Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thankful for answers.

share|improve this question
    
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
add comment

1 Answer

up vote 2 down vote accepted

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 $K((X))$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.