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.

Suppose $R=k[[x_1,...,x_n]]$ is a formal power series ring over the field $k$, what can we say about the structure of $R/p$ if $p$ is a prime ideal of $R$ such that dim$(R/p)=1$. In particular, are these subrings of power series rings in one variable?

Motivation: Given a subring of a power series ring in one variable which can be written as, $k[[x^{a_1},...,x^{a_n}]]$ where $a_1,...,a_n$ are distinct positive integers, we have a homomorphism $k[[x_1,...,x_n]]\to k[[x^{a_1},...,x^{a_n}]]$ which sends $x_i\to x^{a_i}$. The kernel of this homomorphism must be a prime ideal since $k[[x^{a_1},...,x^{a_n}]]$ is a domain. Moreover, the quotient modulo the kernel must be one dimensional since the image ring is one dimensional. I was wondering if we have a (partial) converse to this result. (I do realize the second paragraph would analogously work with polynomial rings, but it seems power series rings and much nicer (they are regular and local), so I was hoping for a nicer description of the quotient in the power series case).

share|improve this question

1 Answer 1

up vote 2 down vote accepted

The integral closure of a complete Noetherian local domain is also complete, local and Noetherian. So the integral closure of $R/p$ is a complete regular local ring containing a field (normal rings in dimension $1$ are regular by Serre criterion), and you win!

EDIT: add more details, by Tymothy's comments (now deleted): So let $S$ be the integral closure of $R/p$ in its quotient field. Then by definition, $S$ is integrally closed (normal), thus regular because $\dim S=1$. Clearly $R/p$ is a subring of $S$. By Cohen Structure theorem, $S$ is a power series ring with one variable.

The fact about integral closure in my first sentence is well-known. Check out Theorem 4.3.4 of the book available online by Huneke-Swanson, or Eisenbud Chapter 13, or Matsumura somewhere (probably the finiteness of integral closure is true for excellent rings as well).

share|improve this answer
    
Thanks, that seems perfect. Cohen's structure theorem was the missing link for me. –  Timothy Wagner Nov 22 '10 at 2:25
1  
@Timothy: Note though that if $k$ is not algebraically closed, then the coefficient field for your power series ring might be bigger than $k$ (although it will be a finite extension of $k$). –  Matt E Nov 22 '10 at 5:15

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.