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I believe it is correct to refer to the complex numbers and their 'native algebra' as a field. See for example Linear Algebra and Matrix Theory, by Evar Nering. I assume the same may be said for the set of quaternions. Please correct me if that is wrong.

Based on that assumption, I ask: is the field of complex numbers a subset (sub-field?) of the quaternions in the same sense that the real numbers are 'contained' in the complex numbers?

Is there a term for such a subordination of number fields?

I admit that I am reaching well beyond my realm of familiarity, and that my question may be nonsensical to those who are more knowledgeable in these matters.

José Carlos Santos
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Steven Thomas Hatton
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    I'm guessing you actually meant to ask whether the complex field can be embedded in the (Hamilton) quaternions $;\Bbb H;$ . Now, this last is a divison ring, or a non-commutative "integral domain" (very cautious with this as integral domains' definition usually carries commutativity as conditon), sometimes also named "as skew-field, whereas $;\Bbb C;$ is commutative...so you need an embedding into the center of $;\Bbb H;$ : what is it? – DonAntonio Jan 16 '19 at 09:43
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    Yes and no : $\mathbb{C}$ is a field and $\mathbb{H}$ is a skew-field, containing many copies of $\mathbb{C}$ (but only one copy of $\mathbb{R}$), $\mathbb{H}$ is a non-commutative $\mathbb{R}$-algebra with defining relations $j^2 = -1, zj = j\overline{z}, z \in \mathbb{C}$ so it is not a $\mathbb{C}$-algebra. In this sense it extends $\mathbb{R}$, not $\mathbb{C}$. – reuns Jan 16 '19 at 09:48
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    Is it correct to say that $\Bbb N$ is a subset of $\Bbb Z$ which is a subset of $\Bbb Q$ which is a subset of $\Bbb R$ which is a subset of $\Bbb C$? All those, and more, were discussed many times over on the site before. – Asaf Karagila Jan 16 '19 at 12:58
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    https://math.stackexchange.com/questions/1055501/why-are-integers-subset-of-reals https://math.stackexchange.com/questions/1553537/why-mathbbr-is-a-subset-of-mathbbc https://math.stackexchange.com/questions/892426/is-a-0i-in-every-way-equal-to-just-a https://math.stackexchange.com/questions/1462727/mathbbr-is-not-contained-in-mathbbc https://math.stackexchange.com/questions/2637069/n-dimensional-integer-space-or-mathbfx-in-mathbbrn-x-1-x-2 and more you can find from these pages. It's really all the same question as yours, dressed in a different shirt. – Asaf Karagila Jan 16 '19 at 13:02
  • @AsafKaragila Being a subset and being a subset with a common algebraic structure are not the same thing. For example, $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$ but $\mathbb{Z}$ is not a field. And the "reasons" for these set inclusions depend on how things are defined. For example, the integers may be defined as a set of residue classes of pairs of natural numbers under a particular homomorphism. The reason the set of natural numbers is a subset is because it's convenient to define it as such. – Steven Thomas Hatton Jan 16 '19 at 13:44
  • And $\Bbb N$ is not a field either. And $\Bbb Q$ is not complete, and $\Bbb C$ is not an ordered field. But the arithmetic structure as a semi-ring, or whatever you want to call it, is indeed extending to the quaternions, and above. You have "reasonably canonical embeddings", and you can use them. Whether or not it's convenient to define $\Bbb C$ as a subfield of $\Bbb H$ is a different question. It is certainly sometimes convenient to define $\Bbb R$ as a subfield (which is, in part, a subset) of $\Bbb C$, etc. – Asaf Karagila Jan 16 '19 at 15:18

4 Answers4

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The ring of quaternions(it is not a field) contains infinitely many subrings that are fields and are isomorphic to the field of complex numbers.

Emilio Novati
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The field of complex numbers is isomorphic to a subfield of the division ring of the quaternions. Also, the field of real numbers is isomorphic to a subfield of the complex field.

Whether or not we have subsets here depends on the way the complex numbers and the quaternions are defined.

Silvio Mayolo
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José Carlos Santos
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    I don't like this answer. Mathematicians assume integers are a subset of reals are a subset of complex numbers all the time, and it's not exactly wrong unless you're being pedantic about set-theoretic underlying details. More importantly, this answer does not highlight the main problem in the question which was that quaternions do not form a field. – Caleb Stanford Jan 16 '19 at 17:27
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    In Mathematics, any question about what something is is a question about definitions. Yes, I suppose that every working mathematician (me included) feels that $\mathbb{R}\subset\mathbb{C}\subset\mathbb H$, but it is clear from the question that the OP knows that. And I mentioned implicitly “the main problem in the question”: I wrote that $\mathbb C$ is a field, and that $\mathbb H$ is a division ring. – José Carlos Santos Jan 16 '19 at 23:22
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    @6005: Even then it's not necessarily wrong. I also don't understand why it's a problem that $\Bbb H$ is not a field. Can a field be only a subset of another field? – Asaf Karagila Jan 16 '19 at 23:39
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    @6005 I think the distinction here is that there is a canonical embedding of rings $\mathbb{Z} \hookrightarrow \mathbb{R}$ whereas there is no canonical copy of $\mathbb{C}$ in the quaternions. – hunter Jan 16 '19 at 23:44
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    @AsafKaragila True. The only reason that's a problem is that in the question it says "contained in the field of quaternions" – Caleb Stanford Jan 17 '19 at 10:55
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    @hunter Agree with that distinction. $\mathbb{R} \hookrightarrow \mathbb{C}$ is canonical, though. – Caleb Stanford Jan 17 '19 at 10:57
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The quaternion skew field ${\Bbb H} = {\Bbb C}\times {\Bbb C}$ has ${\Bbb C}$ as a subfield by considering the ring monomorphism ${\Bbb C}\rightarrow {\Bbb H}: z\mapsto (z,0)$.

Here the addition is defined component-wise and the multiplication is defined as $$(z,t)(z',t') := (zz' - t^* t' , z^*t' + tz' ),$$ where $*$ means conjugation.

Thus $(z,0)+(z',0) = (z+z',0)$ and $(z,0)(z',0) = (zz',0)$ as required.

Wuestenfux
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2

The Hurwitz-Frobenious theorem is your friend here. It states that the only normed division algebras over the real numbers are the reals, the complex numbers which are obtained by doubling the reals, the quaternions that are obtained by doubling the complex numbers, and the octonions that are obtained by doubling the quaternions.

With each doubling, you lose something. You lose ordering moving to the complex numbers. You lose commutativity moving to the quaternions, and you lose associativity moving to the octonions.

You can continue the process to get the Clifford algebras with dimension $2^n$, but you loose the existence of inverses for all non-zero elements, and so these are no longer normed division algebras.

The doubling process is mechanically isomorphic to constructing $2\times2$ matrices from two elements of the previous algebra with with the diagonal duplicating one value 'a' and the cross diagonal conjugating the other value 'b'.

Geometrically, the quaternions are a four dimensional vector space with Euclidean norm that admits a well defined multiplication with inverses. This multiplication is a representation of the four dimensional rotations (which rotate a plane through the origin and fix the two dimensions orthogonal to that plane). The three dimensional rotations are a proper subset of the four dimensional rotations.

Any plane through the origin within the quaternions is algebraically isomorphic to the complex numbers, which in turn may be thought of as a representation of the two dimensional rotations.

These are the basic facts, consideration of which should serve to clarify your terminology and answer your questions.

José Carlos Santos
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Hurwitz_showed_the_way
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