To be or not to be (an ideal)

1. Sep 24, 2006

matness

let F be a field.x^2/y^2 is not an element of F[x,y](is it?)
(x^2/y^2) can or can not be ideal in F[x,y] ?

2. Sep 24, 2006

Hurkyl

Staff Emeritus
Strictly speaking, the expression

x²/y²

is undefined, because one cannot divide in the ring F[x, y]. And because x²/y² is undefined, so is the expression (x²/y²).

We're usually more generous with notation, though; rather than leave x²/y² undefined, we implicitly shift our attention to the field F(x, y), which does contain an element by that name.

Where did this come from?

3. Sep 24, 2006

matness

the origin of my question:i have to prove F[x,y]/(x^2/y^2) is a vector space it seemed a bit meaningless anddid not remember fraction fields
probably it was F(x,y)/(x^2/y^2) and i misread it
sorry:uhh:

4. Sep 24, 2006

Hurkyl

Staff Emeritus
F(x, y) / (x² / y²) doesn't make much sense either; the only ideals of a field are the zero ideal and the whole field itself.

It probably said F[x, y] / (x², y²)

5. Sep 24, 2006

matness

now it is clear thank you very much