Regularity ascends and descends along a flat local homomorphism

The map is faithfully flat. Since the homomorphism is local, $\mathfrak{m}S\subseteq \mathfrak{n}⊊S$; for every proper ideal $I⊊R$ the ideal $I$ is contained in the maximal ideal $\mathfrak{m}$ by [F20], so $IS\subseteq \mathfrak{m}S$ is proper. Hence $R\to S$ is faithfully flat by [F4].

Descent, the case $d\ge 1$: setting up the resolution. Assume $S$ regular of dimension $d\ge 1$ and put $k:=R\mathrm{/}\mathfrak{m}$. By [F7] choose a resolution $F_{\bullet }\to k\to 0$ by finite free $R$-modules and let $\mathrm{\Omega }$ be its $d$-th syzygy, a finitely generated $R$-module by [F12]. Tensoring with the flat $R$-module $S$ preserves exactness by [F8], and [F9] identifies the tensor of the augmentation with $S\mathrm{/}\mathfrak{m}S$, so $F_{\bullet }{\otimes }_{R}S\to S\mathrm{/}\mathfrak{m}S\to 0$ is a free resolution of the $S$-module $S\mathrm{/}\mathfrak{m}S$ whose $d$-th syzygy is $\mathrm{\Omega }{\otimes }_{R}S$.

Ascent, set-up. Assume $R$ regular of dimension $d$ with regular system of parameters $y_1,\dots ,y_d$, so that $(y_1,\dots ,y_d)=\mathfrak{m}$ by [F2]; if $d=0$ then $\mathfrak{m}=0$ and $S=S\mathrm{/}\mathfrak{m}S$ is regular by hypothesis. For $d\ge 1$, the parameters are an $R$-regular sequence by [F17]; hence [F5] and the faithful flatness of step 1.1 make $y_1,\dots ,y_d$ an $S$-regular sequence, and $S\mathrm{/}(y_1,\dots ,y_d)S=S\mathrm{/}\mathfrak{m}S$.

Descent, the case $d=0$. Assume now that $S$ is regular of dimension $d=0$. Then $S$ is a domain by [F17] and hence a field by [F18]. Because $R\to S$ is faithfully flat by step 1.1, the extension $\mathfrak{m}S$ is proper by [F4], and $\mathfrak{m}S$ is an ideal of the field $S$, so $\mathfrak{m}S=0$; moreover $R\to S$ is injective by [F16], so $\mathfrak{m}=0$. A nonzero element $x$ of the local ring $R$ is then a unit: otherwise $(x)$ would be a proper ideal, hence would lie in a maximal ideal by [F19], necessarily the unique maximal ideal $\mathfrak{m}=0$, forcing $x=0$. Thus $R$ is a field by [F21], in particular regular.

Descent, the case $d\ge 1$: the syzygy is projective. Since $S$ is regular local of dimension $d$, its global dimension is $d$ by [F15], so every $S$-module, in particular $S\mathrm{/}\mathfrak{m}S$, has projective dimension at most $d$; by [F10] applied to the resolution of step 1.2, the $S$-module $\mathrm{\Omega }{\otimes }_{R}S$ is projective, hence flat over $S$ by [F11].

Ascent, induction. For $0\le i\le d$ put $S_i:=S\mathrm{/}(y_1,\dots ,y_i)S$, a nonzero Noetherian local ring, so that $S_0=S$, $S_d=S\mathrm{/}\mathfrak{m}S$ and $S_i\mathrm{/}(y_{i+1}{S}_i)\cong S_{i+1}$ for $i<d$ with the image of $y_{i+1}$ a nonzerodivisor on $S_i$. If $S_{i+1}$ is regular for some $0\le i<d$, then [F3] applied to the nonzero Noetherian local ring $S_i$ and the nonzerodivisor $y_{i+1}$ makes $S_i$ regular. Since $S_d=S\mathrm{/}\mathfrak{m}S$ is regular by hypothesis when $d\ge 1$, downward induction gives that $S=S_0$ is regular; together with the case $d=0$ of step 2.1 this proves the ascent claim.

Descent, conclusion. By [F13] and the faithful flatness of step 1.1, the finite $R$-module $\mathrm{\Omega }$ is flat over $R$, hence finite projective by [F14]. Therefore the resolution $F_{\bullet }\to k$ of step 1.2 has projective $d$-th syzygy, so ${\mathrm{pd}}_{R}k\le d$ by [F10], and [F15] makes the Noetherian local ring $R$ regular. Combined with step 2.2 this proves the descent claim for every $d$.

Both claims are proved: the ascent in step 3.1 and the descent in steps 2.2 and 3.2; the case $d=0$ was separated out in steps 2.2 and 3.2 because [F10] requires $n\ge 1$. ∎