# A Certain Integral Closure

• Mar 30th 2013, 11:49 AM
spudwish
A Certain Integral Closure
I have a bunch of assertions without motivation which I'm trying to sort out.

Let H be a subgroup of S_n, [; A = k[x_1,...,x_n] ;], and [; \sigma_i ;] the elementary symmetric polynomials. The assertions are:

i) A_H is defined as the integral closure of [; k[\sigma_1,...,\sigma_n] ;] in [; k(x_1,...,x_n)^H ;], and [; A_H = k[x_1,...,x_n] \cap k(x_1,...,x_n)^H ;].

ii) [; k(x_1,...,x_n)^H ;] is the field of fractions of A_H, "i.e." [; k(x_1,...,x_n)^H = A_H[1/k[\sigma_1,...,\sigma_n]] ;].

Alright, so about the first assertion, that A_H is the intersection. This intersection contains only polynomials, i.e. we must have [; k[x_1,...,x_n] \cap k(x_1,...,x_n)^H \subset k[x_1,...,x_n] ;], so we can simply consider the intersection [; k[x_1,...,x_n] \cap k[x_1,...,x_n]^H ;], which must be [; k[x_1,...,x_n]^H ;]. I've no idea how to show or see that [; k[x_1,...,x_n]^H ;] is the supposed integral closure, but let's leave it at that for the moment.

Second assertion. What the notation [; A_H[1/k[\sigma_1,...,\sigma_n]] ;] means I have no idea; my guess is that it's the set [; \{ f/g \mid f \in A_H, g \in k[\sigma_1,...,\sigma_n] \} ;]. If this is the case, how do I verify it? Because I would have thought that Frac A_H = k(x_1,...,x_n)^H?
• Mar 30th 2013, 11:56 AM
HallsofIvy
Re: A Certain Integral Closure
Quote:

Originally Posted by spudwish
I have a bunch of assertions without motivation which I'm trying to sort out.

Let H be a subgroup of S_n, $A = k[x_1,...,x_n]$, and $\sigma_i$ the elementary symmetric polynomials. The assertions are:

i) A_H is defined as the integral closure of $k[\sigma_1,...,\sigma_n]$ in $k(x_1,...,x_n)^H$, and $A_H = k[x_1,...,x_n] \cap k(x_1,...,x_n)^H$.

ii) $k(x_1,...,x_n)^H$ is the field of fractions of A_H, "i.e." $k(x_1,...,x_n)^H = A_H[1/k[\sigma_1,...,\sigma_n]]$.

Alright, so about the first assertion, that A_H is the intersection. This intersection contains only polynomials, i.e. we must have $k[x_1,...,x_n] \cap k(x_1,...,x_n)^H \subset k[x_1,...,x_n]$, so we can simply consider the intersection $k[x_1,...,x_n] \cap k[x_1,...,x_n]^H$, which must be $k[x_1,...,x_n]^H$. I've no idea how to show or see that $k[x_1,...,x_n]^H$ is the supposed integral closure, but let's leave it at that for the moment.

Second assertion. What the notation $A_H[1/k[\sigma_1,...,\sigma_n]]$ means I have no idea; my guess is that it's the set $\{ f/g \mid f \in A_H, g \in k[\sigma_1,...,\sigma_n] \}$. If this is the case, how do I verify it? Because I would have thought that Frac A_H = k(x_1,...,x_n)^H?

I have replaced your [; and ;] with "tex" and "/tex".