quotient and lifting regularity across a regular element

For a nonzerodivisor $x\in \mathfrak{m}$, put $d=\dim R$ and $t=\dim R\mathrm{/}(x)$. Both dimensions are finite by the embedding bound. Lifting $t$ radical generators gives $d\le t+1$. Every prime chain containing $x$ can be extended strictly downwards by a minimal prime $\mathfrak{q}$ of $R$. By [F6], $\mathfrak{q}={\mathrm{Ann}}_R(m)$ for some nonzero $m\in R$. If $x\in \mathfrak{q}$, then $xm=0$, contradicting that $x$ is a nonzerodivisor. Thus every minimal prime omits $x$, the chain extends by one, and $t+1\le d$. Hence $t=d-1$.

If $R\mathrm{/}(x)$ is regular, lift its $d-1$ maximal-ideal generators and adjoin $x$. This gives $\mathrm{edim}R\le d$, and the embedding bound makes it equality. If $x$ were in ${\mathfrak{m}}^2$, cotangent reduction would leave dimension unchanged, giving $\mathrm{edim}(R\mathrm{/}(x))=d$, contrary to $d-1$.

In a regular local ring, a nonzero $x$ is a nonzerodivisor: the domain property is exactly F5. Thus the preceding implication applies. In the other direction, $x\mathrm{\notin }{\mathfrak{m}}^2$ makes the quotient regular by the parameter-quotient lemma. There is no $0\mathrm{\ne }x\in \mathfrak{m}$ when the regular ring has dimension zero; in dimension one the regular quotient is a field.