Thread: 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?

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 is not integral over you don't need to use such a strong result! just see that is not integral over : the reason is that if was integral over then there would

exist and an integer such that but then for some which is obviously nonsense.

so the integral closure of in is not in fact is integrally closed in i.e. the integral closure of in is itself: clearly for every there exist and such that

if is integral over then is also integral over so there exist and an integer such that

that would give us for some which is possible only if because, since there is no in therefore

which gives us thus and the proof is complete.