Multivariate polynomial and Laurent rings over commutative rings, domains and fraction fields

Multivariate basis and universal property, by induction on $n$. For $n=0$, $R[x_1,\dots ,x_0]=R$ by [F4], the single monomial $x^{(0)}=1$ is a basis of the free rank-one module $R$ by [F6], and the structure map of $A$ is the unique $R$-algebra homomorphism $R\to A$ by [F11]. For the step, put $S:=R[x_1,\dots ,x_n]$, free over $R$ on the monomials $x^{\alpha }$ by induction, and $R[x_1,\dots ,x_{n+1}]=S[x_{n+1}]$ by [F4]: by [F1] and [F2] (applied over the coefficient ring $S$) an element of $S[x_{n+1}]$ is a finitely supported function $k\mapsto s_k$ with $s_k\in S$, each $s_k={\sum }_{\alpha }{a}_{\alpha ,k}{x}^{\alpha }$ a finite $R$-linear combination, so substituting the unique coefficient expansions gives a unique finite $R$-linear combination of the monomials $x^{\alpha }{x}_{n+1}^k$, and these therefore form an $R$-basis indexed by ${\mathbb{N}}^n\times \mathbb{N}\cong {\mathbb{N}}^{n+1}$; applying [F3] twice, a unital $R$-algebra homomorphism $S[x_{n+1}]\to A$ is exactly a unital $R$-algebra homomorphism $S\to A$ together with an element $a\in A$ (the image of $x_{n+1}$), which by induction is exactly images $a_1,\dots ,a_{n+1}\in A$ of the generators.

If $R$ is an integral domain, $R[x_1,\dots ,x_n]$ is one by [F5].

Construction of ${\mathrm{\Lambda }}_{R,n}$: let ${\mathrm{\Lambda }}_{R,n}:=R^{({\mathbb{Z}}^n)}$ be the free $R$-module with standard basis the monomials $x^{\alpha }$ by [F6], and define multiplication on the basis by $x^{\alpha }{x}^{\beta }:=x^{\alpha +\beta }$, extended $R$-bilinearly, so that $(ab{)}_{\gamma }={\sum }_{\begin{array}{c}\alpha \in \mathrm{supp}a,\beta \in \mathrm{supp}b\\ \alpha +\beta =\gamma \end{array}}{a}_{\alpha }{b}_{\beta }$ is a finite sum by [F8], and $\mathrm{supp}(ab)\subseteq \mathrm{supp}a+\mathrm{supp}b$ is finite. The rule is closed and associative because ${\mathbb{Z}}^n$ has associative, commutative coordinatewise addition by [F12] and reindexing the finite triple sum gives $((ab)c{)}_{\gamma }=(a(bc){)}_{\gamma }={\sum }_{\alpha +\beta +\delta =\gamma }{a}_{\alpha }{b}_{\beta }{c}_{\delta }$ by [F8]; it is commutative because $\alpha +\beta =\beta +\alpha$ and $R$ is commutative; it is distributive over the coefficientwise addition inherited from [F6]; and the basis vector $x^0$, coefficient $1_R$ at $0$ and $0$ elsewhere, is a two-sided identity. Hence ${\mathrm{\Lambda }}_{R,n}$ is a commutative ring, it is an $R$-algebra through $r\mapsto r{x}^0$ by [F11], it is free on the monomials by construction, and each $x_i$ is a unit since the monomial $x^{-e_i}$ with coefficient $1_R$ satisfies $x_i{x}^{-e_i}=x^{e_i-e_i}=x^0=1$.

Universal property of ${\mathrm{\Lambda }}_{R,n}$: let $u_1,\dots ,u_n\in A$ be units and for $\alpha \in {\mathbb{Z}}^n$ put $u^{\alpha }:=u_1^{{\alpha }_1}\cdots u_n^{{\alpha }_n}$, negative exponents denoting powers of the inverses. Every element of ${\mathrm{\Lambda }}_{R,n}$ is a unique finite sum ${\sum }_{\alpha }{a}_{\alpha }{x}^{\alpha }$ by [F6], so $x^{\alpha }\mapsto u^{\alpha }$ extends uniquely to an $R$-linear map $\mathrm{\Phi }:{\mathrm{\Lambda }}_{R,n}\to A$ by [F7]; it is a unital ring homomorphism because $u^{\alpha +\beta }=u^{\alpha }{u}^{\beta }$ and $u^0=1$ in $A$ by [F13], applied in the unit group of $A$, which is abelian because $A$ is commutative, and it is the unique $R$-algebra homomorphism with $x_i\mapsto u_i$ because the monomials span.

