Nonzero divisor in a quotient ring

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Discussion Overview

The discussion revolves around demonstrating that the element x is a nonzero divisor in the quotient ring C[x,y,z,w]/. Participants explore methods to prove this property, including the implications of x being a zero divisor and the structure of elements in the ideal.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests starting the proof by assuming that if x were a zero divisor, there exists a function f not in the ideal such that f*x = g*(yz-xw).
  • Another participant clarifies the goal is to show that the coset of x is not a zero divisor in the quotient ring, proposing that if (x+I)(f+I)=0, then xf must belong to the ideal I.
  • It is noted that elements in the ideal I can be expressed in terms of g(x,y,z,w)(yz-xw), and that x cannot be factored out unless g is zero.
  • Questions arise regarding the meaning of "x-degree of g" and whether g must be homogeneous, with participants discussing the form of g as a polynomial in multiple variables.
  • Concerns are raised about the complexity of inducting on the degree of x, with one participant acknowledging the difficulty of the approach.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approaches to the proof, indicating that there is no consensus on the best method to demonstrate that x is a nonzero divisor. Some participants agree on the structure of g, while others question the implications of the degree of x.

Contextual Notes

There are unresolved assumptions regarding the properties of the polynomials involved and the implications of the degree of g on the proof's structure.

naturemath
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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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naturemath said:
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 \,\,I:=&lt;yz-xw&gt;\,\,, suppose \,\,(x+I)(f+I)=\overline{0}\Longrightarrow xf\in I , but any element in the

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

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

DonAntonio
 
Wow, you are fast...
 
What do you mean by "x-degree of g"? Does g have to be homogeneous?
 
Wouldn't g equal

\sum 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.
 
naturemath said:
Wouldn't g equal

\sum 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 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

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

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