Why is the submodule of K[x,y] generated by x and y not a free module?

Question

If $\mathbb{K}$ is a field, and we consider the module $\mathbb{K}[x,y]$ over itself, I've seen others give the example that the ideal generated by $\langle x, y \rangle$ is not a free module to demonstrate that a submodule of a free module need not be free. However, I don't understand why.

From the definition of a free module, a module is free if it has a linearly independent basis. Since $x,y$ have no relations between them, they seem to form a linearly independent $\mathbb{K}[x,y]-$basis for the ideal. So what am I missing here? Why is this submodule not free?

Similarly, I've seen the example that $\mathbb{Z}_6$ is a free module over itself but apparently $2\mathbb{Z}_6$ is not free for some reason. Again, this modules is generated by $\langle 2\rangle$ so by definition it has to be free, right? I'm confused.

Any explanation would be appreciated, thank you!

Answer 1

They are not independent over $\mathbb{K}[x,y]$, which is the ring in question: using boldface for the module elements, you have $y\mathbf{x} -x\mathbf{y}=\mathbf{0}$, even though neither $x$ nor $y$ are zero. It is not an independent set. So it is not true that "there are no relations between them."

They do span a free $\mathbb{K}$-module, but that isn't what you are considering.

The subgroup $\langle2\rangle$ of the integers modulo $6$ is not free since $3\mathbf{2}=\mathbf{0}$, even though the scalar $3$ is not zero. It is not an independent set. You can also verify it does not satisfy the universal property of free modules: you cannot extend the map $\{2\}\to \mathbb{Z}_6$ that sends $2$ to $1$ to a $\mathbb{Z}_6$-module homomorphism $\langle 2\rangle\to \mathbb{Z}_6$.

Answer 2

$2\Bbb Z_6$ is generated by $2$ but it's not freely generated by $2$. It's not isomorphic to $\Bbb Z_6[X]$ for some set $X$, you can see that just by checking the cardinality. Since $3\cdot2=0$ and $3\neq0$, there is a problem... free modules must in particular be torsion free. In general for a ring $R$, $(\alpha)$ is a free (left/right) $R$-module iff. $\alpha$ is not a (left/right) zero divisor.

With $R=\Bbb K[x,y]$ and $J=(x,y)\subset R$; as a $\Bbb K$-module, $J$ is certainly free, but as an $R$-module it is not. Yes, $x,y$ generate but the generation isn't itself free; for example there is a nontrivial linear dependence $y\cdot x-x\cdot y=0$, which couldn't occur if $x,y$ were free generators (this in itself is not sufficient to conclude $J$ isn’t a free $R$-module, just that it is not free in the obvious ways; however, you can combine this observation with the fact ideals of a domain are free iff. they are principal to make a conclusion).