Is it correct to say the field of complex numbers is contained in the field of quaternions?
$begingroup$
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.
abstract-algebra complex-numbers definition quaternions
asked 19 hours ago
Steven HattonSteven Hatton
794316
$endgroup$
|
show 1 more comment
$begingroup$
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.
abstract-algebra complex-numbers definition quaternions
asked 19 hours ago
Steven HattonSteven Hatton
794316
$endgroup$
$begingroup$
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?
$endgroup$
– DonAntonio
18 hours ago
3
$begingroup$
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 $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
$endgroup$
– reuns
18 hours ago
2
$begingroup$
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.
$endgroup$
– Asaf Karagila♦
15 hours ago
2
$begingroup$
math.stackexchange.com/questions/1055501/… math.stackexchange.com/questions/1553537/… math.stackexchange.com/questions/892426/… math.stackexchange.com/questions/1462727/… math.stackexchange.com/questions/2637069/… and more you can find from these pages. It's really all the same question as yours, dressed in a different shirt.
$endgroup$
– Asaf Karagila♦
15 hours ago
$begingroup$
@AsafKaragila Being a subset and being a subset with a common algebraic structure are not the same thing. For example, $mathbb{N}subsetmathbb{Z}subsetmathbb{Q}subsetmathbb{R}subsetmathbb{C}$ 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.
$endgroup$
– Steven Hatton
14 hours ago
|
show 1 more comment
$begingroup$
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.
abstract-algebra complex-numbers definition quaternions
asked 19 hours ago
Steven HattonSteven Hatton
794316
$endgroup$
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.
abstract-algebra complex-numbers definition quaternions
abstract-algebra complex-numbers definition quaternions
asked 19 hours ago
Steven HattonSteven Hatton
794316
asked 19 hours ago
Steven HattonSteven Hatton
794316
edited 18 hours ago
Steven Hatton
asked 19 hours ago
Steven HattonSteven Hatton
794316
asked 19 hours ago
Steven HattonSteven Hatton
794316
asked 19 hours ago
Steven HattonSteven Hatton
794316
794316
$begingroup$
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?
$endgroup$
– DonAntonio
18 hours ago
3
$begingroup$
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 $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
$endgroup$
– reuns
18 hours ago
2
$begingroup$
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.
$endgroup$
– Asaf Karagila♦
15 hours ago
2
$begingroup$
math.stackexchange.com/questions/1055501/… math.stackexchange.com/questions/1553537/… math.stackexchange.com/questions/892426/… math.stackexchange.com/questions/1462727/… math.stackexchange.com/questions/2637069/… and more you can find from these pages. It's really all the same question as yours, dressed in a different shirt.
$endgroup$
– Asaf Karagila♦
15 hours ago
$begingroup$
@AsafKaragila Being a subset and being a subset with a common algebraic structure are not the same thing. For example, $mathbb{N}subsetmathbb{Z}subsetmathbb{Q}subsetmathbb{R}subsetmathbb{C}$ 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.
$endgroup$
– Steven Hatton
14 hours ago
|
show 1 more comment
$begingroup$
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?
$endgroup$
– DonAntonio
18 hours ago
3
$begingroup$
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 $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
$endgroup$
– reuns
18 hours ago
2
$begingroup$
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.
$endgroup$
– Asaf Karagila♦
15 hours ago
2
$begingroup$
math.stackexchange.com/questions/1055501/… math.stackexchange.com/questions/1553537/… math.stackexchange.com/questions/892426/… math.stackexchange.com/questions/1462727/… math.stackexchange.com/questions/2637069/… and more you can find from these pages. It's really all the same question as yours, dressed in a different shirt.
$endgroup$
– Asaf Karagila♦
15 hours ago
$begingroup$
@AsafKaragila Being a subset and being a subset with a common algebraic structure are not the same thing. For example, $mathbb{N}subsetmathbb{Z}subsetmathbb{Q}subsetmathbb{R}subsetmathbb{C}$ 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.
$endgroup$
– Steven Hatton
14 hours ago
$begingroup$
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?
$endgroup$
– DonAntonio
18 hours ago
$begingroup$
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?
