# Examples of fields of characteristic 1

I'm a beginner in this topic, so this question seems stupid, but I think it's a doubt many beginners can have.

I realized that I can't find examples of fields with characteristic 1, and as 1 is a prime number we can try to find examples of such fields.

I find a little bit strange this kind of field, it seems that it can have only the zero element, then it can't exist such fields

I said something wrong?

Thanks

• Presumably characteristic $1$ would mean $1=0$. But the specification of a field says $1\ne 0$. – André Nicolas Dec 19 '12 at 8:08
• this should help: en.wikipedia.org/wiki/Field_with_one_element – Holdsworth88 Dec 19 '12 at 8:08
• @PatrickDaSilva Which specifications are these? Just about anywhere I look we require that $K\backslash\{0\}$ is an abelian group, and therefore contains a neutral element $1$. – Cocopuffs Dec 19 '12 at 8:18
• @PatrickDaSilva: I believe that most definitions of field do not allow the one element ring. To check, one would have to go through a collection of standard books. For what it's worth, the Wikipedia entry for "field with one element" explicitly says it is not a field. – André Nicolas Dec 19 '12 at 8:19
• @RafaelChavez: The word "prime," like other mathematical words, is subject to definition. Very few people in the field have called $1$ a prime. The lone serious exception is Legendre. – André Nicolas Dec 19 '12 at 8:24

## 2 Answers

The fields of characteristic $p$ are such that "$p=0$" by handwaving. Therefore, if $1=0$, the only field you can expect is the zero field, which is indeed, as you stated, a bit strange, for it is the only field with this property. For every other field, $1 \neq 0$.

(EDIT : You can interpret my word "expect" in "the only field you can expect" this way : the definition of field that allows $1=0$ only adds the zero field to the possible fields, even though it is non-standard to do so, so we usually assume $1 \neq 0$ to get rid of this case. See the discussion in the comments for more details.)

Usually people do not consider $1$ as a prime, for it does not generate a prime ideal in the ring $\mathbb Z$. Now this again is a matter of definition ; we define the prime ideals those who satisfy some property and are not the whole ring. There are many other reasons why $1$ is usually not a prime, and you just found one of them. $1$ behaves significantly differently than the non-$1$ primes, so it is natural to leave it aside.

Hope that helps,

• The uniqueness of the factorization in prime numbers can be another reason why 1 is not a prime. – user42912 Dec 19 '12 at 8:12
• @Rafael : There are a thousand reasons not to consider $1$ as a prime, I didn't want to make a list. =P – Patrick Da Silva Dec 19 '12 at 8:13
• Maybe can be a subject of another question on MSE :P – user42912 Dec 19 '12 at 8:14
• Dear Patrick, I don't know why you have been downvoted, but I'll upvote you as a compensation. – Georges Elencwajg Dec 19 '12 at 10:36
• Most likely the downvote was for the suggestion that $\{ 0 \}$ is a field... – Zhen Lin Dec 19 '12 at 12:08

According to Alain Connes, characteristic 1 only really shows itself clearly when you deal with semirings (only a commutative monoid under addition) and semifields (semirings in which non-zero elements form a group under multiplication). The only finite semifields are the finite fields and the 2-element Boolean algebra B, with "or" and "and" for + and *. (Note - this is not the 2-element field F_2, which has exclusive or for +.)

Although B does not satisfy 1=0, it does satisfy 1+1 = 1. Hence any B-algebra is a semilattice under addition.

• Can you provide a reference for your quote of Alain Connes? – Aweygan Nov 14 '18 at 19:38