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Let $R$ be a topological ring (commutative with $1$) and let $R((t))$ the ring of Laurent power series. So, it is the ring containing the formal power series:

$$\sum_{i\ge m} a_i t^i,\quad m\in\mathbb Z,\;a\in R$$

It is easy to see that $R((t))$ can be obtained as an ind-pro objects in the following way:

$$R((t))=\varinjlim_{m\in\mathbb Z}\left (\varprojlim_{n\in\mathbb N} \frac{t^m R[[t]]}{t^{m+n} R[[t]]}\right)\quad (\ast)$$

If we topologize $\frac{t^m R[[t]]}{t^{m+n} R[[t]]}\cong \bigoplus^n_{i=1} R$ with the product topology, we then obtain a topology on $R((t))$ by equation $(\ast)$.

One can show that: $$\cdots\supset t^m R[[t]]\supset t^{m+1}R[[t]]\supset\cdots\quad (\ast\ast)$$

Is a local basis (of open sets) at the point $0$. Here my question:

By the above construction it seems that we obtain always the same topology on $R((t))$, independently from the starting topology on $R$! This because the local basis at $0$ is always $(\ast\ast)$. How is it possible? There should be some mistake in my reasoning.

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    How do you show that (**) gives a local basis? – 2017-02-07
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    You give to $R((t))$ the strongest topology such that $t^mR[[t]]\subset R((t))$ is open for any $m$. This the definiton of the direct limit topology. – 2017-02-07
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    @manifold: That is not definition of the direct limit topology. The direct limit topology is the strongest topology such that the inclusion $t^mR[[t]]\to R((t))$ is continuous for all $m$ (and in this case you have given each $t^mR[[t]]$ has the inverse limit topology). – 2017-02-07
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    I thought this was exactly the topology I defined in $(\ast)$. What are the differencies? – 2017-02-07
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    Oh yes. Now I see it. The embedding should be open for *any* index. – 2017-02-07
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    None of the maps have to be open, they just have to be continuous. – 2017-02-07
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    Yes, I meant continuous (I was thinking to the open subsets). Thank you. – 2017-02-07

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It is simply not true that $(**)$ is a local basis at $0$ in general. Indeed, these sets are not even open if $R$ is not discrete.

For instance, let us consider $tR[[t]]$. If it is open in $R((t))$, then it must be open in $R[[t]]$ with the inverse limit topology. But the inverse limit topology on $R[[t]]$ is just the product topology, identifying $R[[t]]$ with $R^\mathbb{N}$ in the obvious way (indeed, you can prove the inverse limit topology has the same universal property as the product topology, since we are using the product topology on each quotient $R[[t]]/t^nR[[t]]$). In this identification, $tR[[t]]$ is the set of elements of $R^\mathbb{N}$ whose $0$th coordinate is $0$. This won't be open in the product topology unless $\{0\}$ is open in $R$, i.e. unless $R$ is discrete.