Localisation of Modules and Support

Reflexivity holds because $1(sm-sm)=0$, so $(m,s)\sim (m,s)$ for every $(m,s)\in M\times S$.

Symmetry holds because $u(tm-sn)=0$ implies $u(sn-tm)=0$, so $(m,s)\sim (n,t)$ implies $(n,t)\sim (m,s)$.

If $(m,s)\sim (n,t)$ via $u$ and $(n,t)\sim (p,v)$ via $w$, then $uwv(tm-sn)=0$ and uws$(vn-tp)=0$; adding these equalities gives $uwt(vm-sp)=0$, so $(m,s)\sim (p,v)$.

Steps 1.1, 1.2, and 1.3 prove that $\sim$ is an equivalence relation.

By [L1], choose $u,v\in S$ with $u(s^{\mathrm{\prime }}m-s{m}^{\mathrm{\prime }})=0$ and $v(t^{\mathrm{\prime }}n-t{n}^{\mathrm{\prime }})=0$.

Multiplying the first equality by $vt{t}^{\mathrm{\prime }}$ and the second by $us{s}^{\mathrm{\prime }}$ and then adding gives $uv(s^{\mathrm{\prime }}{t}^{\mathrm{\prime }}(tm+sn)-st(t^{\mathrm{\prime }}{m}^{\mathrm{\prime }}+s^{\mathrm{\prime }}{n}^{\mathrm{\prime }}))=0$.

Step 2.1 is exactly the relation witnessing $(tm+sn)\mathrm{/}(st)=(t^{\mathrm{\prime }}{m}^{\mathrm{\prime }}+s^{\mathrm{\prime }}{n}^{\mathrm{\prime }})\mathrm{/}(s^{\mathrm{\prime }}{t}^{\mathrm{\prime }})$, so the addition formula is independent of representatives.

By [L1] and [L2], choose $v,w\in S$ with $v(u^{\mathrm{\prime }}a-u{a}^{\mathrm{\prime }})=0$ and $w(s^{\mathrm{\prime }}m-s{m}^{\mathrm{\prime }})=0$.

Multiplying the module-fraction equality by $auv$ and the ring-fraction equality by $us{m}^{\mathrm{\prime }}w$, then adding, gives $uvw(u^{\mathrm{\prime }}{s}^{\mathrm{\prime }}am-us{a}^{\mathrm{\prime }}{m}^{\mathrm{\prime }})=a{u}^{\mathrm{\prime }}vw(s^{\mathrm{\prime }}m-s{m}^{\mathrm{\prime }})+us{m}^{\mathrm{\prime }}vw(u^{\mathrm{\prime }}a-u{a}^{\mathrm{\prime }})=0.$ So $uvw(u^{\mathrm{\prime }}{s}^{\mathrm{\prime }}am-us{a}^{\mathrm{\prime }}{m}^{\mathrm{\prime }})=0$.

Step 2.1 is exactly the relation witnessing $am\mathrm{/}(us)=a^{\mathrm{\prime }}{m}^{\mathrm{\prime }}\mathrm{/}(u^{\mathrm{\prime }}{s}^{\mathrm{\prime }})$, so the scalar action is independent of representatives.

Define $\tilde{f}(m\mathrm{/}s):=(1\mathrm{/}s)f(m)$. If $m\mathrm{/}s=m^{\mathrm{\prime }}\mathrm{/}{s}^{\mathrm{\prime }}$, choose $u\in S$ with $u(s^{\mathrm{\prime }}m-s{m}^{\mathrm{\prime }})=0$; applying $f$ gives $(u\mathrm{/}1)((s^{\mathrm{\prime }}\mathrm{/}1)f(m)-(s\mathrm{/}1)f(m^{\mathrm{\prime }}))=0$, and multiplying by the units $(u\mathrm{/}1{)}^{-1}(s\mathrm{/}1{)}^{-1}(s^{\mathrm{\prime }}\mathrm{/}1{)}^{-1}$ from [L4] gives $(1\mathrm{/}s)f(m)=(1\mathrm{/}{s}^{\mathrm{\prime }})f(m^{\mathrm{\prime }})$.

