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I started reading about monoids (and semigroups in general) and came across of the example of some non-commutative monoids which cannot be endowed with some addition turning it into a ring (the monoid is constructed such that $x^2=x$ and so the ring would be boolean and thus commutative). My question is:

Do we have (checkable) criteria to decide whether a monoid can be turned into a ring?

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    $\begingroup$ One obvious condition is that there must be an element $\mathbf 0$ satisfying $\mathbf 0\cdot x=x\cdot \mathbf 0=\mathbf 0 $ for all $x$ $\endgroup$
    – Gregory Arone
    Jan 4, 2021 at 9:25
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    $\begingroup$ If $p$ is prime the only possible ring structures are zero (semigroup with 0 and constant law $xy=0$) and a cyclic group of order $p-1$ plus an absorbing element. This discards most monoid structures on a set with $p$ elements. $\endgroup$
    – YCor
    Jan 4, 2021 at 9:39

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This was a classical question in semigroup theory. I suggest looking at this paper http://www.numdam.org/article/SD_1969-1970__23_2_A12_0.pdf which is interested in finite semigroups. It gives a reference to a Russian paper that in some sense says it impossible to axiomatize such rings but I don't have access to the Russian paper to say what sense.

A key obstruction is that commuting idempotents in a ring have joins and relative complements in rings but not semigroups. Another problem is rings with no zero divisors are cancelative but not semigroups.

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  • $\begingroup$ Thanks, this is exactly what I was looking for! $\endgroup$
    – Severin Schraven
    Jan 4, 2021 at 16:58
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I am not sure how checkable this is, but one formal answer is that the forgetful functor from rings to monoids has a left adjoint: the monoid ring functor. Rings are the same thing as algebras over the monad resulting from this adjunction. So the set of ring structures on a monoid M is the same as the set of monoid maps $\mathbb Z[M]\to M$ that satisfy the axioms of algebra over a monad.

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  • $\begingroup$ Thanks for the answer! Took me a moment to digest (I am not that fluent in category theory). $\endgroup$
    – Severin Schraven
    Jan 4, 2021 at 18:52

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