Let $R$ be an integral domain with field of fractions $K$, and let $f \in R$ be a non-zero non-unit. Prove that the subring $S=R[1/f]$ of $K$ is not finitely-generated as an $R$-module, using the fact that every element of $S$ may be written in the form $r/f^n$ with $r\in R$ and $n\geq0$, and any finite set of elements of $S$ can be written like this with a common denominator.
Asked
Active
Viewed 2,054 times
3
-
1Well, what have you tried? – Tobias Kildetoft Feb 06 '13 at 20:25
3 Answers
3
Let $K$ be the quotient field of $A$. Let be a subring of $K$ containing $A$ and $f^{-1}$.
We have to show that $A[f^{-1}]$ is not finitely generated.
On contrary, suppose $A[f^{-1}]$ is finitely generated. Then $f^{-1}$ is integral over $A$ by proposition of the book Commutative Algebra by Atiyah. $i.e.,$ there are $a_{1}, a_{2},...,a_{n}\in A$ such that \begin{equation} f^{-n}+ a_{n}f^{-(n-1)}+...+ a_{1}=0 \end{equation} By multiplying with $f^{n}$, we have \begin{equation} 1+ a_{n}f+...+ a_{1}f^{n}=0 \end{equation} $i.e.,$ \begin{equation} -f(a_{n}+...+a_{1}f^{n-1})=1 \end{equation} Thus $f$ is a unit that yield a contradiction of the hypothesis.
Amit
- 676
2
Easy Answer: $fx-1$ is not monic. Edit: This is wrong unless $R$ is integrally closed.
Hard Answer: Suppose $S=R[1/f]$ is finitely generated and let $a_1,\dots,a_n$ be a generating set of $S$. By the fact we're allowed to use we may write the generators as $r_1/f^{e_1},\dots, r_n/f^{e_n}$. Let $e=e_1\dots e_n $ and write $1/f^{e+1}$ with respect to this generating set,
$$\frac{1}{f^{e+1}}=\sum_{i=1}^n \frac{b_i r_i}{f^{e_i}}$$
for some $b_i \in R$. Multiplying by $f^e$ we deduce that $1/f \in R$ and so $f$ is a unit which is a contradiction.
JSchlather
- 15,427
-
It's not clear what proof the "easy answer" denotes. Could you please elaborate. – Math Gems Feb 06 '13 at 20:59
-
@MathGems In an extension of rings $R \subset S$ an element $\alpha \in S$ is integral over $R$ if it satisfies a monic polynomial over $R$, in particular if its minimum polynomial over $R$ is monic. We have the very nice equivalence $\alpha$ is integral over $R$ if and only if $R[\alpha]$ is a finitely generated $R$-module. – JSchlather Feb 06 '13 at 21:08
-
But if $R$ is not integrally closed, then minimal polynomials of elements integral over $R$ need not be monic, e.g. the minimal polynomial of $\zeta_3$ over $,\Bbb Z[\sqrt{-3}],$ is $\rm:2, x+1! +!\sqrt{-3}\ \ $ – Math Gems Feb 06 '13 at 21:29
-
@MathGems, Good point I get too comfortable working over $\mathbb Z$. – JSchlather Feb 06 '13 at 21:56
-
@YACP It was a tongue-in-cheek comment in contrast to my original one-line solution, which of course turned out to be wrong in general. – JSchlather Feb 07 '13 at 01:02
1
Hint $\ $ One conceptual way to view this is as a generalization of the familiar case when $\rm\,R = \Bbb Z,\,$ where the Rational Root Test implies that proper fractions cannot be roots of polynomials $\rm\in \Bbb Z[x]\,$ that are monic (i.e. leading coefficient $= 1).$
Let $\rm\,e = 1/f.\:$ If $\rm\:R[e] = R\left<1,e,\ldots,e^n\right>\,$ then $\rm\,e^{n+1}$ is an $\rm\,R$-linear combination of lower powers of $\rm\,e,\,$ i.e. the proper fraction $\rm\,e\,$ is a root of a monic polynomial over $\rm\,R.\,$ Therefore the proof of the Rational Root Test (RRT) shows that the denominator $\rm\,f\,$ divides the leading coefficient $ = 1,\, $ i.e. $\rm\,1/f\in R,\,$ contra hypothesis.
Remark $\ $ The RRT requires that the fraction be in lowest terms, i.e. that the denominator and numerator are coprime, which is true here since the numerator $= 1$.
A domain $\rm\,D\,$ is called integrally closed (in its fraction field) if none of its proper fractions are integral over $\rm\,D,\,$ i.e. they are not roots of monic polynomials over $\rm\,D.\:$ The usual proof of the Rational Root Test works in any domain where gcds exist (gcd-domain), e.g. in any UFD. Therefore gcd-domains and UFDs are integrally-closed.
