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An exercise in my algebra course book asks to prove that if $R$ is a PID, then $R[[x]]$ is a UFD, where $R[[x]]$ is the ring of formal power series over $R$.
After some failed attempts at proving the ACC I visited Wikipedia, which comments:

If $R$ is Noetherian, then so is $R[[x]]$; this is a version of the Hilbert basis theorem.

This is very useful, as $R$ is a PID and hence Noetherian. Unfortunately we only saw (without proof) Hilbert's basis theorem in the form

If $R$ is Noetherian, then so is $R[x]$.

I'm not sure how to conclude the Noetherianity of $R[[x]]$ from this. I know that $R[x]$ is a UFD because $R$ is (we saw this without proof), and have been trying to conclude the ACC in $R[[x]]$ from the ACC in $R[x]$, without success.
I can prove the ACC if $R$ is a field, because then every $a_kx^k+a_{k+1}x^{k+1}+\cdots$ with $a_k\neq0$ is associate with $x^k$, hence every ideal is of the form $(x^k)$. (In fact this readily proves that $R[[x]]$ is a UFD.) If $R$ is not a field, say it has some non-invertible element $r$, then there are many more ideals such as $(r)$ or $(r+x)$.

I'm also facing difficulties at identifying the irreducible elements of $R[[x]]$. (I wish to prove that they are all primes in order to conclude the uniqueness of factorisation.) I have figured that they either take the form $ux+Px^2$ for some unit $u\in R^\times$ and some $P\in R[[x]]$ or have a non-zero constant term which is not invertible in $R$. The elements of the form $ux+Px^2$ are indeed primes, the difficulty lies in those with non-zero constant term.

Any hints at proving $R[[x]]$ is a UFD?

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  • $\begingroup$ You can prove that if $R$ is noetherian, then $R[[x]]$ is noetherian using the same strategy that one would use to prove the Hilbert basis theorem (of which there is a fairly short proof in Atiyah & MacDonald's "Commutative Algebra"). $\endgroup$ – msteve Jan 8 '15 at 14:52
  • $\begingroup$ I don't think we're supposed to do it that way because we didn't see a proof of Hilbert's basis theorem. There should be another way. $\endgroup$ – punctured dusk Jan 8 '15 at 14:53
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    $\begingroup$ Have you found an irreducible in $\Bbb Z[[x]]$ that’s not a unit times an irreducible from $\Bbb Z[x]$? $\endgroup$ – Lubin Jan 8 '15 at 14:55
  • $\begingroup$ @Lubin Interesting question. I haven't yet. Proving this would show the ACC. $\endgroup$ – punctured dusk Jan 8 '15 at 15:01
  • $\begingroup$ In fact we dont have to find them all, we only have to show that every irreducible is prime. $\endgroup$ – punctured dusk Jan 16 '15 at 13:18
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Another answer to the problem given by the OP:

If $R$ is a PID, then $R[[X]]$ is a UFD.

The key to prove the above result is to use a theorem known as Kaplansky's criterion for UFDs:

Theorem: An integral domain $D$ is a UFD if only if every nonzero prime ideal of $D$ contains a prime element.

I'm not going to write all the details because I don't want to write a long post. In order to use Kaplansky's criterion we need the following lemma:

Lemma: Let $A$ be a commutative unitary ring. Let $\phi\colon A[[X]]\rightarrow A$ be the map given by $$f=\sum_{i=0}^{\infty}a_iX^i\mapsto a_0.$$ Then $\phi$ is a surjective ring homomorphism. Moreover if $\mathfrak{P}$ is a prime ideal of $A[[X]]$, then $\mathfrak{P}$ is finitely generated in $A[[X]]$ if only if $\phi(\mathfrak{P})$ is finitely generated in $A$.

A proof of the above lemma can be found in Sharp's book "Steps in Commutative Algebra" or in Watkins' book "Topics in Commutative Ring Theory". In such proofs it's shown the following:

If $X\in \mathfrak{P}$, then $\mathfrak{P}$ is generated by the same elements that generate $\phi(\mathfrak{P})$ plus $X$; and if $X\notin \mathfrak{P}$, then the number of generators of $\mathfrak{P}$ is the same as that of $\phi(\mathfrak{P})$.

Finally, to prove the theorem given at the beginning of the post we proceed as follows. We first note that $X$ is a prime element of $R[[X]]$. Next, given a nonzero prime ideal $\mathfrak{P}$, if $X\in \mathfrak{P}$, then $\mathfrak{P}$ contains the prime element $X$. Otherwise, by the lemma $\mathfrak{P}$ is f.g. with the same number of generators as the ideal $\phi(\mathfrak{P})$, but $R$ being a PID implies that $\phi(\mathfrak{P})$ is principal, so $\mathfrak{P}$ is principal too, i.e., there is some $f\in R[[X]]$ such that $\mathfrak{P}=(f)$. Since $\mathfrak{P}$ is a prime ideal is easy to see that $f$ is a prime element, so $\mathfrak{P}$ contains the prime $f$. Hence, by Kaplansky's criterion $R[[X]]$ is a UFD.

As a side remark, the lemma that we've used is also applied (together with Cohen's theorem) to prove the analogue of Hilbert basis' theorem for formal power series.

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The set of irreducible elements of $\mathbb{Z}[[x]]$ is not known, and is more complicated as it seems.

You might find some interesting families of irreducible elements in this paper :

http://www.ijma.info/index.php/ijma/article/view/2357

However, it does not answer the original question.

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To show that $R[[X]]$ is a UFD you can use Theorem 20.3 in Commutative Ring Theory by Matsumura which states that if $A$ is a regular UFD so is $A[[X]]$ and show that every PID is regular which is rather easy.

See also this paper: http://people.brandeis.edu/~buchsbau/miscpapers/10052.pdf

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    $\begingroup$ Given that the OP has not even seen the proof the Hilbert's basis theorem, using the concept of regularity is quite likely to not be helpful... $\endgroup$ – Mariano Suárez-Álvarez Jul 10 '17 at 23:46

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