Induced map between localizations

Consider a ring $R$, commutative ring with unity, and noetherian. Call $S\subset R$ the multiplicative set of the elements of $R$ that are not zero-divisors. Take also another multiplicative set $U\subset R$. I must prove that there is a canonical map $U^{-1}K(R)\to K(U^{-1}R)$, where for a generic ring $A$, we define $K(A)$ as the localization of $A$ to the multiplicative set of all elements that don't divide zero.

1. $K(R)=S^{-1}R$ by definition, $K(U^{-1}R)=S^{-1} U^{-1}R$.

2. the map $R\to U^{-1}R$ induces a map $S^{-1}R\to S^{-1}U^{-1}R$, that sends every $u/s$ to an invertible element. So the previous map factorizes through another map $f: U^{-1}S^{-1}R\to S^{-1}U^{-1}R$.

Now, this map should be defined $(r/s)/u\mapsto r/u/s$; can you make me an example of when it's not injective? I've been thinking for a while and I can't see how this could happen, thank you

Solution 1:

Let's work this out carefully. Let $\ell_U : R\to U^{-1}R$ denote the localization morphism, and for any ring $S,$ let $\ell_S^\times$ denote the localization map $\ell_S^\times : S\to K(S).$ We want to look at the morphism $$ \varphi : \ell_R^\times(U)^{-1}K(R)\to K(U^{-1}R) $$ induced by the universal property of localization.

Elements of the ring $\ell_R^\times(U)^{-1}K(R)$ are equivalence classes of pairs $(\alpha,\beta),$ where $\alpha\in K(R)$ and $\beta\in\ell_R^\times(U),$ and $(\alpha,\beta)\sim(\alpha',\beta')$ if there exists $\gamma\in\ell_R^\times(U)$ such that $\gamma(\beta'\alpha - \beta\alpha') = 0\in K(R).$

Unwrapping further, this means that $\alpha$ is an equivalence class of the form $(r,x),$ where $r\in R$ and $x\in R$ is a non-zero divisor, and $\beta$ is an equivalence class of the form $(u,1),$ where $u\in U,$ and this time, we have $(r,x)\sim (r',x')$ if there exists a non-zero divisor $y\in R$ such that $y(x'r - xr') = 0.$ Since $y$ is a non-zero divisor, this is the same as asking that $x'r = xr'.$

On the other hand, elements of $K(U^{-1}R)$ are given by equivalence classes of pairs $(\delta,\epsilon),$ where $\delta\in U^{-1}R$ and $\epsilon$ is a non-zero divisor in $U^{-1}R.$ Similarly to before, $(\delta,\epsilon)\sim(\delta',\epsilon')$ if $\epsilon'\delta = \epsilon\delta'.$

And elements of $U^{-1}R$ are equivalence classes of pairs $(r,u),$ where $r\in R$ and $u\in U,$ subject to the condition that $(r,u)\sim (r',u')$ if there exists $v\in U$ such that $v(u'r - ur') = 0.$

The map $\varphi$ is then defined as follows: \begin{align*} \varphi : \ell_R^\times(U)^{-1}K(R)&\to K(U^{-1}R)\\ ((r,x),(u,1))&\mapsto ((r,u),(x,1)). \end{align*} This is the map you described in your comment, just written out a little more laboriously. First, one should check that this map does actually make sense: that $(x,1)$ is indeed a non-zero divisor in $U^{-1}R.$ If $(r,u)(x,1) = 0,$ then it follows that there exists $v\in U$ such that $vrx = 0$ in $R.$ Since $x$ is a non-zero divisor in $R,$ this means that $vr = 0.$ But this implies that $(r,u)$ is $0$ in $U^{-1}R,$ as desired.

Now, let's make sure this is well-defined. Suppose that $((r,x),(u,1)) \sim ((r',x'),(u',1)).$ We need to show that $((r,u),(x,1))\sim ((r',u'),(x',1)).$ That is, we must show that there exists a non-zero divisor $(s,v)\in U^{-1}R$ such that $$(s,v)((x',1)(r,u) - (x,1)(r',u')) = 0,$$ which is equivalent to showing that $(x',1)(r,u) = (x,1)(r',u')$ in $U^{-1}R,$ which in turn amounts to showing that there exists $v\in U$ such that $v(u'x'r - uxr') = 0.$

But if $((r,x),(u,1)) \sim ((r',x'),(u',1)),$ then there exists $(v,1)$ such that $(v,1)((u',1)(r,x) - (u,1)(r',x')) = 0$ in $K(R),$ and one can (and should) check that this $v$ is precisely what is desired.

Why is this $\varphi$ the desired map? Well, by the universal property of localization, there is a unique map $\ell_R^\times(U)^{-1}K(R)\to K(U^{-1}R)$ which factors the map \begin{align*} f : K(R)&\to K(U^{-1}R)\\ (r,x)&\mapsto ((r,1),(x,1)) \end{align*} through the localization map \begin{align*} \ell_{\ell^\times_R(U)}:K(R)&\to \ell^\times_R(U)^{-1}K(R)\\ (r,x)&\mapsto ((r,x),(1,1)). \end{align*} We can check that $\varphi$ is such a factorization: \begin{align*} \varphi\circ\ell_{\ell^\times_R(U)}(r,x) &= \varphi((r,x),(1,1))\\ &= ((r,1),(x,1))\\ &= f((r,x)) &\checkmark. \end{align*}

With all that verification out of the way, we can now prove the main result.

Claim: With all notation as above, $\varphi : \ell_R^\times(U)^{-1}K(R)\to K(U^{-1}R)$ is injective.

Proof: Suppose that $\varphi(((r,x),(u,1))) = ((r,u),(x,1))$ is $0\in K(U^{-1}R),$ represented as $0 = ((0,1),(1,1)).$ Then there exists some non-zero divisor $(s,v)\in U^{-1}R$ such that $$(s,v)((1,1)(r,u) - (x,1)(0,1)) = 0$$ in $U^{-1}R.$ Again, since $(s,v)$ is a non-zero divisor, this is equivalent to saying that $$ (r,u) = (0,1) $$ in $U^{-1}R.$ That is, there exists $v\in U$ such that $vr = 0.$

I claim that the existence of such a $v$ implies that $((r,x),(u,1))$ is already $0$ in $\ell_R^\times(U)^{-1}K(R).$ Indeed, we have that $(v,1)(r,x) = (vr,x) = (0,x) = 0$ in $K(R),$ so that there does exist $(v,1)\in \ell_R^\times(U)$ such that $$ (v,1)((1,1)(r,x) - (u,1)(0,1)) = 0, $$ which precisely means that $$ ((r,x),(u,1))\sim ((0,1),(1,1)); $$ i.e., that $((r,x),(u,1))$ is $0$ in $\ell_R^\times(U)^{-1}K(R).$ $\square$