Characteristic of a Monoid

In ring theory, the characteristic is defined as the min $n$ such that $n1 = 0$ , or 0 if no such $n$ exists; judging from https://en.wikipedia.org/wiki/Characteristic_(algebra), it seems that no concept of characteristic has been considered for structures more general than rings. Here, we generalize the concept to any monoid $M$ .

To do this, we note that the definition of characteristic of a ring is equivalent to: the min nonzero $n$ such that $nx = 0$ for all $x$ , or equivalently the max of the additive orders of the elements, or equivalently the LCM of the additive orders of the elements. (This equivalence follows immediately when we write it out; however, for the sake of completeness, we include proofs below.) These latter definitions don’t require 1 or even any multiplication, thus they can be generalized to algebraic structures with just one operation. We’d still need 0 (identity) and associativity (to write out $nx = x + \ldots + x$ ) for either definition, so we can generalize to a monoid $M$ . Following the standard convention of writing monoids multiplicatively if commutativity isn’t assumed, we generalize each definition as:

The min nonzero $n$ such that $x^{n} = e$ for all $x \in M$
The max order of an element
The LCM of the orders of the elements

In either case, if such a number doesn’t exist, we can say that $M$ has characteristic zero.

With this formulation, the characteristic of a ring is the same concept as the characteristic of a ring’s additive group. Unlike for rings, it is not immediately clear whether these definitions are equivalent for general monoids. Thus, we will adopt the first definition, and we investigate equivalence to the other two:

Lemma. In a monoid $M$ , the characteristic is zero if and only if there is no max order of an element.

(Note here that in our notation we do not allow an element’s order to be zero; for a given $x$ , if no positive $n$ exists such that $x^{n} = e$ , then we say that the order of $x$ doesn’t exist, and in that case we would have in particular that the max order doesn’t exist.)

Proof. This is equivalent to saying that the characteristic is nonzero if and only if the max order of an element exists. For one direction, assume that there is $n \neq 0$ such that $x^{n} = e$ for all $x$ . Then, for any element $x$ , the order of $x$ must exist and be less than or equal to $n$ . Hence, a max order exists. Conversely, if a max order $m$ exists, then for any $x$ there must exist nonzero $n(x) \leq m$ with $x^{n(x)} = e$ . Defining $N$ to be the LCM of all such $n(x)$ , we have that $x^{N} = e$ for all $x$ , as desired. (Written out explicitly, even if the monoid is infinite, we may define $S = \left\{ n(x):x \in M \right\}$ ; then, $S$ is guaranteed finite, and we can define $N$ to be the LCM of $S$ .) $\blacksquare$

Conjecture (Max Order). If the characteristic of a monoid is nonzero, then it is equal to the max order.

I tried proving this, but it seems to be a bit more involved than I first thought. We will come back to this conjecture later.

Lemma. In a monoid $M$ , the characteristic is zero if and only if the LCM of the orders doesn’t exist.

Proof. This is equivalent to saying that the characteristic is nonzero if and only if the LCM exists. For one direction, assume the characteristic is nonzero; then, by a previous lemma, the max order exists, thus the set $S = \left\{ |x|:x \in M \right\}$ is finite, hence $lcm(S)$ exists. Conversely, if $lcm(S)$ exists, then certainly $\max S$ does, thus the characteristic is nonzero. $\blacksquare$

Lemma. If the characteristic is nonzero, then it is equal to the LCM of the orders.

Proof. Say the characteristic is $n$ . Then for any element $x$ we have $x^{n} = e$ . If $n$ isn’t a multiple of the order of $x$ , then $n$ must be greater than the order of $x$ (otherwise it’d contradict the definition of order), so we can subtract a multiple of the order (call it $o$ ) from $n$ until we have an exponent $n^{'}$ with $x^{n^{'}} = e$ and $n^{'} < o$ . This contradicts the definition of $o$ , so $n$ must be a multiple of the order of $x$ .

Thus, $n$ must be a multiple of the order of any element. As a result, $n$ must be a multiple of the LCM. But we also see that the LCM of $S$ actually works in the definition of characteristic: calling this LCM $l$ , we have that $x^{l} = e$ for all $x$ . Thus, $n = lcm S$ . $\blacksquare$

Thus, we have that the concept of characteristic is equivalent to the LCM of the orders.

Regarding the max order, let’s look at the proof of equivalence for rings (below) to see why it seems harder to establish for monoids. Assuming the existence of min nonzero $n$ such that $nx = 0$ for all $x$ , everything was translatable to monoids except that $n1 = 0$ . Basically, this identifies an element (1) for which $n$ is an order. We already have in a general monoid that the order of any element is less than or equal to the characteristic, so identifying one element that has the characteristic as an order is the missing piece (achievement of the max.)

