Denote by $|\cdot |$ the $p$-adic norm, and let $$\mathbb Z_p \{z\}=\left\{\sum_{n=0}^\infty a_nz^n;\ a_n\in \mathbb Z_p ,\ |a_n|\underset {n\rightarrow \infty} {\longrightarrow 0} \right\}$$ the ring of converging power series over the p-adic integers with indeterminate $z$. I need to show that this ring is a unique factorization domain.

What I know is that $\ \mathbb Q_p\{z\}$ is a principal ideal domain, and in fact every element $\ f\in\mathbb Q_p\{z\}$ has a unique representation $f=qu$ where $\ q\in\mathbb Q_p[z],\ u\in \mathbb Q_p\{z\}^\times$. Another thing is that for archimedean valuations in general this isn't true. For example, the ring of converging power series over $\ \mathbb C$ with the usual absolute value is not a UFD, as $\sin z$ which is holomorphic in the unit disc don't have a unique representation as finite product of irreducibles (the Laurent series of $\sin z$ around the origin is an infinite product of irreducibles).
I think this is supposed to be true for $\mathbb Z_p\{z\}$, and the fact that $\mathbb Q_p\{z\}$ is a PID is crucial for the proof, but I haven't been able to see how.

  • $\begingroup$ I’d be happy to discuss this via e-mail. $\endgroup$ – Lubin Jul 2 '15 at 13:06

I can only give an inefficient answer; others will do much better.

I hope you know about and are thinking of Newton Polygon. Remember that it’s described by using the additive valuation on $\Bbb Q_p$, $v(p)=1$, $v(ab)=v(a)+v(b)$, $v(a+b)\ge\min(v(a),v(b))$, $v(0)=\infty$. For $f=\sum_ia_ix^i$, you plot all points $(i,v(a_i))$ in the plane, erect a vertical halfline above each of these, and take the convex hull of the union of these lines.

Now, an irreducible $\Bbb Q_p$-polynomial has a polygon consisting of only one nonvertical segment. Same for irreducible $\Bbb Z_p$-polynomials. But the irreducible polynomial $1-px$, whose polygon has only the two vertices $(0,0)$ and $(1,1)$, is not irreducible in $\Bbb Z_p\{x\}$, because it’s a unit, with reciprocal $1+px+p^2x^2+\cdots\,$.

You should be able to prove, from what you have been given, that the only irreducibles in $\Bbb Z_p\{x\}$ are the irreducible $\Bbb Z_p$-polynomials whose polygon has a single nonvertical segment, of nonpositive slope. The rest I leave to you.

(Maybe I should add that the Polygon is not an essential port of the proof here; in this case, it just concentrates the mind wonderfully.)

  • $\begingroup$ Thank you for your answer. I wasn't thinking about Newton Polygon, actually this question came from a topic related to galois theory. I read a little bit about Newton Polygon today, and I think that I may be able to prove what you wrote. So currently my idea is to prove the question using Weierstrass preparation theorem, and what you wrote here. I still have difficulty to prove that if an element $u$ is invertible in $\mathbb Q_p \{x\}$ then it is irreducible $\mathbb Z_p \{x\}$. $\endgroup$ – giladude Jul 2 '15 at 12:37
  • $\begingroup$ It’s certainly not true that invertibility in $\Bbb Q_p\{x\}$ implies irreducibility in $\Bbb Z_p\{x\}$. Maybe you meant “irreducible in $\Bbb Q_p\{x\}$”? On another topic, I think I may have missed the point somewhat in my answer. Without thinking very much on the topic, it seemed to me last night that the irreducibles in the two rings should be the same, and your desired result should drop out fairly easily from the information that you have in hand. $\endgroup$ – Lubin Jul 2 '15 at 12:55

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