ℚ上有限维结合除代数相关：其有限维域扩张能否嵌入ℍ？

Hey there! Let's break down your questions about finite-dimensional extensions of ℚ (commonly called number fields) and their ability to embed into the quaternion division algebra ℍ.

Key Properties of ℚ's Finite-Dimensional Field Extensions #

• Core Definition: A finite-dimensional extension of ℚ is a field ( K ) that acts as a finite-dimensional vector space over ℚ. The dimension ( [K:\mathbb{Q}] ) is called the degree of the extension. Every element in ( K ) is algebraic over ℚ—meaning it's a root of some monic polynomial with rational coefficients (this polynomial is unique for each element and called its minimal polynomial).
• Algebraic Integer Rings: Each number field ( K ) has an associated ring of algebraic integers ( \mathcal{O}_K ), which is all elements in ( K ) that satisfy a monic polynomial with integer coefficients. ( \mathcal{O}_K ) is a Dedekind domain: every non-zero ideal factors uniquely into a product of prime ideals, a generalization of unique factorization in ℤ. Note that unique factorization of elements doesn't always hold here—for example, in ( \mathbb{Q}(\sqrt{-5}) ), the integer ring ( \mathbb{Z}[\sqrt{-5}] ) has ( 6 = 2 \times 3 = (1+\sqrt{-5})(1-\sqrt{-5}) ), two distinct irreducible factorizations.
• Galois Extensions: If ( K ) is a Galois extension of ℚ (normal and separable), the Galois group ( \text{Gal}(K/\mathbb{Q}) ) is a finite group with order exactly equal to the extension degree ( [K:\mathbb{Q}] ). The Galois correspondence applies here: subgroups of the Galois group match up bijectively with intermediate fields between ℚ and ( K ), and normal subgroups correspond to intermediate fields that are also Galois over ℚ.
• Embeddings into ℂ: Every number field can be embedded into the complex numbers ℂ. The number of distinct such embeddings is exactly ( [K:\mathbb{Q}] ). For example, ( \mathbb{Q}(\sqrt{2}) ) has two embeddings: one that leaves ( \sqrt{2} ) unchanged, and another that maps ( \sqrt{2} ) to ( -\sqrt{2} ).
• Local-Global Link: For every prime number ( p ), we can "localize" ( K ) to get a finite extension of the ( p )-adic field ( \mathbb{Q}_p ). This local field theory is foundational in number theory, enabling the local-global principle—many problems over ( K ) can be solved by first solving them over all corresponding local fields.

Can All These Extensions Embed into ℍ? #

Short answer: Yes, every finite-dimensional extension of ℚ can be embedded into the quaternion division algebra ℍ. Here's the breakdown:

1. First, any number field ( K ) can be embedded into ℂ (as we noted earlier—ℂ contains all algebraic numbers, so this is guaranteed).
2. ℂ itself can be embedded into ℍ via the map ( a + bi \mapsto a + bi + 0j + 0k ). This is a valid field homomorphism that preserves all addition and multiplication operations.
3. Composing these two embeddings gives a field homomorphism ( K \to \mathbb{C} \to \mathbb{H} ), which is exactly an embedding of ( K ) into ℍ.

Even for non-real number fields like ( \mathbb{Q}(i) ), this works: the embedding maps ( \mathbb{Q}(i) ) to the subset ( { a + bi \mid a,b \in \mathbb{Q} } \subseteq \mathbb{H} ), which is a subfield of ℍ. For real number fields like ( \mathbb{Q}(\sqrt[3]{2}) ), we can embed them directly into ℝ (a subfield of ℍ) instead. And for number fields with mixed real/complex embeddings, we can always choose either a real embedding (guaranteed for odd-degree extensions by the Artin-Schreier theorem) or the complex-to-ℍ embedding path.

It's worth clarifying: this only applies to field extensions of ℚ. Finite-dimensional division algebras over ℚ are a different story—many can't embed into ℍ—but your question is specifically about field extensions, so all of them work.

内容的提问来源于stack exchange，提问作者Mar