# Formal power series ring over a valuation ring of dimension $\geq 2$ is not integrally closed.

I recently tried exercise 10.4 in Matsumura's Commutative Ring Theory, but got stuck. The question is:

If $R$ is a valuation ring of Krull dimension $\geq 2$, then the formal power series ring $R[[X]]$ is not integrally closed.

Reading the solution: "Let $0 \subset p_1 \subset p_2$ be a strictly increasing chain of prime ideals of $R$ and let $0 \neq b \in p_1$, $a \in p_2-p_1$, thus $ba^{-n} \in R$ for all $n > 0$. Take $f=\sum_{i=1}^\infty u_i X^i$ to be a root of $f^2+af+X=0$. Then $u_1 = a^{-1}$ and for all i we have $u_i \in a^{-2i+1}R$ so $bf(x) \in R[[X]]$ but $f(X) \not \in R[[X]]$."

I understand all the steps of the proof in some sense, but I don't really see why one should think of f to be a root of $f^2+af+X=0$. What is the motivation for considering this equation and roots here? What I am after is as such, the idea of the proof, which I can't seem to find at the moment.

• I believe the motivation is just that we need $f$ to be integral over $R[[x]],$ but not inside $R[[x]],$ in order to conclude that $R[[x]]$ is not integrally closed. The simplest way to find such an element is via roots of (monic) quadratic polynomials, though roots of monic polynomials of any degree would work. – Andrew Sep 25 '12 at 16:51
What Andrew says is correct. However the element / root $f$ must be in the field of fractions of $R[[X]]$, which is the field of Laurent series $K((X))$, $K$ the fraction field of $R$. To this end the form of the polynomial helps: $Y^2+aY+X$ modulo the maximal ideal $XK[[X]]$ of the discrete valuation ring $K[[X]]$ is a polynomial having the roots $0$ and $-a$ in $K$. Hence for $a\neq 0$ this polynomial is separable and Hensel's lemma assures the existence of a root $f$ in $K((X))$.