Let $K$ be a field with a valuation $\mathrm{val}$ and residue field $\mathbb{k}$. Show that if $K$ is algebraically closed then $\mathbb{k}$ is. Give a counterexample for the converse.

Obviously, the residue field is union of zero and the set of elements in the valuation ring with valuation zero, $\mathbb{k} = R/m = \{ a\in R\ :\ val(a)=0\} \cup \{0\}$.

Now consider an arbitrary univariate polynomial $f = \sum_{i=0}^n c_i x^i \in \mathbb{k}[x]$ of degree n with coefficients in the residue field. Since K is algebraically closed, f has a root in K, say $a$, ie $0 = f(a) = \sum_{i=0}^n c_i a^i$.

Assume $\mathrm{val}(a)>0$.

Case 1: $c_0\neq 0$, ie $f(0)\neq 0$. Then $\mathrm{val}(\sum_{i=1}^n c_i a^i) = \mathrm{val}(-c_0)=0$ since $c_0\in\mathbb{k}^{*}$. For the left-hand side of the equation we get

(1)which is a contradiction since by combine both lines we get $0>0$. Therefore, $\mathrm{val}(a)=0$ or, equivalently, $a\in\mathbb{k}$, ie $\mathbb{k}$ is algebraically closed.

Case 2: If zero is a root of f, then there exists a positive integer k such that $f = \sum_{i=k}^n c_i x^i = x^k \cdot \sum_{i=0}^{n-k} c_{k+i} x^i$. We get

(2)Subtract $\mathrm{val}(x^k)$ on both sides and proceed as in case 1.

**Counterexample for the converse**. ( may be there is something wrong, check this)

The field of formal Laurent series over the complex numbers (ie formal power series with finitely many negative exponents), denoted by $\mathbb{C}((t))$ is not algebraically closed. Let f be an element in $\mathbb{C}((t))$, $f = \sum_{i\geq k} c_i t^i$ for an integer k with $c_k\neq 0$.

Consider the valuation $\mathrm{val}(f) = \mathrm{val}\left( \sum_{i\geq k} c_i t^i \right) = k$. Then the valuation ring R and the maximal ideal m of it are $R = \langle \bigcup_{n\geq 0} t^n \rangle$ and $m = \{ \bigcup_{n>0} t^n\}$, respectively. Hence, the residue field $\mathbb{k} = R / m = \mathbb{C}$ is the field of complex numbers, which is alg. closed.