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I am currently exploring the fundamental of field theory, especially, in its connection to monoid and group.

One way we can describe a field $\mathbb{F}$ is using the following axioms:
F1. $(\mathbb{F}, +)$ forms an abelian group.
F2. $(\mathbb{F}, \times)$ forms an abelian monoid.
F3. There exists some $F \subseteq \mathbb{F}$ such that $(F, \times)$ forms a nontrivial abelian group.
F4. $\forall a, b, c \in \mathbb{F}:\; a \cdot (b + c)= (a \cdot b) + (a \cdot c)$

My first question: Please confirm whether these axioms are valid for describing a field $\mathbb{F}$.

And if we recall the well-known fields such as $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$, axioms F3 is satified by these fields since $\mathbb{Q} \setminus \{0\}$, $\mathbb{R} \setminus \{0\}$ and $\mathbb{C} \setminus \{0\}$ form multiplicative abelian groups. Perhaps, it is worth noting that the subset of each of these fields which forms an abelian multiplicative group excludes only the additive identity (the number $0$). My second question: Is this property mandatory, or we can stick to axiom F3 I mentioned above?

I would really appreciate your feedbacks. Thank you.

rp23
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    Your $F_3$ is not sufficient, you need to have $\mathbb{F} \setminus {0}$ is an abelian group. Otherwise it is a ring. The trivial group verifies your property – julio_es_sui_glace Apr 13 '24 at 09:01
  • @julio_es_sui_glace, you are right about the trivial group will verifies my property. So, let me revise the question a little bit. – rp23 Apr 13 '24 at 09:12
  • Still not correct, for $\mathrm{char} \neq 2$ the group ${-1,1}$ verifies your property. Also for a ring the group of units $R^\times$ will always verify your property, you can’t modify this axiom like that. Also if you kept this axiom, then $\mathbb{F}_2$ would not be a field – julio_es_sui_glace Apr 13 '24 at 09:16
  • @julio_es_sui_glace You are right, so the requirement that only the additive identity in the exclusion of the multiplicaitve group, i. e., $\mathbb{F} \setminus {0}$ is mandatory. So my axiom F3 is not sufficient. The revised axiom F3 shall be "$(\mathbb{F} \setminus {0}, \times)$ forms an abelian group", as you've mentioned earlier. Thank you very much for your response. I'd appreciate it more if you would write your response in the answer section. But, thank you. – rp23 Apr 13 '24 at 09:54

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As we discussed in the comment section, your axiom $F_3$ needs to be $(\mathbb{F}\setminus \{0\},\times)$ is an abelian group.

  • For your first version of $F_3$ you wanted the existence of a submonoïd $F$ of $\mathbb{F}$ that happens to be a group, but this is clearly not sufficient since this is always the case with $F = \{1\}$, and all rings are not fields, worse you allow non-integral domains such as $\mathbb{Z}/mn\mathbb{Z}$, which don't have a field of fractions.
  • After your revision, you wanted to have $F$ non-trivial. This has two main issues:
    Firstly: $\mathbb{F}_2$ would not be a field since the only submonoid that happens to be a group is the trivial group.
    Secondly: If $\operatorname{char} \mathbb{F} \neq 2$, there is the non trivial group $\{-1,1\}$ (which only works for $\mathbb{F}_3$), so every ring of characteristic different of $2$ would be a field.

Even if you wanted to make this harsher, the group of units $\mathbb{F}^\times$ is the largest submonoid that happens to be a group, and this for any ring, so you must impose $\mathbb{F}^\times = \mathbb{F} \setminus \{0\}$ or much simpler $\mathbb{F}\setminus\{0\}$ is a group. In fact this is at the classical definition of a field: a ring which every non-zero element is invertible.

julio_es_sui_glace
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