Let $k$ be a ring, put $k[\epsilon] := k[t]/(t^2)$. What is the cotangent complex of $k[\epsilon] \to k$?
I know $\Omega^1_{k/k[\epsilon]]}$ is going to be zero. But I don't see any way around explicitly writing down a cofibrant replacement to get the whole complex. I tried to do something like a bar construction for this, but I can't get it to work.
If it helps, I'm happy to assume $k$ is a field.
edit: I've gotten an idea, would still appreciate a confirmation. Look at $k[T] \to k[\epsilon] \to k$. Then the fundamental triangle gives \begin{align*} \mathcal{L}_{k[\epsilon]/k[T]} \otimes_{k[\epsilon]} k \to \mathcal{L}_{k / k[T]} \to \mathcal{L}_{k/k[\epsilon]} \end{align*}
We know that $f:k[T] \to k[\epsilon]$ is regular since $T^2$ is not a zero-divisor in $k[T]$. Let $I$ be the kernel of $f$, i.e. $I=(T^2)$. Then we know that the cotangent complex of $f$ is the conormal sheaf of $\mathrm{Spec} k[\epsilon] \to \mathrm{Spec}k[T]$ concentrated in degree 1. We can thus compute the first term in the above triangle as \begin{align*} \mathcal{L}_{k[\epsilon]/k[T]} \otimes_{k[\epsilon]} k \simeq (I/I^2)[1] \otimes_{k[\epsilon]} k \end{align*} which I think comes out to be $k[1]$.
Since we also know that $\mathcal{L}_{k / k[t]}$ is zero, this gives us $\mathcal{L}_{k/k[\epsilon]} = k[2]$.
Is this answer believable?