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Nonzero divisor in a quotient ring

  1. May 8, 2012 #1
    How do you show that x is a nonzero divisor in C[x,y,z,w]/<yz-xw>?

    Here's how one can start off on this problem but I would like a nice way to finish it:

    If x were a zero divisor, then there is a function f not in <yz-xw> so that

    f*x = g*(yz-xw).

    Here's another question which is slightly more interesting:

    prove that x is a nonzero divisor in C[x,y,z,w]/<yw-z^2, yz-xw>.
    Last edited: May 8, 2012
  2. jcsd
  3. May 8, 2012 #2

    Do you mean to prove that the coset of x is NOT a zero divisor in that quotient ring?

    Putting [itex]\,\,I:=<yz-xw>\,\,[/itex], suppose [itex]\,\,(x+I)(f+I)=\overline{0}\Longrightarrow xf\in I[/itex] , but any element in the

    ideal I has the form [itex]\,\,g(x,y,z,w)(yz-xw)\Longrightarrow \,\,[/itex] x cannot be factored out in this product if [itex]\,\,g(x,y,z,w)\neq 0\,\,[/itex] (you may

    try to prove this by induction on the x-degree of g), and then it can't be [itex]\,\,xf\in I\,\,[/itex] unless [itex]\,\,f=0\,\,[/itex]

  4. May 8, 2012 #3
    Wow, you are fast....
  5. May 8, 2012 #4
    What do you mean by "x-degree of g"? Does g have to be homogeneous?
  6. May 8, 2012 #5
    Wouldn't g equal

    [itex]\sum[/itex] a_I x^i y^j z^k w^l

    where I = (i,j,k,l) ?

    How would one induct on the degree of x? It seems quite messy.
  7. May 9, 2012 #6

    Didn't say it'd be easy...:) . But it is: we can always write as follows an element in that ring [tex]f\in\mathbb{C}[x,y,z,w]\Longrightarrow f=a_0+a_1x+a_2x^2+...+a_nx^n\,\,,\,\,a_n\in\mathbb{C}[y,z,w]\,\,,\,\,so\,\,deg_x(f)=n[/tex]

  8. May 10, 2012 #7
    Thank you. I spent earlier today reading about this technique. =)
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