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This has been asked here but marked as answered and I don’t feel like the question was ever answered, or at least was not clear to me.

I don’t understand why the set consisting only of the element $\{0\}$ along with the usual $+$ and $×$ does not satisfy the criteria, since $0$ acts as both the additive and multiplicative identity.

That is, letting $G = \{0\}$, then

  • $∀ g ∈ G, 0+g = g$ and

  • $∀ g ∈ G, 0·g = g$ (Since $0·0 = 0$ )

Similarly, it is both its own additive and multiplicative inverse. What is the problem at only the field level, without wishing it satisfy some additional properties for category theory or algebraic/arithmetic geometry?

PrincessEev
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Furrier Transform
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    One of the axioms of the field is that $0\ne 1$. This is part of the definition. We don't want the ${0}$ ring to be a field. – Mark Dec 06 '20 at 09:11
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    I wouldn't say $0 \ne 1$ is necessarily part of the definition. I've seen it framed that way sometimes, sure, but other framings I've seen simply imply that, which is just as good IMO. – PrincessEev Dec 06 '20 at 09:20
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    It's analogous to why we don't consider $1$ to be a prime number: prime numbers have exactly two divisors, fields have exactly two ideals. These are instances of the "too simple to be simple" phenomenon. – pregunton Dec 06 '20 at 09:53
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    For conveneince, most common field axiomatizations exclude models of cardinality $= 1,$ (either explicitly vua $0\neq 1$ or implicitly as a consequence of other axioms). But this has little to do with the hypothetical field with one element. – Bill Dubuque Dec 06 '20 at 10:56
  • See Is {0} a field?. – Alexey Dec 06 '20 at 10:58
  • These reasons seem to be “because it was too simple to be interesting” rather than any sort of fundamental reason. I find this rather unsatisfying off it were true (no offence to the bearers of bad news). It’s akin to saying that the trivial solution, y = 0, of differential equations isn’t a “real” solution because it’s boring, so adjusting the definition to include the clause “other than y = 0”. – Furrier Transform Dec 06 '20 at 21:04

2 Answers2

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Let $K := \{0\}$. Then $K \setminus \{0\}$ cannot be a multiplicative group, since there is no identity element contained in it.

Marktmeister
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So, let's review: $(F,+,\cdot,0,1)$ is a field if

  • $(F,+,0)$ is an abelian group
  • $(F \setminus \{0\}, \cdot, 1)$ is an abelian group

What happens if $0 = 1$ and $F$ is the singleton containing that element? Then the latter characteristic is not satisfied, for $F \setminus \{ 0 \} = \varnothing$ yet all groups are nonempty by assumption. (Namely, the axioms of group imply the existence of an element in it, the identity element, so a group is always nonempty.)

PrincessEev
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    I was under the impression that this (F \ {0}, · ) criterion was in place to address the problem that 0 did not have a multiplicative inverse in the standard examples (ℝ, ℚ, ℂ...) that fields were modelled after, and that it was either difficult or impossible to define a field in which the additive identity 0, was not absorptive when multiplied, sending everything to 0 ...rather than because 1 was desired to be different from 0.

    Is this not the case?

     – Furrier Transform Dec 06 '20 at 09:33
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    Maybe, maybe not, I don't know for sure. All I know is that within this definition, it's obvious no singleton field exist. – PrincessEev Dec 06 '20 at 19:30