1
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ Find a suitable multiplicative set. $\endgroup$ Commented Feb 26, 2021 at 22:46
  • 1
    $\begingroup$ 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 $t$... $\endgroup$ Commented Feb 26, 2021 at 22:46
  • $\begingroup$ I think I am in localization over modules @Paul $\endgroup$
    – user889267
    Commented Feb 26, 2021 at 22:59
  • 1
    $\begingroup$ @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! $\endgroup$ Commented Feb 26, 2021 at 23:01
  • 2
    $\begingroup$ $\mathbb C[t,t^{-1}]$ is not localization at the complement of a prime. $\endgroup$
    – D_S
    Commented Feb 27, 2021 at 1:26

1 Answer 1

2
$\begingroup$

"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?

$\endgroup$

You must log in to answer this question.