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Nonzero divisor in a quotient ring 
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#1
May812, 10:27 PM

P: 31

How do you show that x is a nonzero divisor in C[x,y,z,w]/<yzxw>?
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 <yzxw> so that f*x = g*(yzxw). Here's another question which is slightly more interesting: prove that x is a nonzero divisor in C[x,y,z,w]/<ywz^2, yzxw>. 


#2
May812, 10:42 PM

P: 606

Do you mean to prove that the coset of x is NOT a zero divisor in that quotient ring? Putting [itex]\,\,I:=<yzxw>\,\,[/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)(yzxw)\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 xdegree of g), and then it can't be [itex]\,\,xf\in I\,\,[/itex] unless [itex]\,\,f=0\,\,[/itex] DonAntonio 


#3
May812, 10:49 PM

P: 31

Wow, you are fast....



#4
May812, 10:53 PM

P: 31

Nonzero divisor in a quotient ring
What do you mean by "xdegree of g"? Does g have to be homogeneous?



#5
May812, 10:58 PM

P: 31

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. 


#6
May912, 05:39 AM

P: 606

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] DonAntonio 


#7
May1012, 04:14 AM

P: 31

Thank you. I spent earlier today reading about this technique. =)



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