# Completion of $\mathbb{F}_2[x]$

I was studying valuations, and I found the topic of completion of a space over a valuation quite challenging. If we have an space $$F$$ and a valuation $$v:F\to\mathbb{Z}$$ we can define a distance $$d_v(p,q)=2^{-v(p-q)}$$ which gives us a topology. In this context the topology can be completed as in the usual way for normed spaces.

Whith this I was able to understand the basic example of completion of $$\mathbb{Q}$$ with respect to the p-adic valuation, which is the field of p-adic integers and some other "easy" examples. But I found one more difficult exercise that I do not understand:

Let $$f=x^2+x+1$$ and $$v_f:\mathbb{F}_2[x]\to\mathbb{Z}$$ such that $$v_f(p)=n$$ if $$f^n$$ divides $$p$$ and $$f^{n+1}$$ does not divide $$p$$. Apparently the completion of this space is $$\mathbb{F}_4[[t]]=\lbrace\sum_{i=0}^{\infty}a_i t^i\mid a_i\in\mathbb{F}_4\rbrace$$ But I dont understand how this is related with the original space. Why $$\mathbb{F}_4$$?, honestly I am quite lost.

Any hint or help will be appreciated. (Avoid using inverse limits if possible)

• $\Bbb F_4$ is the field $\Bbb F_2[x]{/}(f)$ so I think that's the reason for that field. Oct 7, 2021 at 7:43

If $$f$$ is an irreducible polynomial over $$\mathbb F_2$$, then the elements of valuation $$0$$ are all those of degree less than $$\deg f$$. If $$f=x$$, or $$x+1$$ say, then there are two elements of valuation $$0$$, namely $$0$$ and $$1$$. If $$f=x^2+x+1$$ though, $$0$$, $$1$$, $$x$$ and $$x+1$$ are valuation $$0$$. The elements of valuation $$0$$ are intended to be the base field of the completion, so that is why you obtain $$\mathbb F_4$$ rather than $$\mathbb F_2$$. If $$f$$ were $$x^3+x+1$$ instead, the completion would be $$\mathbb F_8[[t]]$$.
• @MarcosEscartínFerrer All of the elements of $\mathbb F_2[x]$ can be thought of as elements of $\mathbb F_8[f]$. Does that help with the start of the proof? Oct 7, 2021 at 21:24
• Thanks, I almost have the proof, but for some reason I am struggling in to prove this isomorphism $$\left\lbrace\sum_{i=0}^{\infty} a_if^i\mid a_i\in\lbrace 0,1,x,x+1\rbrace\right\rbrace\cong\mathbb{F}_4[[t]]$$ Oct 8, 2021 at 14:58
The element $$\zeta=x+f+f^2+f^4+...$$ in the completion satisfies $$\zeta^2+\zeta+1=0$$ so $$\mathbb{F}_4 \cong \mathbb{F}_2[\zeta]$$ and the completion $$\mathbb{F}_2[\zeta][[f]] \cong \mathbb{F}_4[[t]]$$.
• Right, but why the completition is $\mathbb{F}_2[\zeta][[f]]$? for what I have understood the completiton would be $\lbrace\sum a_if^i\mid a_i\in\lbrace 0,1\rbrace\rbrace$ or maybe I have misunderstood something. Oct 7, 2021 at 17:51
• $a_i \in \mathbb{F}_2[x]/(f)=\{0,1,x,1+x\}$ and $x=\zeta+f+f^2+f^4...$ Oct 7, 2021 at 19:20