*4.1. On the Characteristic*

In classical field theory, the characteristic of a field, if finite, can only be a prime number. In contrast, we demonstrate below that any positive integer larger than 1 can be realized as the characteristic of some hyperfield. For a group *G* and *x*1, ..., *xn* ∈ *G*, we will denote by &*x*1, ..., *xn*' the subgroup of *G* generated by *x*1, ..., *xn*.

**Theorem 3.** *For every natural number n* ∈ N>1*, there exists an infinite quotient hyperfield F, such that* char *F* = *n.*

**Proof.** We are going to demonstrate that for every natural number *n* ∈ N, the quotient hyperfield *F* = Q&−*n*' has characteristic *n* + 1. Observe that

$$0 = \underbrace{1 + \ldots + 1}\_{n \text{ times}} - n \text{, so } [0]\_{\langle -n \rangle} \in (n+1) \times [1]\_{\langle -n \rangle}.$$

Hence, char *F* ≤ *n* + 1. If *n* = 1, then char *F* = 2. Let *n* > 1 and suppose that char *F* = *k* < *n* + 1, i.e.,

$$[0]\_{\langle -n \rangle} \in k \times [1]\_{\langle -n \rangle'} \text{ where } 1 < k \le n.$$

Then, there exist *xi* ∈ Z, *i* ∈ {1, ..., *k*}, such that

$$0 = (-n)^{x\_1} + \dots + (-n)^{x\_k}.$$

Let *N* = min{*x*1, ..., *xk*} and denote *mi* := *xi* − *N*. Then

$$0 = (-n)^{m\_1} + \dots + (-n)^{m\_k},\tag{2}$$

where *mi* ∈ N ∪ {0} and 1 < *k* ≤ *n*. Observe that

$$(-n)^{2l} \equiv 1 \pmod{n+1} \text{ and } (-n)^{2l+1} \equiv 1 \pmod{n+1}$$

for every *l* ∈ N. Thus, the left side of the Equation (2) is congruent to 0 modulo *n* + 1, while the right side is congruent to *k* modulo *n* + 1. Hence,

$$0 \equiv k \pmod{n+1} \,\_r$$

which is a contradiction, since 1 < *k* ≤ *n*. As a consequence, we obtain that *F* has characteristic *n* + 1.

Let us now demonstrate that finite hyperfields of even cardinality must have characteristic 2.

**Proposition 6.** *Let F be a finite hyperfield of even cardinality. Then,* char *F* = 2*.*

**Proof.** Assume that *F* is a finite hyperfield, such that char *F* > 2. Then, 1 = −1 and thus *a* = −*a* for all *a* ∈ *F*×. Therefore, |*F*×| must be even, and hence |*F*| is odd.

If we restrict our attention to hyperfields obtained with the quotient construction, then we can make the following observations.

**Lemma 2.** *Let K be a field and G a subgroup of K*×*. Then,* char *KG* ≤ char *K.*

If we consider quotient hyperfields *KG*, constructed with respect to a finite group *G*, then by looking at the proper divisors of the cardinality of *G*, it is possible to bound from above the characteristic of *KG*.

**Proposition 7.** *Let K be a field and G be a finite subgroup of K*×*. If n* ∈ N><sup>1</sup> *divides the cardinality of G, then* char *KG* ≤ *n.*

**Proof.** The group *G* is a finite subgroup of the multiplicative group of a field; thus, it is cyclic. Therefore, if *n* divides |*G*|, then there exists an element *g* ∈ *G* of order *n*. Hence, &*g*' = {1, *g*, ..., *gn*−1} and |&*g*'| = *n*. Since

$$0 = \mathbb{g}^n - 1 = (\mathbb{g} - 1)(1 + \mathbb{g} + \mathbb{g}^2 + \dots + \mathbb{g}^{n-1})$$

and *g* = 1, we obtain that

$$1 + \mathfrak{g} + \mathfrak{g}^2 + \dots + \mathfrak{g}^{n-1} = 0.$$

We conclude that 0 ∈ *n* × [1]*KG* , so char *F* ≤ *n*.

Conversely, sometimes from the characteristic of *KG* it is possible to deduce information on the divisors of |*G*|.

**Lemma 3.** *Let K be a field such that* 1 = −1 *and G are a finite subgroup of K*×*. If* char *KG* = 2*, then the cardinality of G is even.*

**Proof.** Since *G* is a finite subgroup of the multiplicative group of a field, it is cyclic. Let *g* be a generator of *G*. If char *KG* = 2, then 0 ∈ [1]*<sup>G</sup>* + [1]*G*. Hence [1]*<sup>G</sup>* = −[1]*<sup>G</sup>* = [−1]*G*, so −1 ∈ *G*. We conclude that there is a positive integer *k* ∈ N, such that *k* < |*G*| and *g<sup>k</sup>* = −1. Then, 1 = (−1)<sup>2</sup> = *g*2*k*, hence |*G*| = 2*k* is even.

**Remark 8.** *The above result does not hold if* 1 = −1 *in K, since in that case the quotient hyperfield KG would have characteristic 2 for any multiplicative subgroup G of K*×*.*

Combining Proposition 7 and Lemma 3, we derive the following result.

**Corollary 1.** *Let K be a field such that* 1 = −1 *and G are a finite subgroup of K*×*. Then, the quotient hyperfield KG has characteristic 2 if and only if* |*G*| *is even.*
