Presentation base change and transport of explicit bases to commutative specializations

The map $\mathrm{\Theta }$. Let $\rho :R⟨X⟩\to S⟨X⟩$ be the $R$-algebra homomorphism extending the generator map $X\to S⟨X⟩$ (which need not be injective when $S$ is the zero ring), existing by [F1]. The pairing $\beta :S\times R⟨X⟩\to S⟨X⟩$, $\beta (s,y):=s\rho (y)$, is additive in each variable and satisfies $\beta (s\phi (r),y)=s\phi (r)\rho (y)=s\rho (ry)=\beta (s,ry)$ because $\rho$ is $R$-linear [F3, F1], so by [F2] it induces an $R$-linear map $\mathrm{\Theta }:S{\otimes }_{R}R⟨X⟩\to S⟨X⟩$ with $\mathrm{\Theta }(s\otimes y)=s\rho (y)$; it is $S$-linear because $\mathrm{\Theta }(s^{\mathrm{\prime }}(s\otimes y))=s^{\mathrm{\prime }}s\rho (y)=s^{\mathrm{\prime }}\mathrm{\Theta }(s\otimes y)$. The map $S\to S{\otimes }_{R}R⟨X⟩$, $s\mapsto s\otimes 1$, is unital and central by [F3], so the tensor product is an $S$-algebra with $(s\otimes y)(s^{\mathrm{\prime }}\otimes y^{\mathrm{\prime }})=s{s}^{\mathrm{\prime }}\otimes y{y}^{\mathrm{\prime }}$, and $\mathrm{\Theta }((s\otimes y)(s^{\mathrm{\prime }}\otimes y^{\mathrm{\prime }}))=s{s}^{\mathrm{\prime }}\rho (y{y}^{\mathrm{\prime }})=s{s}^{\mathrm{\prime }}\rho (y)\rho (y^{\mathrm{\prime }})=\mathrm{\Theta }(s\otimes y)\mathrm{\Theta }(s^{\mathrm{\prime }}\otimes y^{\mathrm{\prime }})$ and $\mathrm{\Theta }(1\otimes 1)=1$; so $\mathrm{\Theta }$ is a unital $S$-algebra homomorphism, and $\mathrm{\Theta }(s\otimes w)=sw$ on a word $w$.

Part 2. The quotient map $\pi :R⟨X⟩\to R⟨X⟩\mathrm{/}I$ is a surjective $R$-algebra homomorphism with kernel $I$, so the sequence $I\to R⟨X⟩\to R⟨X⟩\mathrm{/}I\to 0$ is exact, and tensoring with $S$ over $R$ gives an exact sequence whose middle map is ${\mathrm{i}\mathrm{d}}_S\otimes \pi$ with kernel $J:=\mathrm{im}(S{\otimes }_{R}I)$ by [F4]. The set $J$ is a two-sided ideal of the $S$-algebra $S{\otimes }_{R}R⟨X⟩$: it consists of finite sums ${\sum }_i{s}_i\otimes e_i$ with $e_i\in I$ and is an additive subgroup, left multiplication by an elementary tensor gives $(s\otimes y)(s_i\otimes e_i)=s{s}_i\otimes y{e}_i$ with $y{e}_i\in I$, right multiplication gives $(s_i\otimes e_i)(s\otimes y)=s_{i}s\otimes e_{i}y$ with $e_{i}y\in I$, and additivity extends both closures to all of $S{\otimes }_{R}R⟨X⟩$. Moreover $J$ is generated as an ideal by the images $1\otimes e$ of the relations, since $s_i\otimes e_i=(s_i\otimes 1)(1\otimes e_i)$. By [F5] the surjective $S$-algebra homomorphism ${\mathrm{i}\mathrm{d}}_S\otimes \pi$, which kills $J$, factors through a surjective $S$-algebra homomorphism $\bar{\mathrm{\Theta }}:(S{\otimes }_{R}R⟨X⟩)\mathrm{/}J\to S{\otimes }_R(R⟨X⟩\mathrm{/}I)$ with kernel $J\mathrm{/}J=0$, hence an isomorphism. Its inverse is induced on the quotient by the pairing $(s,y+I)\mapsto s\otimes y+J$, which is well defined because $y^{\mathrm{\prime }}\in y+I$ gives $s\otimes y^{\mathrm{\prime }}-s\otimes y=s\otimes (y^{\mathrm{\prime }}-y)\in J$ and is bilinear by [F2]; the two maps are inverse because they are inverse on the spanning elements $s\otimes (y+I)$ and $s\otimes y+J$. The inverse is the unique $S$-algebra map with these prescribed values, since the tensors $s\otimes (y+I)$ span its source.