$endgroup$
– DonAntonio
18 hours ago
3
3
$begingroup$
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 $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
$endgroup$
– reuns
18 hours ago
$begingroup$
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 $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
$endgroup$
– reuns
18 hours ago
2
2
$begingroup$
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.
$endgroup$
– Asaf Karagila♦
15 hours ago
$begingroup$
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.
$endgroup$
– Asaf Karagila♦
15 hours ago
2
2
$begingroup$
math.stackexchange.com/questions/1055501/… math.stackexchange.com/questions/1553537/… math.stackexchange.com/questions/892426/… math.stackexchange.com/questions/1462727/… math.stackexchange.com/questions/2637069/… and more you can find from these pages. It's really all the same question as yours, dressed in a different shirt.
$endgroup$
– Asaf Karagila♦
15 hours ago
$begingroup$
math.stackexchange.com/questions/1055501/… math.stackexchange.com/questions/1553537/… math.stackexchange.com/questions/892426/… math.stackexchange.com/questions/1462727/… math.stackexchange.com/questions/2637069/… and more you can find from these pages. It's really all the same question as yours, dressed in a different shirt.
$endgroup$
– Asaf Karagila♦
15 hours ago
$begingroup$
@AsafKaragila Being a subset and being a subset with a common algebraic structure are not the same thing. For example, $mathbb{N}subsetmathbb{Z}subsetmathbb{Q}subsetmathbb{R}subsetmathbb{C}$ 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.
$endgroup$
– Steven Hatton
14 hours ago
$begingroup$
@AsafKaragila Being a subset and being a subset with a common algebraic structure are not the same thing. For example, $mathbb{N}subsetmathbb{Z}subsetmathbb{Q}subsetmathbb{R}subsetmathbb{C}$ 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.
$endgroup$
– Steven Hatton
14 hours ago
|
show 1 more comment
4 Answers
4
active
oldest
votes
$begingroup$
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.
edited 13 hours ago
Silvio Mayolo
20017
answered 18 hours ago
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
7
$begingroup$
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.
$endgroup$
– 6005
11 hours ago
$begingroup$
In Mathematics, any question about what something is is a question about definitions. Yes, I suppose that every working mathematician (me included) feels that $mathbb{R}subsetmathbb{C}subsetmathbb H$, 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.
$endgroup$
– José Carlos Santos
5 hours ago
$begingroup$
@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?
$endgroup$
– Asaf Karagila♦
5 hours ago
$begingroup$
@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.
$endgroup$
– hunter
4 hours ago
add a comment |
$begingroup$
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.
answered 18 hours ago
Emilio NovatiEmilio Novati
51.8k43474
$endgroup$
add a comment |
$begingroup$
The quaternion skew field ${Bbb H} = {Bbb C}times {Bbb C}$ has ${Bbb C}$ as a
subfield by considering the ring monomorphism ${Bbb C}rightarrow {Bbb H}: zmapsto (z,0)$.
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.
answered 17 hours ago
WuestenfuxWuestenfux
4,1451411
$endgroup$
add a comment |
$begingroup$
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 loose something. You loose ordering moving to the complex numbers. You loose commutativity moving to the quaternions, and you loose 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 2x2 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.
answered 5 hours ago
Hurwitz_showed_the_wayHurwitz_showed_the_way
1
New contributor
Hurwitz_showed_the_way is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
edited 13 hours ago
Silvio Mayolo
20017
answered 18 hours ago
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
7
$begingroup$
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.
$endgroup$
– 6005
11 hours ago
$begingroup$
In Mathematics, any question about what something is is a question about definitions. Yes, I suppose that every working mathematician (me included) feels that $mathbb{R}subsetmathbb{C}subsetmathbb H$, 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.
$endgroup$
– José Carlos Santos
5 hours ago
$begingroup$
@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?
$endgroup$
– Asaf Karagila♦
5 hours ago
$begingroup$
@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.
$endgroup$
– hunter
4 hours ago
add a comment |
$begingroup$
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.
edited 13 hours ago
Silvio Mayolo
20017
answered 18 hours ago
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
7
$begingroup$
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.
$endgroup$
– 6005
11 hours ago
$begingroup$
In Mathematics, any question about what something is is a question about definitions. Yes, I suppose that every working mathematician (me included) feels that $mathbb{R}subsetmathbb{C}subsetmathbb H$, 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.
