Is the residue field of an algebraically closed valued field $K$ with valuation ring $A$, algebraically closed?
If I take a polynomial $f(x)$ of $k_A$ since $K$ is algebraically closed it has a root in $K$, say $b$, so the valuation of $f(b)$ is infinity and consequently the valuation of $b$ must be infiinity …
Best Answer
You want to find a root of $f$ where $\deg f \geq 1$ is a polynomial in $k_A$. It is not a polynomial in $K$ so you cannot say that it has a root in $K$. However, each coefficient $c_i$ of $f=\sum \limits_{i=0}^d c_i X^i$ is the residue of some element $a_i$ of $A$. Let's pick such elements $a_i$ and consider $P=\sum \limits_{i=0}^d a_i X^i$. This has degree $\deg f \geq 1$ so it must have roots in $K$. Can you justify that every root of $P$ lies in $A$? Assuming this, fixing such a root $b$, it is easy to see that $f_A(\overline{b})=\overline{P(b)}=\overline{0}$ where for $x \in A$, I denote the residue of $x$ by $\overline{x}$. So $k_A$ is indeed algebraically closed.