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)
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 $\mathbb F_8[[t]]$.