Also, let’s take a closer look at how we yielded that in a ring the characteristic is equivalent to the min $n$ such that $nx = 0$ for all $x$ . The main thing there that wouldn’t be directly translatable to monoids was that $m1 = 0 \rightarrow mx = 0$ for all $x$ . Adapted to a general monoid, if we have an element $y$ such that $y^{m} = e \rightarrow x^{m} = e$ for all $m$ and $x$ , then we can use this $y$ the same way the argument for rings used 1. In fact, we’d have then that the characteristic is the order of $y$ .

Let’s write this out more slowly. But we need to be careful, since if $y$ doesn’t have an order then “ $y^{m} = e \rightarrow x^{m} = e$ for all $x$ ” will be true vacuously (take for example $y = 1$ in additive $\mathbb{Z}$ .) We define:

Definition. For a monoid $M$ , a characteristic generating element (or CGE for short) is a $y \in M$ whose order exists and such that, for any $m \in \mathbb{Z}_{+}$ and $x \in M$ , we have $y^{m} = e \rightarrow x^{m} = e$ .

Lemma. An alternative definition of a CGE is an element whose order is the characteristic. Or, more precisely:

A CGE exists if and only if there is a nonzero characteristic $n$ and an element whose order is $n$ .
If $y$ is a CGE, then its order is equal to the characteristic, and any element whose order is equal to the characteristic must be a CGE.

Proof. First, assume that a CGE exists, call it $y$ ; its order must exist, so we have $y^{|y|} = e \rightarrow x^{|y|} = e$ for all $x$ , thus the characteristic is nonzero and at most $|y|$ . In fact, since $|y|$ is the order of $y$ , the characteristic must be equal to $|y|$ .

Conversely, assume that the characteristic $n \neq 0$ and there is an element $y$ of order $n$ . We show that $y$ is a CGE, simultaneously establishing the remainder of statement 1 and half of statement 2. Assume that $y^{m} = e$ . Since the order of $y$ is $n$ , we must have $m = nk$ . But then $x^{m} = x^{nk} = e$ for all $x$ , so $y$ is a CGE.

Finally, assume $y$ is a CGE; we show that its order must be equal to the characteristic. We have $y^{|y|} = e$ , but since $y$ is a CGE we have $x^{|y|} = e$ for all $x$ . If there existed $n < |y|$ such that $x^{n} = e$ for all $x$ then we’d have $y^{n} = e$ , contradicting definition of order. Hence, $|y|$ is the characteristic. $\blacksquare$

As we suspected, we now show that the existence of a CGE is what is needed for the equivalence of the characteristic to max order:

Lemma. If $M$ has a CGE and its characteristic $n \neq 0$ , then $n$ is equal to the max order.

Proof. From previous lemmas, $n$ is equal to the LCM of the orders (call the set of orders $S$ .) Let $y$ be a CGE; then $n = |y| \in S$ . Since $n$ is an upper bound for $S$ , it follows that $n = \max S$ . $\blacksquare$

Lemma. If $M$ has nonzero characteristic $n$ that is equal to the max order, then $M$ has a CGE.

Proof. Again, denote the set of orders by $S$ . We have $n = \max S \rightarrow n \in S$ , thus there exists $y \in M$ with order $n$ , hence $y$ is a CGE. $\blacksquare$

We suspect that CGEs can be used like unity in rings to produce analogous results for monoids.

Given the result that a field has prime characteristic, must a monoid with a CGE have a prime characteristic too? Actually, rings themselves that aren’t integral domains may have composite characteristics; simply take for example the additive abelian group $\mathbb{Z}_{n}$ and the CGE 1. Thus, even an abelian group with a CGE may have a composite characteristic.

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Switching gears, we consider some other questions and conjectures about monoid characteristics:

Like for -algebras over fields of characteristic 2, we suspect there will be a difference between characteristic 2 or not when studying module-like structures over a monoid .
1. At least can we show that: if $M$ has inverses (aka $M$ is a group), then $x = x^{- 1} \rightarrow x = e$ if $M$ has characteristic not 2, and otherwise $x = x^{- 1}$ for all $x$ ? (This seemed to be one source of differences.) Indeed, in general $x = x^{- 1} \leftrightarrow x^{2} = e$ , and if $M$ has characteristic 2 then we always have $x^{2} = e$ , thus we always have $x = x^{- 1}$ . If $M$ has characteristic not 2, then must $x = e$ ? Not necessarily — it’s possible that a particular element $x$ has order 2 but other elements have orders greater than 2, making the characteristic greater than 2 and providing a counterexample $x$ . In fact, just look for example at the additive group $\mathbb{Z}_{4}$ and the element 2. We have that $2(2) = 0$ but $2 \neq 0$ . Thus, this isn’t fully true.
Can we study something involving isomorphisms depending on characteristic of $M$ ? Maybe classification of monoids of a given characteristic? Will have to look into this.

