Let , is this ring normal (integral closed in its field of fraction)?
Edit (see comments below): Because is of height 1, then there exists one among say such that for any . Otherwise, if there are polynomials in one variable for each of the variables contained in , then is of height . Thus becomes invertible in for any .
7 thoughts on “Checking the ring k[x,y,u,v]/(xy-uv) is a normal domain”
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.
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])
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.
![A= k[x,y,u,v]/(xy=uv)](https://s0.wp.com/latex.php?latex=A%3D+k%5Bx%2Cy%2Cu%2Cv%5D%2F%28xy%3Duv%29&bg=ffffff&fg=000000&s=0&c=20201002)
![f(x)\in k[x]](https://s0.wp.com/latex.php?latex=f%28x%29%5Cin+k%5Bx%5D&bg=ffffff&fg=000000&s=0&c=20201002)
Thinking Bear 
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.
LikeLike
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
LikeLike
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])
LikeLike
Thanks for the comment! I should have said that we may assume p is contained in the maximal ideal (x,y,u,v).
LikeLike
Then I think you should assume that k is algebraically closed to make sure all maximal ideals have similar form.
LikeLike
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.
LikeLike
That makes a lot of sense. Thanks
LikeLike