Part 3. Let $A$ be free with basis $(a_i{)}_{i\in I}$, so every $y\in A$ has a unique expansion $y={\sum }_i{c}_i{a}_i$ by [F6]. The pairing $\gamma :S\times A\to S^{(I)}$, $\gamma (s,{\sum }_i{c}_i{a}_i):={\sum }_i(s\phi (c_i))e_i$, is additive in each variable and satisfies $\gamma (s\phi (r),y)=\gamma (s,ry)$ because the coordinates of $ry$ are $r{c}_i$, and $\phi (r{c}_i)=\phi (r)\phi (c_i)$; by [F2] it induces an $S$-linear map $\mathrm{\Gamma }:S{\otimes }_{R}A\to S^{(I)}$ with $\mathrm{\Gamma }(s\otimes y)={\sum }_{i}s\phi (c_i)e_i$, in particular $\mathrm{\Gamma }(1\otimes a_i)=e_i$, and the $S$-linear map $\mathrm{\Delta }:S^{(I)}\to S{\otimes }_{R}A$ with $\mathrm{\Delta }(e_i)=1\otimes a_i$, existing by [F6], satisfies $\mathrm{\Gamma }\mathrm{\Delta }=\mathrm{i}\mathrm{d}$ on the standard basis and $\mathrm{\Delta }\mathrm{\Gamma }(1\otimes a_i)=1\otimes a_i$; since elementary tensors and basis elements span, $\mathrm{\Gamma }$ and $\mathrm{\Delta }$ are mutually inverse isomorphisms. Hence $(1\otimes a_i)$ is an $S$-basis of $S{\otimes }_{R}A$, and an explicitly constructed basis isomorphism of a presented $R$-algebra may be tensored with $S$ to transport the basis to every commutative specialization.

Inverse for part 1. The map $X\to S{\otimes }_{R}R⟨X⟩$, $x\mapsto 1\otimes x$, extends by the universal property of [F1] applied over $S$ to an $S$-algebra homomorphism $\mathrm{\Lambda }:S⟨X⟩\to S{\otimes }_{R}R⟨X⟩$, and for a word $w=x_{i_1}\cdots x_{i_n}$ one has $\mathrm{\Lambda }(\rho (w))=\mathrm{\Lambda }(x_{i_1}\cdots x_{i_n})={\prod }_k\mathrm{\Lambda }(x_{i_k})={\prod }_k(1\otimes x_{i_k})=1\otimes w$ by multiplicativity and [F3]; hence $\mathrm{\Lambda }\mathrm{\Theta }(s\otimes w)=\mathrm{\Lambda }(sw)=s(1\otimes w)=s\otimes w$ for every $s\in S$ and word $w$, so $\mathrm{\Lambda }\mathrm{\Theta }=\mathrm{i}\mathrm{d}$ on the spanning elementary tensors $s\otimes w$. Conversely $\mathrm{\Theta }\mathrm{\Lambda }$ and the identity of $S⟨X⟩$ are unital $S$-algebra homomorphisms agreeing on the generators $x\in X$, so by the uniqueness in [F1] they are equal. Thus $\mathrm{\Theta }$ is an $S$-algebra isomorphism, unique because it is determined on the spanning elementary tensors, and $\mathrm{\Lambda }$ sends a word to $1\otimes w$.

Collecting: step 1.1 constructs $\mathrm{\Theta }$ and step 2.1 proves it is the unique $S$-algebra isomorphism of part 1 with the stated values; step 1.2 proves the quotient identification of part 2 with the image ideal generated by the relations and no flatness hypothesis; step 1.3 proves the free-basis transport of part 3.