Integral Closure and the Incompatibility Property

• Nov 4th 2009, 05:08 AM
robeuler
Integral Closure and the Incompatibility Property
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?
• Nov 4th 2009, 10:01 PM
NonCommAlg
Quote:

Originally Posted by robeuler
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 $\displaystyle R'$ is not integral over $\displaystyle R,$ you don't need to use such a strong result! just see that $\displaystyle y$ is not integral over $\displaystyle R$: the reason is that if $\displaystyle y$ was integral over $\displaystyle R,$ then there would

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

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

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

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

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