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Let’s go back to rings. Note that a ring also induces a multiplicative monoid, and we could talk about the characteristic of this monoid. To disambiguate, from now on in my writings I will use the terms “additive characteristic” and “multiplicative characteristic” explicitly for rings.

Consider a ring $R$ . There is no $n$ for which $0^{n} = 1$ , thus the multiplicative characteristic would never exist. Instead, for a more interesting concept, we can define the multiplicative characteristic to be for the monoid $R - \left\{ 0 \right\}$ . But we still don’t think this is interesting enough: consider the multiplicative group $\mathbb{Z}_{n}$ for $n$ composite. If $n = ab$ , then there exists no $n$ such that $a^{n} = 1$ or $b^{n} = 1$ (indeed, for any $x$ not relatively prime to $n$ .) But given Euler’s Totient Theorem, saying that the multiplicative characteristic of $\mathbb{Z}_{n}$ is zero for $n$ composite doesn’t seem to “capture the richness” of the theory we could study. Instead, we will define the multiplicative characteristic of a ring to correspond to the monoid of multiplicative units. (This is in fact a group.)

Then, based off of Euler’s totient theorem and Fermat’s little theorem, we know that the multiplicative characteristic of $\mathbb{Z}_{n}$ is nonzero and at most $\phi(n)$ . We can make the following conjectures:

The multiplicative characteristic of $\mathbb{Z}_{n}$ is $\phi(n)$ .
The multiplicative characteristic of $\mathbb{Z}_{p}$ is $p - 1$ .

Clearly, the first of these conjectures implies the second.

An interesting avenue to consider in studying these would be the Carmichael numbers. If $n$ is composite, then $\phi(n) < n - 1$ , but we still have that $x^{n - 1} = 1$ for all $x$ . Do we have that $\phi(n)|(n - 1)$ for all Carmichael numbers $n$ ? Otherwise, can we derive that the GCD of $\phi(n)$ and $n - 1$ (call it $g$ ) would satisfy $x^{g} = 1$ for all $x$ ? Would the characteristic actually be $g$ ?

To build evidence for the second conjecture, we can quickly check that the multiplicative characteristic of $\mathbb{Z}_{5}$ is $5 - 1 = 4$ : it clearly must be at most 4, so we only need to check up to 3, and we have that neither of ${2,2}^{2},2^{3}$ is equal to 1.

An offshoot question here would be to consider multiplicative characteristics of fields. Can we classify the set of all numbers that could be the multiplicative characteristic of some field? What is the multiplicative characteristic of the field of order $p^{k}$ ? (By the classification of finite fields this number must only depend on $p$ and $k$ .)

We suspect that we can approach this by using the minimal subfield / prime subfield, either $\mathbb{Q}$ or $\mathbb{Z}_{p}$ (see the PlanetMath article.) We have that the multiplicative characteristic of $\mathbb{Q,R,C}$ , and in fact any extension of $\mathbb{Q}$ is 0 (just look at the element 2: there is no $n$ for which $2^{n} = 1$ .) Is the converse true, that if a field has multiplicative characteristic 0 then it must contain $\mathbb{Q}$ as a subfield (and thus have $\mathbb{Q}$ as its prime subfield)? For inspiration, we can look at the corresponding proofs for additive characteristics. We have:

Lemma. The following are equivalent for a field $K$ : (1) $K$ contains $\mathbb{Q}$ as a subfield; (2) the prime subfield of $K$ is $\mathbb{Q}$ ; (3) the additive characteristic of $K$ is 0.

Proof. The equivalence of (1) and (2) is well-known, and that (1) or (2) implies (3) is immediate. Thus, it only remains to show that (3) implies (1).

Since the additive characteristic is 0, the elements $n1$ for $n \geq 0$ are all different. In fact, the elements $n1$ for all integers $n$ are all different.

Since we’re in a field, we can divide, and since the additive characteristic is 0, $n1$ is not 0 for nonzero $n$ . Thus, for nonzero $n$ , we can form the number $\frac{m1}{n1}$ .

