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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)

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

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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 $\mathbb F_8[[t]]$.

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  • $\begingroup$ Thanks, that is a very nice an intuitive way to understand this. However I am still having problems with making a formal proof. $\endgroup$
    – Marcos
    Oct 7, 2021 at 17:53
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    $\begingroup$ @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? $\endgroup$ Oct 7, 2021 at 21:24
  • $\begingroup$ Thanks, I almost have the proof, but for some reason I am struggling in to prove this isomorphism $$\left\lbrace\sum_{i=0}^{\infty} a_if^i\mid a_i\in\lbrace 0,1,x,x+1\rbrace\right\rbrace\cong\mathbb{F}_4[[t]]$$ $\endgroup$
    – Marcos
    Oct 8, 2021 at 14:58
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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 $\mathbb{F}_2[\zeta][[f]] \cong \mathbb{F}_4[[t]]$.

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  • $\begingroup$ 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. $\endgroup$
    – Marcos
    Oct 7, 2021 at 17:51
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    $\begingroup$ $a_i \in \mathbb{F}_2[x]/(f)=\{0,1,x,1+x\}$ and $x=\zeta+f+f^2+f^4...$ $\endgroup$ Oct 7, 2021 at 19:20

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