$M =\mathbb{C}[x]/(x^2)$ is not a free module over $\mathbb{C}[x]$

Question

I'm trying to understand this example:

$M= \mathbb{C}[x]/(x^2)$ is a module over $\mathbb{C}[x]$, (since if R is any ring and I is any ideal of R, then R/I is always an R-module.)

Then

(i) $M$ is cyclic

(ii) $M$ is not a free module over $\mathbb{C}[x].$

I understand the first claim. $M$ is cyclic because the element $1 \in \mathbb{C}[x]$ generates $M:$ $\mathbb{C}[x]/(x^2)= \mathbb{C}[x]\cdot 1$.

In class, the explanation given for why $M$ is not a free module is that $x^2\cdot1= 0,$ and $x^2 \neq 0,$ so the module can't be free.

I understand that for a module to be free, it must have a basis. So if $X$ were a basis, and $1\in X,$ then the above statement would make sense. But why is $1$ necessarily in the basis? What am i missing?

Answer 1

Note that if $M$ would be free then it would be isomorphic to $\mathbb{C}[x]$ since it is cyclic. But this is not the case since in $\mathbb{C}[x]$, any nonzero element has trivial annihilator whereas in $M$, $1$ has $x^2$ in its annihilator.

Answer 2

A free $R$ module is isomorphic to $\oplus_{i\in I}R_i$ where $R_i$ is isomorphic to $R$, since $\mathbb{C}[X]$ does not have torsion element, a free $\mathbb{C}[X]$ module cannot have nilpotent elements.

Answer 3

Consider a commutative ring $R$ and an ideal $I$. Then $R/I$ projective implies $I$ is a direct summand of $R$, hence of the form $eR$, for $e$ an idempotent element.

Also the converse is true, because a direct summand of a projective module (in particular, free) is projective.

Since $\mathbb{C}[x]$ is a domain, it has no nontrivial idempotents, so for no proper and nonzero ideal $I$, $\mathbb{C}[x]/I$ is projective and, a fortiori, free.

Answer 4

Yet another proof: any free $\mathbb C[x]$-module is an infinite dimensional $\mathbb C$-vector space, but your module has dimension two, which is finite.