Prove that $(S^{-1}R)[x] \cong S^{-1}(R[x])$.

Question

Let $R$ be a ring and $S\subset R$ is multiplicatively closed set. Then prove that $\displaystyle \left(S^{-1}R\right)[x] \cong S^{-1}\left(R[x]\right)$.

I'm thinking to prove through universal property.Define a map $f:R[x]\to S^{-1}(R[x])$ by $f\mapsto \frac{f}{1}$. Then for the map $g:R[x]\to (S^{-1}R)[x]$ by $x\mapsto x$ & $r\mapsto \frac{r}{1}$ there exists a map $h:S^{-1}(R[x])\to (S^{-1}R)[x]$ such that $h\circ f=g$. Now , how I can define the map $h$ and what's the reverse map to show that $h^{-1}$ exists ?

Can anyone help me ?

Answer 1

You can do this with the universal properties of localization and of polynomial rings.

For $S$ a multiplicative set in the ring $R$, let's denote by $\lambda_{R,S}$ the map $$ \lambda_{R,S}\colon R\to S^{-1}R,\qquad \lambda_{R,S}(r)=r/1 $$ For a ring $R$, let's denote by $j_R$ the embedding map $$ j_R\colon R\to R[x] $$

Universal property of localization. (UPL) If $S$ is a multiplicative set in $R$ and $\varphi\colon R\to A$ is a ring homomorphism so that, for each $s\in S$, $\varphi(s)$ is invertible in $A$, there exists a unique ring homomorphism $\varphi_S\colon S^{-1}R\to A$ such that $$ \varphi_S\colon\lambda_{R,S}=\varphi $$

This homomorphism is $\varphi_S(r/s)=\varphi(r)\varphi(s)^{-1}$.

Universal property of polynomial rings. (UPPR) For every ring homomorphism $\varphi\colon R\to A$ and every element $a\in A$, there exists a unique ring homomorphism $\varphi^a\colon R[x]\to A$ such that $\varphi^a\colon j_R=\varphi$ and $\varphi^a(x)=a$.

The homomorphism is $\varphi^a(r_0+r_1x+\dots+r_nx^n)=r_0+r_1a+\dots+r_na^n$.

With these properties we can first define a homomorphism $$ \alpha\colon S^{-1}R\to S^{-1}(R[x]) $$ using the composition of $j_R$ followed by $\lambda_{R[x],S}$ and using UPL. Then we get $\alpha^x\colon(S^{-1}R)[x]\to S^{-1}(R[x])$.

We can also consider the homomorphism $(j_{S^{-1}R}\circ\lambda_{R,S})^x\colon R[x]\to (S^{-1}R)[x]$ given by the UPPL and then define $$ \beta\colon S^{-1}(R[x])\to (S^{-1}R)[x] $$ using the UPL.

Now it's just a matter of checking that the compositions $\alpha^x\circ\beta$ and $\beta\circ\alpha^x$ are the identities on the respective domains.

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The above may sound complicated, but it's basically just a formal way of expressing that $$ \frac{r_0}{s_0}+\frac{r_1}{s_1}x+\dots+\frac{r_n}{s_n}x^n= \frac{r_0'+r_1'x+\dots+r_n'x^n}{s_1\dots s_n} $$ (using a common denominator) and $$ \frac{r_0+r_1x+\dots+r_nx^n}{s}= \frac{r_0}{s}+\frac{r_1}{s}x+\dots+\frac{r_n}{s}x^n $$

Answer 2

There is an universal property of localization of a ring. Given any map $R[x] \to T$. where every elements of $S$ goes to the unit. Show that there is a map $ (S^{-1}R)[x] \to T$ making the diagram commutative. (Here you need to define an obvious map from $R[x] \to (S^{-1}R)[x] $ and show the commutativity). This satisfies the universal property of localization and thus we shall have $(S^{-1}R)[x] \cong S^{-1}(R[x]) $