# First-order characterisations of free and flat…projective?

Here is an interesting question involving free, projective, and flat modules that I will leave to the readers of this blog for now.

First, consider free modules. If $R$ is a ring, then every $R$-module is free if and only if $R$ is a division ring. The property of $R$ being a division ring can be expressed in terms of first-order logic in the language of rings: $\forall x[x\not=0 \rightarrow \exists y(xy = 1)]$.

The meat of this first-order statement is the equation $xy = 1$. Now, multiply by $x$ on the right to get the equation $xyx = x$. Now we can put this in a first-order sentence: $\forall x\exists y[xyx = x]$. Notice how we removed the condition $x\not=0$ from this one. That's because $x=0$ satisfies $xyx = x$ for any $y$ in all rings. Rings that model $\forall x\exists y[xyx = x]$ are called von Neumann regular. More importantly, these are exactly the rings for which every $R$-module is flat.

By weakening the statement that $R$ is a division ring, we got a statement equivalent to the statement that every $R$-module is flat. One might wonder: where did the projective modules go? Is there a first-order sentence (or set of sentences perhaps) in the language of rings whose models are exactly those rings $R$ for which every $R$-module is projective? Diagrammatically:

Can we replace the question mark with a first-order sentence, or a set of them?

My initial thoughts are no because of ultraproducts, but I have not yet come up with a rigorous argument.

• Erik Crevier says:

Rings over which every module is projective are exactly semi-simple rings.

For each n, you can say "R has n pairwise distinct commuting idempotents" with a first order sentence P_n. Let R_n be an n-fold product of fields. R_n satisfies P_n, so an ultraproduct R_n's will have arbitrarily large families of commuting idempotents. Such a ring cannot be semi-simple by Artin-Wedderburn. Each R_n certainly is semi-simple, so it follows that this cannot be characterized by first order sentences.

• Erik Crevier says:

Sorry, that condition should be "n non-zero central idempotents which sum to 1", the point being that such families govern non-trivial direct product decompositions into n factors.

• Jason Polak says:

Very good! Thanks for explaining this.