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I am trying to prove its not UFD. I started by assuming

x=ab in K[v] , where v=v(xw-yz) then x-ab=(xw-yz)f, for some f in K[x,y,w,z]

I tried to say that deg of a, and b is less or equal 1, and then try to do a non-matching degree argument, but I could not see why a,and b should have deg leess or equal 1, how should I define degree in K[v]. Any help on how to prove this? Thanks.

  • This is what I am trying to do, via proving that each x,w,y,z is irreducible. the issue is if I have x=ab in K[v], Can I talk about degree of a,b being less or equal one in K[x,y,z,w] –  Sep 23 '14 at 21:18
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    Because the ideal you divided with is homogeneous the quotient ring is a graded ring. Therefore the "obvious" definition of degree makes sense. Also because the ring is a domain (have you proved that?) the usual rule that the degree of a product is the sum of the degrees of factors holds. – Jyrki Lahtonen Sep 23 '14 at 21:25
  • Dear Jyrki, Thanks a lot, it makes more sense now, and can continue my argument. I have not proved its a domain, I am working on it. Now its much easier. –  Sep 23 '14 at 21:41
  • I have one only little detail. If the we have a, and b is of degree one in the ring R[x,y,z,t]/(xw-yz) how can pass this information to the original ring R[x,y,z,w] –  Sep 24 '14 at 03:12
  • The definition of "degree" in the quotient ring would be = "minimum of degrees of polynomials in the coset". You can probably use that lowest degree element in the coset. – Jyrki Lahtonen Sep 24 '14 at 04:18

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