessed over various proofs of R UFD => R[x] UFD over the last few days, I reached the following conclusions:

1. The "traditional" proof seems very unpedagogic and unhelpful to me. One problem is its use of uniqueness of factorization in F[x]. It's uneconomical to appeal to a host of separate results, F[x] Euclidean => PID => UFD, that aren't really illuminating for the task at hand. But that's the lesser problem. My bigger problem with it is that it uses Gauss's Lemma twice, not once, in ways that are quite difficult for a student to catch. Once to establish that f(x) irreducible in R[x] => irreducible in F[x]. After that is done, and the non-constant factors in two factorizations are paired off up to F*, using uniqueness in F[x], we still need a second application of Gauss's Lemma to "clear off" the accumulated factor a/b \in F*, from all these pairings, and to show that in fact it's a unit in R. The strong impression I have left after consuming this proof is that it was a "trick" that didn't really show me what was going on.

2. Gauss's Lemma seems (almost) unescapable in all proofs, including putatively "direct" ones which usually turn out to use it covertly. This became much clearer to me after I internalized properly that it's trivially equivelant to "irreducible elements of R are prime in R[x]". In fact, given an R in which factorizations exist, the following are easily equivalent:

a) Irreducibles of R remain prime in R[x].
b) cont(fg) = cont(f)cont(g). In other words, "content is multiplicative".
c) f,g primitive => fg primitive. In other words, "content is not created from nothing".
d) If we write f = cont(f)*f~, f~ primitive, then from f | g we can conclude cont(f) | cont(g), f~ | g~. In other words, the content and the primitive part act separately as factors.

Note that nearly every proof of R[x] being UFD uses one of a)-d) one way or another. For example, your "lemmaless" proof on page 247 actually ends up using a). The proof via GCD properties on page 246 due to Haible still uses c).

Usually only b), c) are called Gauss's Lemma. But I think introducing a) explicitly as a variant of it is an excellent idea, because that makes it really easy to see Gauss's Lemma as a stepping stone towards proving R[x] UFD:

R[x] is UFD <==> every irreducible element is prime. So as a first step towards proving that, it makes very good sense to prove "every *constant* irreducible is prime". This is much more intuitive than c) which seems to have no relation to what we want to prove. Thus the most lucid proof from my perspective is this:

- show, trivially, that R[x] UFD <==> every irreducible element of R[x] is prime.
- start by proving that every irreducible element of R is prime in R[x], variant a) of Gauss's Lemma.
- introduce content, uniqueness up to units of the representation cont(f) f~, and variant d) of Gauss's Lemma that says the content and the primitive part act separately as factors.
- now prove that every f(x) irreducible non-constant is prime, by induction on degree of f(x), and for a given f(x) induction on sum of degrees of g,h, such that f(x) | g(x) h(x).
Given f | gh, if deg(g) < deg(f), take g' an irreducible factor of g, then g' | f s for some s(x), and by inductive hypothesis, either g' | f, impossible, or g' | s, then factor out g' from the original relation f | gh and get a lower deg(g)+deg(h).
Otherwise, if deg(g) >= deg(f), let f_n be the leading coefficient of f, then f | f_n*g*h, and f | (f_n*g - A*f) h, which is
of lower degree for a suitable A. So either f|h, or f | f_n*g, and then because by Gauss's Lemma content and primitive part factor separately, and f is primitive, in fact f | g.

This may be a bit long, but not longer than the traditional proof I think, and it's easy to see what exactly is going on.

Finally, of all the proofs I've seen, Borofsky's proof you reference in your notes is the only one that really avoids a use of Gauss's Lemma. Your "lemmaless" proof is similar to Borofsky, but the details are a bit different, and because of that in the end you need the lemma (in its a) form above) to separate out the constant and the primitive parts of factorizations. The difference is that you reduce the degree using the largest degree factor, while Borofsky uses the smallest degree factor (including constant if necessary). If proofs via Gauss's Lemma generally go

- prove that irreducibles of R are prime in R[x], or another equivalent formulation of Gauss's Lemma
- prove that non-constant irreducibles are prime, using induction on degree, or using factorisations in F[x], or GCD properties, or...

Borofsky's proof goes

- prove that all irreducibles are prime in R[x], by common induction on degree and length of content. Given factorizations f(x) = f_1*...*f_n = g_1*....*g_n, we use the smallest degree factor, say f_1, and find an associate in g_1*...g_n. If f_1 is constant, we find a g_i with a leading monomial that f_1 divides, and reduce g_i by a multiple of f_1. By contrast with Gauss's Lemma, which would firmly find a *constant* associate g_i for us in this case, this does not necessarily home in on a constant g_i, but when g_i is non-constant, it still lowers the degree. It may make sense to adjust your proof along these lines to make it, well, I guess, more truly "lemmaless". I don't know if this is important enough to do, just a thought.
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