域的另类定义：能否将域定义为与其自同构群同构的群？

Great question! Let's break this down into two clear parts to address your query thoroughly.

1. 能否用与自同构群同构的群G定义域？ #

First, we need to clarify: when you say "automorphism group of G", we're talking about the automorphism group of G as an additive group (since you want to use G's group operation as the field's addition). For G to be the additive group of a field, it must be abelian (field addition is always commutative), so we're looking at abelian groups G where G ≅ Aut(G) (Aut(G) is the group of additive automorphisms of G).

But if we look at all common classes of fields, this is impossible:

• Finite fields GF(pⁿ): Their additive group is (ℤ/pℤ)ⁿ, whose automorphism group is GL(n, ℤ/pℤ) (the group of n×n invertible matrices over ℤ/pℤ). The order of the additive group is pⁿ, while GL(n, ℤ/pℤ) has order (pⁿ-1)(pⁿ-p)...(pⁿ-pⁿ⁻¹) — these are never equal (except for the trivial group, which can't form a field).
• Characteristic 0 fields: Take ℚ (the rationals) as an example. The additive group (ℚ, +) has automorphism group isomorphic to (ℚ*, ·) (the multiplicative group of non-zero rationals). But (ℚ, +) is torsion-free (no elements of finite order), while (ℚ*, ·) has elements like -1 (order 2) — they can't be isomorphic. More generally, characteristic 0 field additive groups are vector spaces over ℚ, whose automorphism groups are linear groups (often non-abelian), which can't match the abelian structure of the additive group.

Are there any abelian groups G where G ≅ Aut(G)? Maybe some exotic infinite groups, but none of these can be the additive group of a field. The additive groups of fields are either torsion (characteristic p) or torsion-free (characteristic 0), and neither type has an automorphism group isomorphic to itself. So the short answer here is: No, you can't define a field using such a group G, because no valid field additive group satisfies this isomorphism condition.

2. 若强行按此定义，是否等价于标准域？ #

Suppose we set aside the existence issue and try to define operations as you proposed:

• Addition: Use G's group operation (denoted +)
• Multiplication: For a, b ∈ G, define a*b = σₐ(b), where σₐ is the automorphism corresponding to a (via the isomorphism G ≅ Aut(G))

This definition fails to meet core field axioms in multiple ways:

(1) Contradiction with multiplicative identity #

Fields require a multiplicative identity 1 ≠ 0 (the additive identity) such that 1b = b for all b. But since G ≅ Aut(G) is a group isomorphism, the additive identity 0 of G maps to the identity automorphism id_G (the identity element of Aut(G)). That means 0b = id_G(b) = b for all b — which directly contradicts the field axiom that 0 multiplied by any element equals 0. The only way this works is if G is trivial, which isn't a field.

(2) Multiplicative associativity isn't guaranteed #

Field multiplication must be associative: (ab)c = a(bc). Expanding both sides:

• Left side: (a*b)c = σ_{ab}(c) = σ_{σₐ(b)}(c)
• Right side: a*(b*c) = σₐ(σ_b(c)) = (σₐ ∘ σ_b)(c)

For these to be equal, we need σ_{σₐ(b)} = σₐ ∘ σ_b for all a, b ∈ G. This is a very strong extra condition that isn't satisfied by a general isomorphism between G and Aut(G). Only special homomorphisms (not just any isomorphism) would meet this requirement.

(3) Inverse elements don't behave as required #

Fields demand that every non-zero element a has an inverse a' such that aa' = 1. But we already saw the only possible "identity" here is 0, so we'd need σₐ(a') = 0. Since σₐ is an automorphism, this implies a' = 0 — but 0a = 0 ≠ 1 (if a valid 1 existed). This completely breaks the inverse axiom for non-zero elements.

(4) Multiplication isn't commutative (usually) #

Fields require commutative multiplication, but here ab = σₐ(b) and ba = σ_b(a). These are equal only if σₐ(b) = σ_b(a) for all a, b ∈ G. Automorphism groups (like GL(n, ℤ) for n ≥ 2) are typically non-abelian, so this equality won't hold in general. Even if G is abelian, its automorphism group might not be, so multiplication would fail commutativity.

All these issues mean this definition is not equivalent to the standard field definition — it doesn't even satisfy the basic axioms of a commutative ring, let alone a field.

内容的提问来源于stack exchange，提问作者Akababa