# Does $A$ a UFD imply that $A[T]$ is also a UFD?

I'm trying to prove that $A$ a UFD implies that $A[T]$ is a UFD.

The only thing I am sure I could try to use is Gauss's lemma.

Also, how can we deduce that the polynomial rings $\mathbb{Z}[x_1,\ldots,x_n]$ and $k[x_1,\ldots,x_n]$ are UFDs?

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Yes (and the proof uses Gauss's Lemma), and induction on $n$. –  Arturo Magidin May 28 '12 at 20:47

You should know that if $k$ is a field, then $k[x]$ is a UFD (even a Euclidean domain). So given $g(x)\in A[x]$, first write it as $g(x) = cG(x)$, where $c$ is a constant and $G(x)$ is primitive. Then show that a primitive polynomial in $A[x]$ is irreducible if and only if it is irreducible when viewed as a polynomial in $k[x]$, where $k$ is the field of fractions of $A$. Then use this to take an arbitrary polynomial in $A[x]$, and factor it by "factoring out the content, then factoring it over $k[x]$, and then "lifting the factorization" back to $A[x]$. The argument is exactly the same as that used in the case of $\mathbb{Z}[x]$ with Gauss's Lemma.

Now, since $\mathbb{Z}$ is a UFD, then by this argument so is $\mathbb{Z}[x_1]$; which means that so is $\mathbb{Z}[x_1][x_2]\cong\mathbb{Z}[x_1,x_2]$. Which means that so is $\mathbb{Z}[x_1,x_2][x_3] = \mathbb{Z}[x_1,x_2,x_3]$. Etc. The same holds for polynomials over a field, since a field is trivially a UFD (or because you know that the first polynomial ring, $k[x_1]$, is a UFD).

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There is a slick general way to do this by localization (usually credited to Nagata). $\$ Suppose that $\rm\:D\:$ is an atomic domain, i.e. nonzero nonunits factor into atoms (irreducibles). If $\rm\:S\:$ is a saturated submonoid of $\rm\:D^*$ generated by primes, then $\rm\: D_S$ UFD $\rm\:\Rightarrow\:D$ UFD. $\$ This yields said slick proof of $\rm\:D$ UFD $\rm\Rightarrow D[x]$ UFD, viz. $\rm\:S = D^*\:$ is generated by primes, so localizing yields the UFD $\rm\:F[x],\:$ $\rm\:F =\:$ fraction field of $\rm\:D.\:$ Therefore $\rm\:D[x]\:$ is a UFD, by Nagata. This yields a more conceptual, more structural view of the essence of the matter (vs. traditional argument by Gauss' Lemma). Moreover, the proof generalizes to rings related to GCD rings, e.g. Riesz/Schreier rings, which provide refinement-based views of UFDs (essential for noncommutative generalizations).

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The fact that $A$ is a UFD implies that $A[X]$ is a UFD is very standard and can be found in any textbook on Algebra (for example, it is Proposition 2.9.5 in these notes by Robert Ash). By induction, it now follows that $A[X_1,\ldots,X_n]$ is a UFD for all $n\geq 1$.

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Can you please reproduce the proof here for the sake of completeness? –  James R. May 28 '12 at 20:51
@JamesR. I thought you were trying to prove it. It's a good exercise to do so yourself, instead of asking someone to do it for you; especially such a standard result. –  Arturo Magidin May 28 '12 at 20:54
I just wanted to make sure that I got it down, the concept. I might have messed up on the induction stuff, and wanted to be sure –  James R. May 28 '12 at 20:55
@JamesR. Then say so, instead of facile statements like "for the sake of completeness". You can post your solution (as an answer, even) and ask people to check it for you. –  Arturo Magidin May 28 '12 at 20:57
@JamesR.If you want to check your proof of the result, the above link should be sufficient. And by the way, this is taken from the second chapter of a wonderful set of notes on abstract algebra. I recommend it! –  M Turgeon May 28 '12 at 21:04