cusp local ring not regular

The quotient $A=k[x,y]\mathrm{/}(y^2-x^3)$ has unique representatives $a(x)+yb(x)$ by division by the monic polynomial in $y$. Under $x\mapsto t^2$, $y\mapsto t^3$, the two summands have even and odd powers of $t$, respectively; their vanishing forces both to be zero. Hence $A$ embeds in $k[t]$ and is a domain in every characteristic. The origin ideal remains a proper nonzero maximal ideal after localization.

The ambient local domain $S=k[x,y{]}_{(x,y)}$ has cotangent basis $x,y$: modulo $(x,y{)}^2$, every localized polynomial has a unique constant and linear part because its denominator has nonzero constant term. Hence $\mathrm{edim}S=2$. The chain $(0)⊊(x)S⊊(x,y)S$ gives $\dim S\ge 2$, while [F1] gives $\dim S\le 2$.

The kernel $(f)S$ of $S\to R$ is a nonzero prime by step 1.1. Since $R$ is a nonfield local domain, $(0)⊊(x,y)R$ gives $\dim R\ge 1$. A chain of length two in $R$ would lift to a chain of primes $(0)⊊(f)S⊊\mathfrak{p}⊊(x,y)S$ of length three in $S$, contradicting step 1.2. Thus $\dim R=1$. As $f\in (x,y{)}^{2}S$, the quotient introduces no relation in the cotangent space, so $\mathrm{edim}R=2$. By [F2], $R$ is not regular. The argument works in every characteristic.