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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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    What convention are you using that considers $1$ a prime number?2012-12-19
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    Presumably characteristic $1$ would mean $1=0$. But the specification of a field says $1\ne 0$.2012-12-19
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    this should help: http://en.wikipedia.org/wiki/Field_with_one_element2012-12-19
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    @Amzoti then my mistake was to be considered 1 prime?2012-12-19
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    @Andre : The specifications of a field don't require that. Many authors assume $1 \neq 0$ because the only field satisfying $1 = 0$ is the zero field, so they leave it aside because it behaves differently with the other ones and we don't want to consider it.2012-12-19
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    @AndréNicolas very interesting!!!2012-12-19
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    @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 $1$.2012-12-19
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    @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.2012-12-19
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    @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.2012-12-19
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    @André Nicolas : Damn wiki. :P But seriously ; if you do not ask that $F^{\times}$ is a group, but rather that it satisfies all its axioms beside non-emptiness, you get the same definition of a field except that $\{0\}$ satisfies it. We have discussed this in my Galois theory class, and we arbitrarily chose to remove $\{0\}$ from our definition of field.2012-12-19
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    It is interesting that definitions are arrived at democratically in that course. Luckily, the deinition arrived at was the usual one.2012-12-19
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    Dear @Patrick, there is not one serious book or mathematician on this planet today claiming that the ring with one element is a field: I hereby challenge you to name a counter-example :-)2012-12-19
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    @Georges : I am not saying it is *useful* to do so, I can even say it is *useless*. What I am saying is that you can remove a *very little subtlety* in the definition of a field and this unique zero-field will appear. This is why most people ask for $1 \neq 0$. I think it is nice to understand such things rather than just accepting the definitions mathematicians tell you.2012-12-19
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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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    The uniqueness of the factorization in prime numbers can be another reason why 1 is not a prime.2012-12-19
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    @Rafael : There are a thousand reasons not to consider $1$ as a prime, I didn't want to make a list. =P2012-12-19
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    Maybe can be a subject of another question on MSE :P2012-12-19
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    Thank you it helped a lot!2012-12-19
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    @Rafael : It is possible that this question has already been asked. Otherwise, feel free to do so.2012-12-19
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    I modified the answer so that the downvoter may revise his downvote.2012-12-19
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    Dear Patrick, I don't know why you have been downvoted, but I'll upvote you as a compensation.2012-12-19
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    Most likely the downvote was for the suggestion that $\{ 0 \}$ is a field...2012-12-19
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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