Nonzero divisor in a quotient ring

[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>.

 

[DonAntonio]

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

 

[naturemath]

Wow, you are fast...

 

[naturemath]

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

 

[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.

 

[DonAntonio]

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

 

[naturemath]

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