Timeline for Factorizing polynomials in $\mathbf{Z}[[x]]$
Current License: CC BY-SA 2.5
7 events
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| Feb 25, 2011 at 10:43 | comment | added | Tommaso Centeleghe | @David. I will read more carefully your lemmas as soon as I have the chance. Just a comment, $x^2-6$ does indeed factor into the product of two power series $f_2$ and $f_3$, where $f_i$ has constant term equal to $i$ (use the fact that $(2,3)$ are relatively prime and inductively construct the factorization). One has that the power series $f_i$ is "satisfied" by $\sqrt{6}$ in $\mathbf{Q}_i$. | |
| Feb 25, 2011 at 2:29 | comment | added | François Brunault | Sorry, in the comment above I meant : $u \in \mathbb{Z}_{(p)}[[X]]^{\times} \cap \mathbb{Z}[[X]]$. | |
| Feb 25, 2011 at 2:26 | comment | added | François Brunault | @ David : Letting $\mathbb{Z}_{(p)}=\mathbb{Z}_p \cap \mathbb{Q}$, it seems that you prove that $f = (\prod h_i) \cdot u$ for some $u \in \mathbf{Z}[[x]]$. The constant term of $h_i$ is necessarily a power of $p$ (quotient everything by $x$). So $u(0)$ is the prime-to-$p$-part of $f(0)$. It seems that doing this for all $p$ gives the result, or am I missing something? | |
| Feb 24, 2011 at 20:27 | comment | added | Pace Nielsen | I think so too. You might be able to patch it up by inverting the primes in the constant coefficient different from the prime you are considering, which one should intuitively be able to do using elements in $\mathbb{Z}[[x]]$. But it gets complicated, because the image of $\varphi_{\alpha,\nu}$ can be weird. (Try $\alpha=3+3\sqrt{2}$.) | |
| Feb 24, 2011 at 20:22 | comment | added | David E Speyer | I don't have time to think about this right now, but you raise a good point. Also, I wrote above that $\mathbb{Z}_p \cap \mathbb{Q} = \mathbb{Z}$ but of course that isn't true. So it looks like what I have shown is the required statement with $\mathbb{Z}$ replaced by $\mathbb{Z}_p \cap \mathbb{Q}$. I think there is something valuable here though; please feel free to improve this. | |
| Feb 24, 2011 at 20:06 | comment | added | Pace Nielsen | David, I don't think $x-6$ is a counter-example (I tried working that example out myself). The issue is that x-6 factors in the power series ring over the integers. If you identify x with 6, then as a 2-adic integer x looks like $0*1+1*2+1*2^{2}+0*2^{3}+\cdots$. Thus, $x^{n}=0*1+0*2+\cdots + 0*2^{n-1}+1*2^{n}+\cdots$ has its first nonzero term as a coefficient of $2^{n}$. In particular, we can form a power series $2+x+x^3+x^5+x^6+...$ which maps to zero in $\mathbb{Q}_{2}$ under the identification $x=6$. (We choose powers of $x$ so that we can use $2$ to carry infinitely.) | |
| Feb 24, 2011 at 19:58 | history | answered | David E Speyer | CC BY-SA 2.5 |