Polynomial diagonal differences form a regular sequence

Put $S=k[X_1,\dots ,X_n,Y_1,\dots ,Y_n]$. By [F4] the assignments $X_i\mapsto x_i\otimes 1$ and $Y_i\mapsto 1\otimes x_i$ define a $k$-algebra map $S\to R{\otimes }_{k}R$. By [F3] the two polynomial maps from the copies of $R$ into $S$ induce a map $R{\otimes }_{k}R\to S$. The composites fix every polynomial generator, so these maps are inverse; [F1] identifies $u_i$ with $X_i-Y_i$.

For $1\le i\le n$, substitution $Y_r\mapsto X_r$ for $r<i$ gives $S\mathrm{/}(X_1-Y_1,\dots ,X_{i-1}-Y_{i-1})\cong k[X_1,\dots ,X_n,Y_i,\dots ,Y_n]$: the inverse includes the displayed remaining variables, and both composites fix their generators. This quotient is a nonzero polynomial ring over $k$.

Substituting every $Y_r\mapsto X_r$ gives $S\mathrm{/}(X_1-Y_1,\dots ,X_n-Y_n)\cong k[X_1,\dots ,X_n]$ by the same inverse-on-generators check. Under [F1] this is the multiplication quotient $R^e\mathrm{/}(u_1,\dots ,u_n)\cong R$, which is nonzero; for $n=0$ both rings are $k$ and the map is the identity.

In the quotient of step 2.1, write $C_i=k[X_1,\dots ,X_{i-1},X_{i+1},\dots ,X_n,Y_i,\dots ,Y_n]$, so that it is $C_i[X_i]$ and $Y_i\in C_i$. If $0\mathrm{\ne }f\in C_i[X_i]$ has degree $d$ and leading coefficient $c\mathrm{\ne }0$, then $(X_i-Y_i)f$ has degree $d+1$ with the same nonzero leading coefficient $c$. Thus multiplication by $u_i$ is injective on each successive quotient.

Steps 2.1 and 2.2 give every required nonzero successive quotient and the nonzero final quotient; step 3.1 gives injectivity at each position. By [F2] this is precisely regularity on $R^e$. When $n=0$, there are no injectivity conditions and the final quotient $k$ is nonzero, so the empty sequence is regular as well.