Rings are abelian

In this post, we prove the following well-known lemma:

Lemma. Let (R,+,\times,0,1) satisfy all axioms of a ring, except possibly the commutativity a + b = b + a. Then (R,+) is abelian.

That is, additive commutativity of a ring is implied by the other axioms.

Proof. By distributivity, we have 2(a+b) = 2a + 2b, so multiplication by 2 is a homomorphism. By our previous post, this implies R is abelian. \qedsymbol

1 thought on “Rings are abelian

  1. Hilarious. I didn’t even know this.

    Exercise: Find a counterexample for the same statement but for (R, +, •, 0) (i.e., without unit). (It’s actually not hard to find a minimal one, as I know you like to do!)

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