Domain of ${\mathrm{\Lambda }}_{R,n}$: suppose $R$ is an integral domain. Order ${\mathbb{Z}}^n$ lexicographically: distinct tuples are compared at their first differing coordinate. By [F12] the integer order is total, and adding the same tuple preserves that first differing coordinate and its strict comparison; hence this is a translation-invariant total order (for $n=0$, there is just the empty tuple). For nonzero $a,b\in {\mathrm{\Lambda }}_{R,n}$ the finite supports $\mathrm{supp}a,\mathrm{supp}b$ are nonempty and have greatest elements $\alpha ,\beta$. A coefficient $(ab{)}_{\gamma }$ with $\gamma >\alpha +\beta$ is a sum of products $a_{{\alpha }^{\mathrm{\prime }}}{b}_{{\beta }^{\mathrm{\prime }}}$ with ${\alpha }^{\mathrm{\prime }}+{\beta }^{\mathrm{\prime }}=\gamma$, and each such term has ${\alpha }^{\mathrm{\prime }}>\alpha$ (then $a_{{\alpha }^{\mathrm{\prime }}}=0$) or ${\beta }^{\mathrm{\prime }}>\beta$ (then $b_{{\beta }^{\mathrm{\prime }}}=0$), since ${\alpha }^{\mathrm{\prime }}\le \alpha$ and ${\beta }^{\mathrm{\prime }}\le \beta$ would give ${\alpha }^{\mathrm{\prime }}+{\beta }^{\mathrm{\prime }}\le \alpha +\beta <\gamma$; so $(ab{)}_{\gamma }=0$ for $\gamma >\alpha +\beta$. For $\gamma =\alpha +\beta$ every term indexed within the supports with ${\alpha }^{\mathrm{\prime }}\mathrm{\ne }\alpha$ has ${\alpha }^{\mathrm{\prime }}<\alpha$ and then ${\beta }^{\mathrm{\prime }}>\beta$ by strict order invariance, hence vanishes, and the remaining term is $a_{\alpha }{b}_{\beta }\mathrm{\ne }0$ because $R$ has no zero divisors [F9]; so $(ab{)}_{\alpha +\beta }\mathrm{\ne }0$ and $ab\mathrm{\ne }0$. Since $1_R\mathrm{\ne }0$ in $R$, the unit of ${\mathrm{\Lambda }}_{R,n}$ differs from $0$, so ${\mathrm{\Lambda }}_{R,n}$ is an integral domain.

