Integral Closure and the Incompatibility Property

robeuler · #1 $Old$ November 4th, 2009, 06:08 AM

Let R'=K[x,y] (K is a field) and R=K+xR'

Even though (x)<(x,y) are prime ideals of R', (x) intersect R = (x,y) intersect R. So R' is not integral over R. What is the integral closure of R in R'.

First I do not see clearly why (x) intersect R equals (x,y) intersect R. I think of R as a polynomial ring over K with every nonconstant term having an x. If you mod out by the x's you are left with K[x,y], if you mod out by (x,y) you get something else.

Second I am not sure what the integral closure could be. Perhaps it is K[x,y] itself?

NonCommAlg · #2 $Old$ November 4th, 2009, 11:01 PM

Quote:

Originally Posted by robeuler $View Post$

Let R'=K[x,y] (K is a field) and R=K+xR'

Even though (x)<(x,y) are prime ideals of R', (x) intersect R = (x,y) intersect R. So R' is not integral over R. What is the integral closure of R in R'.

First I do not see clearly why (x) intersect R equals (x,y) intersect R. I think of R as a polynomial ring over K with every nonconstant term having an x. If you mod out by the x's you are left with K[x,y], if you mod out by (x,y) you get something else.

Second I am not sure what the integral closure could be. Perhaps it is K[x,y] itself? No!

first of all to see that $R'$ is not integral over $R,$ you don't need to use such a strong result! just see that $y$ is not integral over $R$ : the reason is that if $y$ was integral over $R,$ then there would

exist $f_i \in R', \ a_i \in K$ and an integer $n \geq 1$ such that $y^n + (a_1+xf_1)y^{n-1} + \cdots + a_n + xf_n=0.$ but then $y^n + a_1y^{n-1} + \cdots + a_n = xf,$ for some $f \in R',$ which is obviously nonsense.

so the integral closure of $R$ in $R'$ is not $R'.$ in fact $R$ is integrally closed in $R',$ i.e. the integral closure of $R$ in $R'$ is $R$ itself: clearly for every $f \in R',$ there exist $g \in K[y]$ and $h \in R$ such that

$f=yg + h.$ if $f$ is integral over $R,$ then $yg=f-h$ is also integral over $R.$ so there exist $p_i \in R', \ b_i \in K$ and an integer $m \geq 1$ such that $y^mg^m + (b_1 + xp_1)y^{m-1}g^{m-1} + \cdots + b_m + xp_m=0.$

that would give us $y^mg^m + b_1y^{m-1}g^{m-1} + \cdots + b_m = xp,$ for some $p \in R',$ which is possible only if $p=0,$ because, since $g \in K[y],$ there is no $x$ in $y^mg^m + b_1y^{m-1}g^{m-1} + \cdots + b_m.$ therefore

$y^mg^m + b_1y^{m-1}g^{m-1} + \cdots +b_m=0,$ which gives us $g=0.$ thus $f=yg+h=h \in R,$ and the proof is complete.