For $m\mathrm{/}s,n\mathrm{/}t\in S^{-1}M$, $\tilde{f}((m\mathrm{/}s)+(n\mathrm{/}t))=(1\mathrm{/}st)f(tm+sn)=(1\mathrm{/}s)f(m)+(1\mathrm{/}t)f(n)=\tilde{f}(m\mathrm{/}s)+\tilde{f}(n\mathrm{/}t)$, and for $a\mathrm{/}u\in S^{-1}R$ one has $\tilde{f}((a\mathrm{/}u)(m\mathrm{/}s))=(1\mathrm{/}us)f(am)=(a\mathrm{/}u)\tilde{f}(m\mathrm{/}s)$.

For every $m\in M$, $\tilde{f}({\lambda }_M(m))=\tilde{f}(m\mathrm{/}1)=f(m)$.

If $g:S^{-1}M\to N$ is $S^{-1}R$-linear and $g{\lambda }_M=f$, then for every $m\mathrm{/}s$ one has $g(m\mathrm{/}s)=g((1\mathrm{/}s)(m\mathrm{/}1))=(1\mathrm{/}s)g(m\mathrm{/}1)=(1\mathrm{/}s)f(m)=\tilde{f}(m\mathrm{/}s)$, so $g=\tilde{f}$.

Steps 1.1, 2.1, 2.2, and 3.1 prove the stated unique $S^{-1}R$-linear factorisation.

The pairing $q((a\mathrm{/}s),m)=am\mathrm{/}s$ is balanced because it is additive in each variable and $q(((a\mathrm{/}s)r),m)=arm\mathrm{/}s=q(a\mathrm{/}s,rm)$ for every $r\in R$.

The map $i:M\to (S^{-1}R){\otimes }_{R}M$, $i(m)=(1\mathrm{/}1)\otimes m$, is $R$-linear, and every $s\in S$ acts invertibly on the target because $(s\mathrm{/}1{)}^{-1}=1\mathrm{/}s$ in $S^{-1}R$.

By [L1], step 1.1 induces a unique homomorphism $\mathrm{\Phi }:(S^{-1}R){\otimes }_{R}M\to S^{-1}M$ with $\mathrm{\Phi }((a\mathrm{/}s)\otimes m)=am\mathrm{/}s$.

By [L3], step 1.2 induces a unique $S^{-1}R$-linear map $\mathrm{\Psi }:S^{-1}M\to (S^{-1}R){\otimes }_{R}M$ with $\mathrm{\Psi }(m\mathrm{/}s)=(1\mathrm{/}s)\otimes m$.

For every $m\mathrm{/}s\in S^{-1}M$, $(\mathrm{\Phi }\mathrm{\Psi })(m\mathrm{/}s)=\mathrm{\Phi }((1\mathrm{/}s)\otimes m)=m\mathrm{/}s$.

For every elementary tensor $(a\mathrm{/}s)\otimes m$, $(\mathrm{\Psi }\mathrm{\Phi })((a\mathrm{/}s)\otimes m)=\mathrm{\Psi }(am\mathrm{/}s)=(1\mathrm{/}s)\otimes am=(a\mathrm{/}s)\otimes m$ by the tensor scalar action of [L2].

Steps 3.1 and 3.2 show that $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are inverse $S^{-1}R$-linear isomorphisms.

For a submodule $N\le M$, the sequence $N\to M\to M\mathrm{/}N\to 0$ is right exact, so [L2] and [L1] give a right-exact sequence $S^{-1}N\to S^{-1}M\to S^{-1}(M\mathrm{/}N)\to 0$. Therefore $S^{-1}(M\mathrm{/}N)$ is the quotient of $S^{-1}M$ by the image of $S^{-1}N\to S^{-1}M$, namely by the submodule $S^{-1}N$.

For a family $(M_i{)}_{i\in I}$, [L3] and [L1] give $S^{-1}({⨁}_i{M}_i)\cong (S^{-1}R){\otimes }_R({⨁}_i{M}_i)\cong {⨁}_i((S^{-1}R){\otimes }_R{M}_i)\cong {⨁}_i{S}^{-1}{M}_i$.

Thus $S^{-1}(M\mathrm{/}N)\cong (S^{-1}M)\mathrm{/}(S^{-1}N)$ naturally in $M$ and $N$.

Steps 2.1 and 1.2 prove the quotient and direct-sum claims.

If $m\mathrm{/}s=0\mathrm{/}1$, then [L1] gives $u(1m-s0)=um=0$ for some $u\in S$.

