Is $C$ projective as $C[x]$ module?

Question

I'm asking myself if $C = C[X]/(x)$ is a projective/flat $C[X]$ module. I found on the internet that this isn't the case, but on the other hand we have $ C[X] = C \oplus xC[X]$, so $C$ is a direct summon of a free $C[X]$ module. Did I make a mistake?

Answer 1

$\mathbb C[x]$ does not have direct summands as a $\mathbb C[x]$ module. There are simply no nonzero submodules intersecting to $\{0\}$.

The illusion I think you are under is that that the constant term $\lambda$ in the polynomials can be exchanged for the coset $\lambda +(x)$, which of course it cannot. In $\mathbb C[x]$, $1\cdot x$ is nonzero, whereas considered as the module $\mathbb C[x]/(x)$ multiplying anything nonzero by $x$ results in the zero coset. They just aren't isomorphic.

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But How could I proof now in general, that this isn't a projective Moudle

I think you mean $\mathbb C[x]/(x)$(?) If it were projective the exact sequence $0\to (x)\to \mathbb C[x]\to \mathbb C[x]/(x)\to 0$ would split, so that $(x)$ is a summand of $\mathbb C[x]$. It isn't, for the reason mentioned above.

Answer 2

The claim that $C[X]\cong C[x]/(x)\oplus xC[x]$ as $C[x]$-modules is false. I imagine you’re thinking the isomorphism would look like \begin{align*} \phi:\,&C[x]/(x)\oplus xC[x]\to C[x] \\ &(c,xd)\mapsto c+xd. \end{align*} But the issue is that this isn’t an isomorphism of $C[x]$-modules. In particular, consider multiplication by $x$: we get $\phi(x(1,0))=x\phi(1,0) = x(1) = x$, if this is to be a $C[x]$-module homomorphism. But in the direct sum, $x(1,0) = (x,0) = (0,0)$ since $x=0$ in $C[x]/(x)$, and $\phi(0,0) = 0$. So this map doesn’t give an isomorphism of $C[x]$-modules.

Writing $C=C[x]/(x)$ obscures the fact that as a $C[x]$-module, multiplication by $x$ takes everything to $0$.