Thread: Free Modules

Let the ring of polynomials in two variables with complex coefficients. Let be the ideal generated by and . is an -module.

Is a free module?

I am still in the habit of associating free with finitely generated, but I think the notion is more subtle than that. My instinct says yes: something along the lines of elements in looking like

Originally Posted by robeuler

Let the ring of polynomials in two variables with complex coefficients. Let be the ideal generated by and . is an -module.

Is a free module?

I am still in the habit of associating free with finitely generated, but I think the notion is more subtle than that. My instinct says yes: something along the lines of elements in looking like

in general suppose is an ideal of a commutative domain then, as an module, is free if and only if it's cyclic. the reason is that if has a basis with more than one elements, then

we can choose two elements of the basis, say but which is impossible because belong to a basis of conversely if then would be a basis for because is

a domain. so, in your question, is not a free module because it cannot be generated by one element. (prove it!)