If $um=0$ for some $u\in S$, then $u(1m-s0)=0$, so [L1] gives $m\mathrm{/}s=0\mathrm{/}1=0$ in $S^{-1}M$.

Steps 1.1 and 1.2 prove the equivalence.

Suppose $(S^{-1}f)(m^{\mathrm{\prime }}\mathrm{/}s)=0$. Then $f(m^{\mathrm{\prime }})\mathrm{/}s=0$, so [L1] gives $uf(m^{\mathrm{\prime }})=0$ for some $u\in S$.

By [L2], $uf(m^{\mathrm{\prime }})=f(u{m}^{\mathrm{\prime }})$, so injectivity of $f$ gives $u{m}^{\mathrm{\prime }}=0$.

Applying [L1] again, step 2.1 gives $m^{\mathrm{\prime }}\mathrm{/}s=0$ in $S^{-1}{M}^{\mathrm{\prime }}$. Hence $S^{-1}f$ is injective.

Let $m^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}s\in S^{-1}{M}^{\mathrm{\prime }\mathrm{\prime }}$. Since $f$ is surjective, choose $m\in M$ with $f(m)=m^{\mathrm{\prime }\mathrm{\prime }}$.

Then $(S^{-1}f)(m\mathrm{/}s)=f(m)\mathrm{/}s=m^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}s$, so $m^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}s$ lies in the image of $S^{-1}f$.

Since every element of $S^{-1}{M}^{\mathrm{\prime }\mathrm{\prime }}$ is hit, $S^{-1}f$ is surjective.

By [L4], the tail $M^{\mathrm{\prime }}\stackrel{f}{\to }M\stackrel{g}{\to }{M}^{\mathrm{\prime }\mathrm{\prime }}\to 0$ is exact. Using [L1] to identify localisation with tensor product, [L2] gives an exact sequence $S^{-1}{M}^{\mathrm{\prime }}\stackrel{S^{-1}f}{\to }{S}^{-1}M\stackrel{S^{-1}g}{\to }{S}^{-1}{M}^{\mathrm{\prime }\mathrm{\prime }}\to 0$.

The map $f$ is injective by [L4], so [L3] makes $S^{-1}f$ injective.

Steps 1.1 and 1.2 show that the localised sequence is short exact.

The standard short exact sequences $0\to \ker f\to M\to \mathrm{im}f\to 0$ and $0\to \mathrm{im}f\to N\to \mathrm{coker}f\to 0$ localise, by [L1], to short exact sequences $0\to S^{-1}(\ker f)\to S^{-1}M\to S^{-1}(\mathrm{im}f)\to 0$ and $0\to S^{-1}(\mathrm{im}f)\to S^{-1}N\to S^{-1}(\mathrm{coker}f)\to 0$.

In the first localised sequence, the middle map is $S^{-1}f$ restricted to $S^{-1}M$, so its kernel is exactly $S^{-1}(\ker f)$ and its image is exactly the embedded copy of $S^{-1}(\mathrm{im}f)$.

In the second localised sequence, the quotient by the image of $S^{-1}f$ is therefore $S^{-1}(\mathrm{coker}f)$, so $S^{-1}(\mathrm{coker}f)\cong \mathrm{coker}(S^{-1}f)$.

Steps 2.1 and 3.1 prove the kernel, image, and cokernel identifications.

Let $\delta :M\to {⨁}_{i=1}^{r}M\mathrm{/}{N}_i$ be the diagonal map $\delta (m)=(m+N_1,\dots ,m+N_r)$. By construction, $\ker \delta ={\bigcap }_{i=1}^r{N}_i$.

By [L1], $S^{-1}({\bigcap }_i{N}_i)\cong \ker (S^{-1}\delta )$. By [L2], the codomain of $S^{-1}\delta$ identifies with ${⨁}_i(S^{-1}M\mathrm{/}{S}^{-1}{N}_i)$, and under this identification $S^{-1}\delta$ is the diagonal map $S^{-1}M\to {⨁}_i(S^{-1}M\mathrm{/}{S}^{-1}{N}_i)$.

