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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

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    Dear @Patrick: I completely agree with you that it is interesting to analyze the axioms for "field", or for that matter of any mathematical structure. At the end of the day, however, we have to adopt the accepted definitions.2012-12-19

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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,

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    @ZhenLin : I added a discussion that explains in which sense I say that. And I obviously not mean it in the standard sense. What's wrong in that?2012-12-19
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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.

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    Can you provide a reference for your quote of Alain Connes?2018-11-14