The polynomial
$$
h(t)=t^3-t-1\in\mathbb F_3[t]
$$
has no root in $\mathbb F_3$ and is therefore irreducible. Define
$$
\left|\frac{a(t)}{b(t)}\right|_h
=c^{\operatorname{ord}_h(b)-\operatorname{ord}_h(a)}
$$
for any fixed $c>1$. Its valuation ring has residue field
$$
\mathbb F_3[t]/(h)\cong\mathbb F_{27}.
$$
The <completion at an irreducible polynomial over a finite field> identifies the completion with $\mathbb F_{27}((T))$, where $T$ corresponds to the uniformizer $h(t)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 136 / 2 / b / ii / Solution

For an irreducible polynomial $h\in\mathbb F_p[t]$ of degree $d$, the $h$-adic completion of $\mathbb F_p(t)$ is isomorphic to $\mathbb F_{p^d}((T))$. Its residue field is $\mathbb F_p[t]/(h)$ and $h$ is a uniformizer.


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