Checking the ring k[x,y,u,v]/(xy-uv) is a normal domain

Let A= k[x,y,u,v]/(xy=uv), is this ring normal (integral closed in its field of fraction)?

Edit (see comments below): Because p is of height 1, then there exists one among x,y,u,v, say x such that f(x)\notin p for any f(x)\in k[x]. Otherwise, if there are polynomials in one variable for each of the variables x,y,u,v, contained in p, then p is of height >1. Thus f(x) becomes invertible in A_p for any f(x)\in k[x].

7 thoughts on “Checking the ring k[x,y,u,v]/(xy-uv) is a normal domain

  1. It seems to me that to conclude A_p=k(x)[u,v]_p , you also need to show that any polynomial in x isn’t conatined in p to make the proof completed. Otherwise, you could only get A_p=k[x,x^{-1}][u,v]_p.

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  2. How could one show that if x isn’t in p, then any polynomial in x can’t be in p? Otherwise, maybe you just have the form A_p=k[x,x^{-1}][u,v]_p

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  3. For the first part, even x isn’t in p, you may have, for example, x+1 is in p=(x+1,y+uv). I think you have to rule out such cases to conculde the coefficients field is indeed frac(k[x])

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    1. Thanks for the comment! I should have said that we may assume p is contained in the maximal ideal (x,y,u,v).

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        1. I don’t think I need to assume that k is algebraically closed. I just need to say a bit more. Because p is of height 1, then there exists one among x,y,u,v, say x such that f(x) not in p for any f(x) in k[x]. Otherwise, if there are polynomials in one variable for each of x,y,u,v, then p is of height >1.

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