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In this question, it is asked when an abelian group admits a ring structure; that is, if $(G,+)$ is an abelian group, then under what conditions is there a binary operation $\cdot$ on $G$ such that $(G,+,\cdot)$ is a ring. (For the purposes of this question, rings are unital.)

I am interested in the "dual" question of when a monoid admits a ring structure; that is, if $(M,\cdot)$ is a monoid, then under what conditions is there a binary operation $+$ on $M$ such that $(M,+,\cdot)$ is a ring?

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    $\begingroup$ Seems likely to be very complicated, and harder than the abelian group case. For instance, if $|M|$ is finite and prime, then it must be a cyclic group on $p-1$ elements together with a $0$ element. That's a rather odd looking condition if you are thinking of $M$ as just some monoid. $\endgroup$ Nov 17, 2023 at 14:17
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    $\begingroup$ I don't know the answer, but one such property that must be required would be an "absorbing element" of the monoid $(M,\cdot)$. When I say absorbing element, I mean an element $x\in M$ such that for any $y\in M$ $xy=yx=x$. As such an element must be unique and you should call this element $0$ as it multiplicatively acts like $0$, and I suspect that it should be the additive identity if you want to put an additive structure on the monoid. $\endgroup$ Nov 17, 2023 at 14:19
  • $\begingroup$ @EricWofsey: Thanks for your response. I see that this is a very difficult question even in the finite case. However, even if a complete characterisation is not possible, I am still interested in conditions which are merely necessary or merely sufficient, or answers which restrict the question to special cases. $\endgroup$
    – Joe
    Nov 17, 2023 at 14:22
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    $\begingroup$ I like this question. It asks for the image of the forgetful functor from rings to monoids. As Steven mentioned, though, it is more natural to work with the category of monoids with zero. $\endgroup$ Nov 17, 2023 at 14:38
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    $\begingroup$ @MartinBrandenburg yes, I asked that question a few years ago! math.stackexchange.com/questions/3367423/… $\endgroup$
    – Kenta S
    Nov 17, 2023 at 14:41

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The question asks for the description of the image of the forgetful functor $\mathbf{Ring} \to \mathbf{Mon}$ that forgets the additive structure. I don't know if such a description actually exists, but there are some interesting non-trivial necessary conditions for a monoid to be contained in the image.

Well, the first observation is that the underlying multiplicative monoid of a ring is actually a monoid with zero. The functor factors as $\mathbf{Ring} \to \mathbf{Mon}_0$, so that it is more natural to look inside of $\mathbf{Mon}_0$.

There is a basic theorem in ring theory stating that every boolean ring is commutative. This is essentially a statement about the underlying monoid, namely if $x^2=x$ is satisfied for all elements $x$, then $xy=yx$ holds for all elements $x,y$. This does not hold for all monoids: take $\{1,x,y\}$ with $ab := a$ for $a,b \neq 1$, and adjoin a zero if you want a zero.

More generally a ring is called an $n$-ring (for $n > 1$) when every element satisfies $x^n=x$. Jacobson has proven (N. Jacobson, Structure theory for algebraic algebras of bounded degree, Ann. of Math 46.4 (1945), 695–707) that every $n$-ring is commutative, i.e. satisfies $xy=yx$ for all elements $x,y$. Again, this is just a statement about the underlying multiplicative monoids. In this context, there also several theorems of the form that every $n$-ring is an $n'$-ring for some smaller $n'< n$. For example, every $10$-ring is actually a $4$-ring because of the equation $$T^4 - T = (T^2+1) (T^{10}-T) + T^2 ((T+1)^{10}-(T+1))$$ in $\mathbb{F}_2[T]$. So this deduction uses the additive structure, and in fact there are $10$-monoids (with zero) that are not $4$-monoids (with zero): just take the universal example $\langle x : x^{10}=x \rangle$ which consists of $1,x,\dotsc,x^9$ (resp. $0,1,x,\dotsc,x^9$). As a consequence, this cannot come from a ring. You can find more information in my recent preprint Equational proofs of Jacobson's Theorem. The introduction also lists lots of other references around this topic.

Wedderburn's Little Theorem can be seen as a special case of Jacobson's Theorem, it is also a restriction on the monoids with zero associated to a ring. Namely, it states that for every such monoid, in case it is finite and every non-zero element is invertible, is commutative.

If $M$ is the multiplicative monoid of a ring, then $M^{\times}$ is the group of units of that ring - some necessary conditions for groups with this property have been discussed at SE/3367423 and SE/384422.

Maybe it helps to also demand an element $-1 \in M$. Then we define $-x := (-1)x$, and of course we require $(-1)^2 = 1$. This way, the additive inverses are already fixed at least.

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  • $\begingroup$ Thanks for all these references. Indeed, it seems that any non-trivial result of ring theory (i.e. a result that uses distributivity in an essential way) that can be stated in terms of the underlying monoid, provides us with a necessary condition for a monoid to admit a ring structure. Are there are any significant sufficient conditions for a monoid to admit a ring structure that come to mind? $\endgroup$
    – Joe
    Nov 19, 2023 at 14:06
  • $\begingroup$ Erm ... well it is sufficient that there is a group structure that satisfies the distributive laws 😅. Even though this is the troll answer, maybe it is in fact the easiest one after all. When we want a property that is "intrinsic" to monoids, it will be much harder to describe and verify that property. Also, there is no classification of all monoids available, so that we can filter out those that admit a ring structure. $\endgroup$ Nov 19, 2023 at 14:52
  • $\begingroup$ By the way, the commutativity of addition is automatic - this is a well-known and cute result. Notice that $x \mapsto (-1) * x$ is a group homomorphism because of the distributive law, so $-(x+y)=(-x)+(-y)$, but $-(x+y)=(-y)+(-x)$ in general, and hence $+$ is commutative. $\endgroup$ Nov 19, 2023 at 14:53

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