$endgroup$
– José Carlos Santos
5 hours ago
$begingroup$
@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?
$endgroup$
– Asaf Karagila♦
5 hours ago
$begingroup$
@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.
$endgroup$
– hunter
4 hours ago
add a comment |
$begingroup$
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.
edited 13 hours ago
Silvio Mayolo
20017
answered 18 hours ago
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
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.
edited 13 hours ago
Silvio Mayolo
20017
answered 18 hours ago
José Carlos SantosJosé Carlos Santos
154k22124227
edited 13 hours ago
Silvio Mayolo
20017
edited 13 hours ago
Silvio Mayolo
20017
edited 13 hours ago
Silvio Mayolo
20017
20017
answered 18 hours ago
José Carlos SantosJosé Carlos Santos
154k22124227
answered 18 hours ago
José Carlos SantosJosé Carlos Santos
154k22124227
answered 18 hours ago
José Carlos SantosJosé Carlos Santos
154k22124227
154k22124227
7
$begingroup$
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.
$endgroup$
– 6005
11 hours ago
$begingroup$
In Mathematics, any question about what something is is a question about definitions. Yes, I suppose that every working mathematician (me included) feels that $mathbb{R}subsetmathbb{C}subsetmathbb H$, 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.
$endgroup$
– José Carlos Santos
5 hours ago
$begingroup$
@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?
$endgroup$
– Asaf Karagila♦
5 hours ago
$begingroup$
@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.
$endgroup$
– hunter
4 hours ago
add a comment |
7
$begingroup$
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.
$endgroup$
– 6005
11 hours ago
$begingroup$
In Mathematics, any question about what something is is a question about definitions. Yes, I suppose that every working mathematician (me included) feels that $mathbb{R}subsetmathbb{C}subsetmathbb H$, 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.
$endgroup$
– José Carlos Santos
5 hours ago
$begingroup$
@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?
$endgroup$
– Asaf Karagila♦
5 hours ago
$begingroup$
@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.
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– hunter
4 hours ago
7
7
$begingroup$
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.
$endgroup$
– 6005
11 hours ago
$begingroup$
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.
$endgroup$
– 6005
11 hours ago
$begingroup$
In Mathematics, any question about what something is is a question about definitions. Yes, I suppose that every working mathematician (me included) feels that $mathbb{R}subsetmathbb{C}subsetmathbb H$, 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.
$endgroup$
– José Carlos Santos
5 hours ago
$begingroup$
In Mathematics, any question about what something is is a question about definitions. Yes, I suppose that every working mathematician (me included) feels that $mathbb{R}subsetmathbb{C}subsetmathbb H$, 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.
$endgroup$
– José Carlos Santos
5 hours ago
$begingroup$
@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?
$endgroup$
– Asaf Karagila♦
5 hours ago
$begingroup$
@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?
$endgroup$
– Asaf Karagila♦
5 hours ago
$begingroup$
@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.
$endgroup$
– hunter
4 hours ago
$begingroup$
@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.
$endgroup$
– hunter
4 hours ago
add a comment |
$begingroup$
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.
answered 18 hours ago
Emilio NovatiEmilio Novati
51.8k43474
$endgroup$
add a comment |
$begingroup$
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.
answered 18 hours ago
Emilio NovatiEmilio Novati
51.8k43474
$endgroup$
add a comment |
$begingroup$
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.
answered 18 hours ago
Emilio NovatiEmilio Novati
51.8k43474
$endgroup$
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.
answered 18 hours ago
Emilio NovatiEmilio Novati
51.8k43474
answered 18 hours ago
Emilio NovatiEmilio Novati
51.8k43474
answered 18 hours ago
Emilio NovatiEmilio Novati
51.8k43474
answered 18 hours ago
Emilio NovatiEmilio Novati
51.8k43474
51.8k43474
add a comment |
add a comment |
$begingroup$
The quaternion skew field ${Bbb H} = {Bbb C}times {Bbb C}$ has ${Bbb C}$ as a
subfield by considering the ring monomorphism ${Bbb C}rightarrow {Bbb H}: zmapsto (z,0)$.
