# Thread: Free Modules

1. ## Free Modules

Let $R=C[x,y]$ the ring of polynomials in two variables with complex coefficients. Let $M=(x,y)$ be the ideal generated by $x$ and $y$. $M$ is an $R$-module.

Is $M$ 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 $M$ looking like

$m=r_{1}x+r_{2}y$

2. Originally Posted by robeuler
Let $R=C[x,y]$ the ring of polynomials in two variables with complex coefficients. Let $M=(x,y)$ be the ideal generated by $x$ and $y$. $M$ is an $R$-module.

Is $M$ 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 $M$ looking like

$m=r_{1}x+r_{2}y$
in general suppose $M \neq (0)$ is an ideal of a commutative domain $R.$ then, as an $R$ module, $M$ is free if and only if it's cyclic. the reason is that if $M$ has a basis with more than one elements, then

we can choose two elements of the basis, say $a,b.$ but $ba - ab = 0,$ which is impossible because $a,b$ belong to a basis of $M.$ conversely if $M=Ra,$ then $\{a \}$ would be a basis for $M$ because $R$ is

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