September | 2019 | Lovely little lemmas

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.

This is the story of Johan Commelin and myself working through the first sections of Hartshorne almost 10 years ago (nothing creates a bond like reading Hartshorne together…). This post is about problem I.1.1(b), which is essentially the following:

Exercise. Let $k$ be a field. Show that $k[x]$ and $k[x,x^{-1}]$ are not isomorphic.

In my next post, I will explain why I’m coming back to exactly this problem. There are many ways to solve it, for example:

Solution 1. The $k$ -algebra $k[x]$ represents the forgetful functor $\mathbf{Alg}_k \to \mathbf{Set}$ , whereas $k[x,x^{-1}]$ represents the unit group functor $R \mapsto R^\times$ . These functors are not isomorphic, for example because the inclusion $k \to k[x]$ induces an isomorphism on unit groups, but not on additive groups. $\qed$

A less fancy way to say the same thing is that all $k$ -algebra maps $k[x,x^{-1}] \to k[x]$ factor through $k$ , while the same evidently does not hold for $k$ -algebra maps $k[x] \to k[x]$ .

However, we didn’t like this because it only shows that $k[x]$ and $k[x,x^{-1}]$ are not isomorphic as $k$ -algebras (rather than as rings). Literal as we were (because we’re undergraduates? Lenstra’s influence?), we thought that this does not answer the question. After finishing all unstarred problems from section I.1 and a few days of being unhappy about this particular problem, we finally came up with:

Solution 2. The set $k[x]^\times \cup \{0\}$ is closed under addition, whereas $k[x,x^{-1}]^\times \cup \{0\}$ is not. $\qed$