# Laurent polynomials $\mathbb C[t,t^{-1}]$ is the localization of $\mathbb {C}[t].$

I want to prove this question:

Show that the ring of Laurent polynomials $$\mathbb C[t,t^{-1}]$$ is the localization of the polynomial ring $$\mathbb {C}[t].$$

Localization is defined as follows: Let $$R$$ be a commutative, $$S\subset R$$ multiplicatively closed. Define $$\sim$$ on $$S \times R$$ by $$\frac{r}{s} \sim \frac{r'}{s'}$$ which is equivalent to $$t(rs' - r's) = 0$$ for some $$t \in S$$ then $$S^{-1}R$$ is a commutative ring.

Still, I do not know how to prove this localization, do I have to find an isomorphism? or what?

• Find a suitable multiplicative set. Commented Feb 26, 2021 at 22:46
• In your example, you also would need to specify the multiplicative subset $S$... But/and with comm rings w/o zero divisors, the equivalence relation doesn't need the extra factor you've called $t$... Commented Feb 26, 2021 at 22:46
• I think I am in localization over modules @Paul
– user889267
Commented Feb 26, 2021 at 22:59
• @MathFear Localization of $R$-modules is done with respect to a multiplicative subset of $R$. You should take a look at your definition of localization of modules to start! Commented Feb 26, 2021 at 23:01
• $\mathbb C[t,t^{-1}]$ is not localization at the complement of a prime.
– D_S
Commented Feb 27, 2021 at 1:26

"Show that the ring of Laurent polynomials $$\mathbb C[t,t^{-1}]$$ is the localization of the polynomial ring $$\mathbb C[t]$$."

The problem does not make sense as you state it because, as others have pointed out in the comments, there are many localizations of $$R =\mathbb C[t]$$. Each localization is done with respect to a multiplicatively closed subset $$S$$ of $$R$$. You must choose a suitable multiplicatively closed subset $$S$$ so that $$S^{-1}R = \mathbb C[t,t^{-1}]$$.

By the way, the abstract definition you have provided of localization is not at all necessary here. The following observation will make the problem much more tractable:

What does a localization of an integral domain look like? If $$A$$ is an integral domain, and $$K$$ is its field of fractions, any localization $$S^{-1}A$$ of $$A$$ is a ring which contains $$A$$ and which is contained in $$K$$. Indeed, $$A \subseteq S^{-1}A = \{ \frac{a}{s} : a \in A, s \in S\} \subseteq K.$$

The abstract definition of $$S^{-1}A$$ as the set of equivalence classes of pairs in $$S \times A$$ is not necessary here and will only confuse you. Sending the class of $$(s,a)$$ to $$as^{-1} \in K$$ gives an isomorphism between the formal definition of $$S^{-1}A$$ and the definition I just provided, as a subring of the field $$K$$.

What multiplicatively closed set $$S$$ should I pick to obtain $$\mathbb C[t,t^{-1}]$$?

The set should obviously contain $$1$$ and $$t$$, as you will want $$t$$ to be invertible in the localization. What other elements does $$S$$ need to contain to be closed under multiplication?