# Defining the most natural map between two modules

Let $R = \mathbb{C}[t]$ be a ring of polynomials in variable $t$ with coefficients in the field of complex numbers $\mathbb{C}$ and let $$M = R[x]/(tx-1).$$

Goal: I need to show that $$M \cong R[t^{-1}]$$ where $R[t^{-1}]$ is the localization of $R$ with respect to the multiplicative set $$S =\{t^{-i} \quad | i \in \mathbb{N}\}$$

For me to show the isomorphism, then I need to know the most natural map between $R[x]$ and $R[t^{-1}]$.

My guess is the following evaluation map $$\phi_t: R[x] \rightarrow R[t^{-1}]$$ that sends $$f(x) \mapsto \frac{f(x)}{t}$$ Is this map good? This map won't give me the kernel $tx-1$ and thats exactly my problem

I can prove the isomorphism if I know the most natural map.

Thanks.

• Hint: Map $x \to t^{-1}$. Apr 5, 2017 at 23:54
• @Ken Duna Got it. Thanks. You should post that as an answer so that I can accept. Please do cos you have just saved me from stressing out lol. Apr 6, 2017 at 0:09
• Apr 21, 2019 at 6:10

Define $$\varphi : R[x] \to R[t^{-1}]$$ by mapping $$x \mapsto t^{-1}$$. Then $$\text{ker}(\varphi) = (tx -1)$$, and by the First Isomorphism Theorem, $$\frac{R[x]}{(tx-1)} \cong R[t^{-1}]_.$$