Studying valuation rings I was struggling with the following problem:
Let $(K,|\cdot|_K)$ be a complete valuation field. Prove that if $[L:K]<\infty$ then $R_L$ is the integral closure of $R_K$ in $L$.
Here $R_L$ and $R_K$ are the closed unit disks (i.e. the set of all elements whose valuation is $\leq 1$ with respect to $|\cdot|_K$ and $|\cdot|_L$)
What I've tried was the following. Let $b\in L$ such that there exists $a_0,\dots,a_{n-1}\in K$ satisfying: $$b^n+a_{n-1}b^{n-1}+\dots+a_1b+a_0=0$$ We have to prove that $b\in R_L$, but I don't know how to proceed. One idea I had was that we can write $a_i=\frac{c_i}{d}~\forall~i=1,\dots,n$ where $c_i,d\in R_L$ and thus, multiplying by $d^n$ I get: $$(db)^n+c_{n-1}(db)^{n-1}+\dots+c_1d^{n-2}(db)+c_0d^{n-1}=0$$ Hence $db$ is in the integral closure of $R_K$ in $L$. But I'm not sure how to use this to prove that $b\in R_L$. Any hint or help will be thanked.
The coefficients $a_i$ in the polynomial relation for $b$ should be in $R_K$ if $b$ is assumed to be integral over $R_K$. You could argue as follows: assume $v(b)>1$. Then the term $b^n$ possesses the largest value among all terms in the polynomial relation. Hence the value of the left hand side of the equation equals $v(b^n)$ -- a contradiction.