nonzero divisor in a quotient ring


by naturemath
Tags: nonzero divisor, quotient ring
naturemath
naturemath is offline
#1
May8-12, 10:27 PM
P: 31
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>.
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DonAntonio
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#2
May8-12, 10:42 PM
P: 606
Quote Quote by naturemath View Post
How do you show that x is a nonzero divisor in C[x,y,z,w]/<yz-xw>?


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]

DonAntonio
naturemath
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#3
May8-12, 10:49 PM
P: 31
Wow, you are fast....

naturemath
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#4
May8-12, 10:53 PM
P: 31

nonzero divisor in a quotient ring


What do you mean by "x-degree of g"? Does g have to be homogeneous?
naturemath
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#5
May8-12, 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.
DonAntonio
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#6
May9-12, 05:39 AM
P: 606
Quote Quote by naturemath View Post
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.


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
naturemath
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#7
May10-12, 04:14 AM
P: 31
Thank you. I spent earlier today reading about this technique. =)


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