# Nonzero divisor in a quotient ring

1. May 8, 2012

### naturemath

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. May 8, 2012

### DonAntonio

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

Putting $\,\,I:=<yz-xw>\,\,$, 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

3. May 8, 2012

### naturemath

Wow, you are fast....

4. May 8, 2012

### naturemath

What do you mean by "x-degree of g"? Does g have to be homogeneous?

5. May 8, 2012

### naturemath

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.

6. May 9, 2012

### DonAntonio

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

7. May 10, 2012