3

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 ?

user26857
  • 52,094
Topo
  • 772
  • 6
  • 17
  • You can also show that for any map $R[x]\rightarrow T$, where $S$ goes to invertible elements, there's a unique extension to $(S^{-1}R)[x]$, which is the characteristic property of $S^{-1}(R[x])$. – Steve D Dec 09 '17 at 07:57
  • Sorry ! Not getting what you are saying ! – Topo Dec 09 '17 at 08:08

2 Answers2

5

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.

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 $$ \frac{r_0+r_1x+\dots+r_nx^n}{s}= \frac{r_0}{s}+\frac{r_1}{s}x+\dots+\frac{r_n}{s}x^n $$

egreg
  • 238,574
  • what happens if we replace polynomial rings by series? Are they still isomorphic ? – user390026 Dec 15 '18 at 17:19
  • @Sara.T No, they aren't. A power series in $(S^{-1}R)[[x]]$ can have whatever denominator you want in each term, and you can't generally collect them as you can do with a polynomial. – egreg Dec 15 '18 at 18:02
  • You are right. But would you provide an actual example? – user390026 Dec 15 '18 at 19:43
  • @Sara.T Just take $S={2^n:n\ge0}$ in $\mathbb{Z}$. – egreg Dec 15 '18 at 21:47
  • When the notation $S^{-1}(R[x])$ is used, we asume that $S=j_R(S)$. These are the same elements but considering them as constant polynomials. Am I correct? – Gerardo Vargas Flores Dec 01 '20 at 13:56
  • 1
    @GerardoVargasFlores Yes, it's a bit sloppy notation. It would only be justified if the homomorphism $R\to S^{-1}R$ is injective. – egreg Dec 01 '20 at 15:45
  • @egreg Hello, could you please give a little more detail on the very last part? That is, why proving that the composition maps are the identity amounts to verifying these two equalities? Also, why do they hold? I’m struggling to understand that step. – lanero Dec 04 '22 at 13:31
  • @lanero The key is uniqueness in the universal properties. When you have found two homomorphisms that have the property, then they're equal. – egreg Dec 04 '22 at 13:33
  • @egreg Oh, you mean that $\alpha^x$ and $\beta$ have to be inverses just because you showed that each of them were unique by construction? – lanero Dec 04 '22 at 13:57
  • @lanero Yes, that's the idea – egreg Dec 04 '22 at 17:27
3

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]) $

Chirantan Chowdhury
  • 1,088
  • Yes..I'm also going through Universal property of localization, taking your $T$ as $S^{-1}(R[x])$. But I'm unable to define the proper maps. ! – Topo Dec 09 '17 at 11:58