Fraction field for a commutative integral domain $D$: on $P:=\{(f,g):f,g\in D,g\mathrm{\ne }0\}$ define $(f,g)\sim (f^{\mathrm{\prime }},g^{\mathrm{\prime }})$ iff $f{g}^{\mathrm{\prime }}=f^{\mathrm{\prime }}g$. This is reflexive and symmetric, and transitive: from $f{g}^{\mathrm{\prime }}=f^{\mathrm{\prime }}g$ and $f^{\mathrm{\prime }}{g}^{\mathrm{\prime }\mathrm{\prime }}=f^{\mathrm{\prime }\mathrm{\prime }}{g}^{\mathrm{\prime }}$ one gets $g^{\mathrm{\prime }}(f{g}^{\mathrm{\prime }\mathrm{\prime }})=(f{g}^{\mathrm{\prime }})g^{\mathrm{\prime }\mathrm{\prime }}=(f^{\mathrm{\prime }}g)g^{\mathrm{\prime }\mathrm{\prime }}=f^{\mathrm{\prime }}(g{g}^{\mathrm{\prime }\mathrm{\prime }})=f^{\mathrm{\prime }}(g^{\mathrm{\prime }\mathrm{\prime }}g)=(f^{\mathrm{\prime }}{g}^{\mathrm{\prime }\mathrm{\prime }})g=(f^{\mathrm{\prime }\mathrm{\prime }}{g}^{\mathrm{\prime }})g=f^{\mathrm{\prime }\mathrm{\prime }}(g^{\mathrm{\prime }}g)=g^{\mathrm{\prime }}(f^{\mathrm{\prime }\mathrm{\prime }}g)$, so cancellation of the nonzero $g^{\mathrm{\prime }}$ in the domain gives $f{g}^{\mathrm{\prime }\mathrm{\prime }}=f^{\mathrm{\prime }\mathrm{\prime }}g$. Sums $(f,g)+(f^{\mathrm{\prime }},g^{\mathrm{\prime }})=(f{g}^{\mathrm{\prime }}+f^{\mathrm{\prime }}g,g{g}^{\mathrm{\prime }})$ and products $(f,g)(f^{\mathrm{\prime }},g^{\mathrm{\prime }})=(f{f}^{\mathrm{\prime }},g{g}^{\mathrm{\prime }})$ have nonzero second components since $D$ has no zero divisors, and they respect $\sim$: if $f{g}^{\mathrm{\prime }}=f^{\mathrm{\prime }}g$ and $h{k}^{\mathrm{\prime }}=h^{\mathrm{\prime }}k$ then $(fk+hg)(g^{\mathrm{\prime }}{k}^{\mathrm{\prime }})=fk{g}^{\mathrm{\prime }}{k}^{\mathrm{\prime }}+hg{g}^{\mathrm{\prime }}{k}^{\mathrm{\prime }}=f^{\mathrm{\prime }}{k}^{\mathrm{\prime }}(gk)+h^{\mathrm{\prime }}{g}^{\mathrm{\prime }}(gk)=(f^{\mathrm{\prime }}{k}^{\mathrm{\prime }}+h^{\mathrm{\prime }}{g}^{\mathrm{\prime }})(gk)$ and $fh{g}^{\mathrm{\prime }}{k}^{\mathrm{\prime }}=f^{\mathrm{\prime }}g{h}^{\mathrm{\prime }}k=f^{\mathrm{\prime }}{h}^{\mathrm{\prime }}gk$, so the operations are well defined on the quotient set $D^{\mathrm{\prime }}=P\mathrm{/}\sim$ of [F10]. Writing $f\mathrm{/}g$ for $[(f,g)]$, addition of $f\mathrm{/}g,h\mathrm{/}k,l\mathrm{/}m$ in either order gives $(fkm+hgm+lgk)\mathrm{/}(gkm)$, multiplication in either order gives $fhl\mathrm{/}(gkm)$, and distributivity gives $f(hm+lk)\mathrm{/}(gkm)$ on both sides. Commutativity follows from that in $D$; $0\mathrm{/}1$ and $1\mathrm{/}1$ are the identities, and $(-f)\mathrm{/}g$ is the additive inverse of $f\mathrm{/}g$. Thus $D^{\mathrm{\prime }}$ is a commutative ring, with $0\mathrm{/}1\mathrm{\ne }1\mathrm{/}1$ because $0\mathrm{\ne }1$ in $D$, and $D^{\mathrm{\prime }}$ is a field: for a class $[(f,g)]\mathrm{\ne }[(0,1)]$ one has $f\mathrm{\ne }0$, so that $(f,g)(g,f)=(fg,fg)\sim (1,1)$ since $fg\cdot 1=1\cdot fg$, and hence $[(g,f)]$ is an inverse. Finally $f\mapsto [(f,1)]$ is a unital ring homomorphism with kernel $\{f:[(f,1)]=[(0,1)]\}=\{f:f=0\}$, so it injects $D$ into $D^{\mathrm{\prime }}$.

Collecting the clauses: step 1.1 gives the monomial basis, the universal property and, with [F5] as used in step 1.2, the domain assertion of the multivariate clause; step 1.3 gives the ring structure, freeness and the unit property of ${\mathrm{\Lambda }}_{R,n}$, step 1.4 its unit-substitution universal property, and step 1.5 its domain assertion. If $R$ is an integral domain then $D:={\mathrm{\Lambda }}_{R,n}$ is a domain by step 1.5 and $R[x_1,\dots ,x_n]$ is a domain by step 1.2, so step 1.6 applies to both and yields the fraction field $K({\mathrm{\Lambda }}_{R,n})$ and the fraction field of $R[x_1,\dots ,x_n]$ with the embedding $f\mapsto [(f,1)]$, which completes all three parts of the claim.