# 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. Anonymous says:

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. Anonymous says:

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. user029 says:

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. Likun Xie (they/them) says:

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. user029 says:

Then I think you should assume that k is algebraically closed to make sure all maximal ideals have similar form.

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1. Likun Xie (they/them) says:

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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1. user029 says:

That makes a lot of sense. Thanks

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