2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Hint: Map $x \to t^{-1}$. $\endgroup$
    – Ken Duna
    Apr 5, 2017 at 23:54
  • $\begingroup$ @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. $\endgroup$
    – Jaynot
    Apr 6, 2017 at 0:09
  • $\begingroup$ See also math.stackexchange.com/questions/1677766/… . $\endgroup$ Apr 21, 2019 at 6:10

1 Answer 1

1
$\begingroup$

Per request:

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, $$ \frac{R[x]}{(tx-1)} \cong R[t^{-1}]_.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.