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.
answered 17 hours ago
WuestenfuxWuestenfux
4,1451411
$endgroup$
add a comment |
$begingroup$
The quaternion skew field ${Bbb H} = {Bbb C}times {Bbb C}$ has ${Bbb C}$ as a
subfield by considering the ring monomorphism ${Bbb C}rightarrow {Bbb H}: zmapsto (z,0)$.
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.
answered 17 hours ago
WuestenfuxWuestenfux
4,1451411
$endgroup$
add a comment |
$begingroup$
The quaternion skew field ${Bbb H} = {Bbb C}times {Bbb C}$ has ${Bbb C}$ as a
subfield by considering the ring monomorphism ${Bbb C}rightarrow {Bbb H}: zmapsto (z,0)$.
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.
answered 17 hours ago
WuestenfuxWuestenfux
4,1451411
$endgroup$
The quaternion skew field ${Bbb H} = {Bbb C}times {Bbb C}$ has ${Bbb C}$ as a
subfield by considering the ring monomorphism ${Bbb C}rightarrow {Bbb H}: zmapsto (z,0)$.
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.
answered 17 hours ago
WuestenfuxWuestenfux
4,1451411
edited 14 hours ago
answered 17 hours ago
WuestenfuxWuestenfux
4,1451411
answered 17 hours ago
WuestenfuxWuestenfux
4,1451411
answered 17 hours ago
WuestenfuxWuestenfux
4,1451411
4,1451411
add a comment |
add a comment |
$begingroup$
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 loose something. You loose ordering moving to the complex numbers. You loose commutativity moving to the quaternions, and you loose 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 2x2 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.
answered 5 hours ago
Hurwitz_showed_the_wayHurwitz_showed_the_way
1
New contributor
Hurwitz_showed_the_way is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
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 loose something. You loose ordering moving to the complex numbers. You loose commutativity moving to the quaternions, and you loose 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 2x2 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.
answered 5 hours ago
Hurwitz_showed_the_wayHurwitz_showed_the_way
1
New contributor
Hurwitz_showed_the_way is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
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 loose something. You loose ordering moving to the complex numbers. You loose commutativity moving to the quaternions, and you loose 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 2x2 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.
answered 5 hours ago
Hurwitz_showed_the_wayHurwitz_showed_the_way
1
New contributor
Hurwitz_showed_the_way is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 loose something. You loose ordering moving to the complex numbers. You loose commutativity moving to the quaternions, and you loose 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 2x2 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.
answered 5 hours ago
Hurwitz_showed_the_wayHurwitz_showed_the_way
1
New contributor
Hurwitz_showed_the_way is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 5 hours ago
Hurwitz_showed_the_wayHurwitz_showed_the_way
1
New contributor
Hurwitz_showed_the_way is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 5 hours ago
Hurwitz_showed_the_wayHurwitz_showed_the_way
1
answered 5 hours ago
Hurwitz_showed_the_wayHurwitz_showed_the_way
1
1
New contributor
Hurwitz_showed_the_way is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Hurwitz_showed_the_way is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hurwitz_showed_the_way is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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$begingroup$
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?
$endgroup$
– DonAntonio
18 hours ago
3
$begingroup$
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 $j^2 = -1, zj = joverline{z}, z in mathbb{C}$ so it is not a $mathbb{C}$-algebra. In this sense it extends $mathbb{R}$, not $mathbb{C}$.
$endgroup$
– reuns
18 hours ago
2
$begingroup$
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.
$endgroup$
– Asaf Karagila♦
15 hours ago
2
$begingroup$
math.stackexchange.com/questions/1055501/… math.stackexchange.com/questions/1553537/… math.stackexchange.com/questions/892426/… math.stackexchange.com/questions/1462727/… math.stackexchange.com/questions/2637069/… and more you can find from these pages. It's really all the same question as yours, dressed in a different shirt.
$endgroup$
– Asaf Karagila♦
15 hours ago
$begingroup$
@AsafKaragila Being a subset and being a subset with a common algebraic structure are not the same thing. For example, $mathbb{N}subsetmathbb{Z}subsetmathbb{Q}subsetmathbb{R}subsetmathbb{C}$ 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.
$endgroup$
– Steven Hatton
14 hours ago