Say that $\frac{p1}{q1} = \frac{r1}{s1}$ . Then, $(ps)1 = (rq)1$ , and since the additive characteristic is 0, we must have that $ps = rq$ . Conversely, clearly if $ps = rq$ then $\frac{p1}{q1} = \frac{r1}{s1}$ . Thus, $\frac{m1}{n1} \rightarrow \frac{m}{n}$ is a desired isomorphism to $\mathbb{Q}$ , and we are done. $\blacksquare$

Now, let us try to adapt this logic to multiplicative characteristics. Given the prevalence of 1 in the additive characteristic proof, we suspect that CGEs may be helpful here.

It remains to expand on this.

Also, while studying this concept I encountered the similarly named concept of a characteristic subgroup. This could potentially result in some ambiguity, especially if we talk about group characteristics, unless there is some deeper connection between characteristic subgroups and group/monoid characteristics in such a way that justifies the similar terminology (for example, a ring of sets in measure theory is actually a ring in the algebraic sense too, under operations of intersection and symmetric difference.) What about connections between characteristic subgroups and ring characteristics, since both concepts have definitely existed for a while in the literature? It remains to learn more about this.

EDIT, 02/10/2023: Based on an answer to my Math StackExchange question, it seems that the ring characteristic is connected to a particular characteristic subring (which would be the natural analogue for rings of characteristic subgroups.) While this does exhibit some connection between the concepts, it also shows that, at least for the particular subgroup considered there, the logic can’t be generalized to monoid characteristics, since that subgroup isn’t fixed under general group automorphisms (and thus the analogous submonoid certainly wouldn’t be fixed under monoid automorphisms.)

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We include aforementioned proofs of equivalencies here. (We don’t include equivalencies that have been generalized to monoids above.)

Lemma. A ring’s characteristic is zero if and only if no nonzero $n$ exists such that $nx = 0$ for all $x$ .

Proof. If a ring’s characteristic is zero, then there can exist no nonzero $n$ such that $n1 = 0$ , and in particular there can exist no nonzero $n$ with $nx = 0$ for all $x$ . Conversely, if there exists no nonzero $n$ with $nx = 0$ for all $x$ , then assume for contradiction the ring has a nonzero characteristic $c$ . Since $c1 = 0$ , we have $cx = 0$ for all $x$ , contradiction. $\blacksquare$

Lemma. If a ring’s characteristic is nonzero, then it is equal to the min nonzero $n$ such that $nx = 0$ for all $x$ .

Proof. Let the characteristic be $c \neq 0$ , so $c1 = 0$ , hence $cx = 0$ for any $x$ ; thus, $c$ is in the set of nonzero $n$ for which $nx = 0$ for all $x$ . If $n \neq 0$ is such that $nx = 0$ for any $x$ , then in particular $n1 = 0$ . If $n < c$ then this contradicts the definition of characteristic, thus we must have $n \geq c$ . $\blacksquare$

Lemma. In a ring, the min nonzero $n$ such that $nx = 0$ for all $x$ is equal to the max additive order, if either exists.

Proof. The equivalence of existence is true for general monoids, as shown above; hence, we can assume the existence of either to yield the existence of both. Thus, assume there is a min nonzero $n$ such that $nx = 0$ for all $x$ ; we show that $n$ is the max additive order. By a previous lemma, $n$ must be the characteristic of the ring, so $n1 = 0$ and $m1 \neq 0$ for $0 < m < n$ . Thus, $n$ is the additive order of 1, so $n$ is in the set of additive orders. To show $n$ is the max, assume there was some element $x$ with additive order $m > n$ . But then we couldn’t have $nx = 0$ , contradiction. $\blacksquare$

Written April 29, 2022. Changed to more standard notation and added proofs on January 26, 2023; realized that I had made a mistake in generalizing the equivalence of the characteristic and the max order from rings to monoids. Revised on February 9, 2023.

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EDIT (later February or March 2023): This concept applied to groups is exactly the exponent. (I guess I hadn’t read the Wikipedia article on characteristic of a ring fully enough, since that article references this concept too.) We could naturally reuse this terminology to discuss the exponent of a monoid.

Correspondingly, it turns out that many of the questions I asked here have been studied before. In particular, I asked about the exponent of the multiplicative group (of units) of $\mathbb{Z}_n$ ; this is in fact given by the Carmichael function. The status of my conjectures above is already known; the Carmichael function $\lambda(n)$ is not in general equal to $\phi(n)$ (of course, it is clear that $\lambda(n)$ divides $\phi(n)$ ), but just for primes we do have $\lambda(p)=p-1$ . Formulas are known to compute this function for all $n$ . It is fitting that this function is named after the same mathematician who studied the Carmichael numbers — this justifies some of the intuition I had discussed earlier about how we could study this function.

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