Rings that are localisations of each other

This is a post about an answer I gave on MathOverflow in 2016. Most people who have ever clicked on my profile will probably have seen it.

Question. If A and B are rings that are localisations of each other, are they necessarily isomorphic?

In other words, does the category of rings whose morphisms are localisations form a partial order?

In my previous post, I explained why k[x] and k[x,x^{-1}] are not isomorphic, even as rings. With this example in mind, it’s tempting to try the following:

Example. Let k be a field, and let K = k(x_1, x_2, \ldots). Let

    \[A = K[x_0,x_{-1},\ldots]\]

be an infinite-dimensional polynomial ring over K, and let

    \[B = A\left[\frac{1}{x_0}\right].\]

Then B is a localisation of A, and we can localise B further to obtain the ring

    \[k(x_0,x_1,\ldots)[x_{-1},x_{-2},\ldots]\]

isomorphic to A by shifting all the indices by 1. To see that A and B are not isomorphic as rings, note that A^\times \cup \{0\} is closed under addition, and the same is not true in B. \qed


Is there a moral to this story? Not sure. Maybe the lesson is to do mathematics your own stupid way, because the weird arguments you come up with yourself may help you solve other problems in the future. The process is more important than the outcome.

Leave a Reply

Your e-mail address will not be published. Required fields are marked *