Is x^2/y^2 an Ideal in the Ring F[x, y]?

[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] ?

 

[Hurkyl]

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?

 

[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

 

[Hurkyl]

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)

 

[matness]

now it is clear thank you very much