An element of $S^{-1}M$ lies in the kernel of that diagonal map exactly when its image in every quotient $S^{-1}M\mathrm{/}{S}^{-1}{N}_i$ is zero, that is, exactly when it lies in every submodule $S^{-1}{N}_i$. Therefore $\ker (S^{-1}\delta )={\bigcap }_i{S}^{-1}{N}_i$.

Combining steps 2.1 and 3.1 gives $S^{-1}({\bigcap }_i{N}_i)={\bigcap }_i{S}^{-1}{N}_i$ inside $S^{-1}M$.

For each $R$-linear map $f:M\to N$, apply [L2] to the composite $M\stackrel{f}{\to }N\to S^{-1}N$ to obtain a unique $S^{-1}R$-linear map $S^{-1}f:S^{-1}M\to S^{-1}N$ with $(S^{-1}f)(m\mathrm{/}u)=f(m)\mathrm{/}u$.

The assignment $f\mapsto S^{-1}f$ is $R$-linear because for $f,g\in {\mathrm{Hom}}_R(M,N)$, $r\in R$, and $m\mathrm{/}u\in S^{-1}M$ one has $(S^{-1}(f+g))(m\mathrm{/}u)=(f(m)+g(m))\mathrm{/}u$ and $(S^{-1}(rf))(m\mathrm{/}u)=rf(m)\mathrm{/}u=(r\mathrm{/}1)(f(m)\mathrm{/}u)$.

By [L1], ${\mathrm{Hom}}_R(M,N)$ is an $R$-module, while ${\mathrm{Hom}}_{S^{-1}R}(S^{-1}M,S^{-1}N)$ is an $S^{-1}R$-module; therefore [L2] applied to the $R$-linear map of step 2.1 gives a unique $S^{-1}R$-linear map ${\theta }_{M,N}$ with ${\theta }_{M,N}(f\mathrm{/}1)=S^{-1}f$.

Since ${\theta }_{M,N}$ is $S^{-1}R$-linear, ${\theta }_{M,N}(f\mathrm{/}s)=(1\mathrm{/}s){\theta }_{M,N}(f\mathrm{/}1)$, so for $m\mathrm{/}u\in S^{-1}M$ one gets ${\theta }_{M,N}(f\mathrm{/}s)(m\mathrm{/}u)=(1\mathrm{/}s)(f(m)\mathrm{/}u)=f(m)\mathrm{/}(su)$.

Steps 3.1 and 4.1 produce the stated natural $S^{-1}R$-linear map.

By [L2] and [L6], every element of $S^{-1}M$ is an $S^{-1}R$-linear combination of $e_1\mathrm{/}1,\dots ,e_n\mathrm{/}1$, so two $S^{-1}R$-linear maps $S^{-1}M\to S^{-1}N$ that agree on all $e_i\mathrm{/}1$ are equal.

If ${\theta }_{M,N}(f\mathrm{/}s)=0$, then $f(e_i)\mathrm{/}s=0$ for each $i$, so [L4] gives $u_i\in S$ with $u_{i}f(e_i)=0$. Let $u=u_1\cdots u_n$. Then $uf(e_i)=0$ for every $i$, hence $uf=0$ by [L3]. Since ${\mathrm{Hom}}_R(M,N)$ is an $R$-module by [L5], [L4] applied there gives $f\mathrm{/}s=0$. Thus ${\theta }_{M,N}$ is injective.

Let $\phi :S^{-1}M\to S^{-1}N$ be $S^{-1}R$-linear. Write $\phi (e_i\mathrm{/}1)=n_i\mathrm{/}{s}_i$ with $n_i\in N$ and $s_i\in S$, set $s=s_1\cdots s_n$, and set $m_i=(s\mathrm{/}{s}_i)n_i$. By [L3], there is a unique $R$-linear map $f:M\to N$ with $f(e_i)=m_i$ for all $i$.

For each basis vector, ${\theta }_{M,N}(f\mathrm{/}s)(e_i\mathrm{/}1)=f(e_i)\mathrm{/}s=(s\mathrm{/}{s}_i)n_i\mathrm{/}s=n_i\mathrm{/}{s}_i=\phi (e_i\mathrm{/}1)$, so step 1.1 gives ${\theta }_{M,N}(f\mathrm{/}s)=\phi$. Thus ${\theta }_{M,N}$ is surjective.

Steps 1.2 and 2.1 prove that ${\theta }_{M,N}$ is an isomorphism.