I'm trying to get some familiarity with projective resolutions and therefore working on this little exercise: Find a projective resolution of a field $k$ considered as a $k[x]/(x^n)$-module (where $x$ acts by zero on $k$).
If $n=1$, this should be easy, as $k$ in this case is isomorphic to the ring $k[x]/(x^n)$ considered as a module over itself. Hence $k$ in this case is free, hence projective.
However, in the case $n>1$, I don't know the answer.
My best attempt is something like $$0\to (x \cdot k[x])/(x^n) \xrightarrow{\text{inclusion}} k[x]/(x^n) \xrightarrow{\text{projection}} k\to 0$$ .
This is sequence is exact and the middle term is projective. But I see no reason why $ (x \cdot k[x])/(x^n)$ should be projective. Any advice on this problem would be appreciated.
Let us write $R=k[x]/(x^n)$, and start as you did with the projection $$ R\to k \to 0.$$
You correctly identified the kernel with $xR$. But instead of using the inclusion $xR\to R$, let us instead use the map $R\to R$ given by $a\mapsto xa$, which has $xR$ as it image. This leads to $$R\xrightarrow{\cdot x} R\to k\to 0.$$
Now let us keep going: the kernel of the leftmost map is $x^{n-1}R$, since $xa=0$ implies that $xa$ is a multiple of $x^n$. Once again we can use the map $R\to R$ sending $1$ to $x^{n-1}$, which has the correct image. This gives $$R\xrightarrow{\cdot x^{n-1}} R\xrightarrow{\cdot x} R\to k\to 0.$$
Now something special happens here: the kernel of the leftmost map is... $xR$ again. Can you see how to keep going?