# The residue field of the algebraic closure of a complete valued field

The following is taken from Cassels' "Local Fields":

Let $k$ be a field which is complete w.r.t. a non-arch valuation and whose residue field $\rho$ is perfect.

On page 123, in Corollary 3, he makes the following claim:

"the residue field of the algebraic closure of $k\;$ = $\;$the algebraic closure of the residue field of $k$"

He proves this as follows:

1) Take any irreducible polynomial $f$ in $\rho[X].$

2) Lift it to a polynomial $\widetilde{f}$ in $k[X]$.

3) Every root of $\widetilde{f}$ lies in the algebraic closure $\bar{k}$ of $k.$

4) Hence every root of $f$ lies in the residue class field of $\bar{k}.$

5) Done.

Question. Why we are done after step 4?

• Algebraic closures are algebraically closed, I'm not sure why you would need to reprove this. Commented Dec 2, 2016 at 18:36

## 1 Answer

Denote the residue field of $\bar k$ by $\Omega$. Then $\Omega/ \rho$ is algebraic. Hence, if $\alpha$ is algebraic over $\Omega$, it is algebraic over $\rho$, i.e. a root of some polynomial $f\in \rho[X]$. So steps 1-4 show $\alpha \in \Omega$.

• How do you know that $\Omega / \rho$ is algebaric? Commented Dec 7, 2017 at 19:02
• @JohnnyT. This should be a consequence of Hensel's lemma. I think this is the statement referred to in the question, but I don't have a copy of the book at hand. Commented Dec 9, 2017 at 10:12
• @JohnnyT. It follows from Hensel's lemma that if $f = \sum_{i=0}^n a_i X^i$ is an irreducible polynomial with coefficients in $k$, then $\max\{|a_i|\} = \max\{|a_0|, |a_n|\}$, see for instance Neukirch's book, page 131. This implies that any element of the valuation ring of $\overline{k}$ is algebraic over the valuation ring of $k$ and any element of $\Omega$ lifts to the valuation ring of $\overline{k}$. Commented Dec 11, 2017 at 16:34