Does the nilpotent extension of a $1$-dimensional algebra always give a projective module?

Question

Let $A$ be a $1$-dimensional reduced Noetherian algebra over an algebraic closed field $k$ with characteristic zero. Let $(B,N)$ be a nilpotent extension of $A$, i.e. $B$ is a Noetherian $k$-algebra, $N$ is the nilpotent radical of $B$ and we have $$ B/N\overset{\cong}{\to}A. $$

Now we have the quotient $N/N^2$, which is an $A$-module. $\textbf{My question}$ is: is it always a projective $A$-module? If not, what is the counter-example?

Answer 1

I think that this is a counterexample: Consider $B = k[x,y]/(xy,y^3)$ so that the nilradical $N$ of $B$ is generated by $y$, $B/N\cong k[x]$, and $N/N^2 = (y)/(y^2)$ is a $1$-dimensional $k$-vector space, certainly not a projective $k[x]$-module.