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Math Help - Integral Closure and the Incompatibility Property

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    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?
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    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.
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