Mixing
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 Jan 16 at 9:37
Steven HattonSteven Hatton
966422
$endgroup$
|
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 Jan 16 at 9:37
Steven HattonSteven Hatton
966422
$endgroup$
$begingroup$
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
Jan 16 at 9:43
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
Jan 16 at 9:48
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♦
Jan 16 at 12:58
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♦
Jan 16 at 13:02
$begingroup$
@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
Jan 16 at 13:44
|
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 Jan 16 at 9:37
Steven HattonSteven Hatton
966422
$endgroup$
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 Jan 16 at 9:37
Steven HattonSteven Hatton
966422
asked Jan 16 at 9:37
Steven HattonSteven Hatton
966422
edited Jan 16 at 9:42
Steven Hatton
asked Jan 16 at 9:37
Steven HattonSteven Hatton
966422
asked Jan 16 at 9:37
Steven HattonSteven Hatton
966422
asked Jan 16 at 9:37
Steven HattonSteven Hatton
966422
966422
$begingroup$
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
Jan 16 at 9:43
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
Jan 16 at 9:48
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♦
Jan 16 at 12:58
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♦
Jan 16 at 13:02
$begingroup$
@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
Jan 16 at 13:44
|
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
Jan 16 at 9:43
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
Jan 16 at 9:48
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♦
Jan 16 at 12:58
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♦
Jan 16 at 13:02
$begingroup$
@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
Jan 16 at 13:44
$begingroup$
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
Jan 16 at 9:43
$begingroup$
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
Jan 16 at 9:43
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
Jan 16 at 9:48
$begingroup$
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
Jan 16 at 9:48
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♦
Jan 16 at 12:58
$begingroup$
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♦
Jan 16 at 12:58
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♦
Jan 16 at 13:02
$begingroup$
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♦
Jan 16 at 13:02
$begingroup$
@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
Jan 16 at 13:44
$begingroup$
@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
Jan 16 at 13:44
|
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 Jan 16 at 15:21
Silvio Mayolo
20017
answered Jan 16 at 9:42
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
10
$begingroup$
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
Jan 16 at 17:27
$begingroup$
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
Jan 16 at 23:22
$begingroup$
@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♦
Jan 16 at 23:39
1
$begingroup$
@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
Jan 16 at 23:44
1
$begingroup$
@hunter Agree with that distinction. $mathbb{R} hookrightarrow mathbb{C}$ is canonical, though.
$endgroup$
– 6005
Jan 17 at 10:57
|
show 1 more 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 Jan 16 at 9:42
Emilio NovatiEmilio Novati
52k43474
$endgroup$
$begingroup$
I've heard it called a "non-commutative field" in many places
$endgroup$
– Pacopaco
Jan 17 at 5:14
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@Pacopaco The source I'm now looking at calls the quaternions a skew-field.
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– Steven Hatton
Jan 29 at 3:29
$begingroup$
Skew field, Non-commutative field, division ring, are synonymous. (en.wikipedia.org/wiki/Division_ring)
$endgroup$
– Emilio Novati
Jan 29 at 8:22
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 Jan 16 at 10:47
WuestenfuxWuestenfux
4,7311513
$endgroup$
1
$begingroup$
For the sake of completeness, I guess we should also demonstrate commutativity of multiplication for $(z,0),$ which seems obvious.
$endgroup$
– Steven Hatton
Jan 29 at 3:48
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 Jan 16 at 23:18
Hurwitz_showed_the_wayHurwitz_showed_the_way
111
$endgroup$
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 Jan 16 at 15:21
Silvio Mayolo
20017
answered Jan 16 at 9:42
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
10
$begingroup$
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
Jan 16 at 17:27
$begingroup$
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
Jan 16 at 23:22
$begingroup$
@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♦
Jan 16 at 23:39
1
$begingroup$
@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
Jan 16 at 23:44
1
$begingroup$
@hunter Agree with that distinction. $mathbb{R} hookrightarrow mathbb{C}$ is canonical, though.
$endgroup$
– 6005
Jan 17 at 10:57
|
show 1 more 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 Jan 16 at 15:21
Silvio Mayolo
20017
answered Jan 16 at 9:42
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
10
$begingroup$
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
Jan 16 at 17:27
$begingroup$
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
Jan 16 at 23:22
$begingroup$
@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♦
Jan 16 at 23:39
1
$begingroup$
@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
Jan 16 at 23:44
1
$begingroup$
@hunter Agree with that distinction. $mathbb{R} hookrightarrow mathbb{C}$ is canonical, though.
$endgroup$
– 6005
Jan 17 at 10:57
|
show 1 more 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 Jan 16 at 15:21
Silvio Mayolo
20017
answered Jan 16 at 9:42
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
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 Jan 16 at 15:21
Silvio Mayolo
20017
answered Jan 16 at 9:42
José Carlos SantosJosé Carlos Santos
163k22130233
edited Jan 16 at 15:21
Silvio Mayolo
20017
edited Jan 16 at 15:21
Silvio Mayolo
20017
edited Jan 16 at 15:21
Silvio Mayolo
20017
20017
answered Jan 16 at 9:42
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 16 at 9:42
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 16 at 9:42
José Carlos SantosJosé Carlos Santos
163k22130233
163k22130233
10
$begingroup$
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
Jan 16 at 17:27
$begingroup$
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
Jan 16 at 23:22
$begingroup$
@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♦
Jan 16 at 23:39
1
$begingroup$
@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
Jan 16 at 23:44
1
$begingroup$
@hunter Agree with that distinction. $mathbb{R} hookrightarrow mathbb{C}$ is canonical, though.
$endgroup$
– 6005
Jan 17 at 10:57
|
show 1 more comment
10
$begingroup$
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
Jan 16 at 17:27
$begingroup$
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
Jan 16 at 23:22
$begingroup$
@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♦
Jan 16 at 23:39
1
$begingroup$
@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
Jan 16 at 23:44
1
$begingroup$
@hunter Agree with that distinction. $mathbb{R} hookrightarrow mathbb{C}$ is canonical, though.
$endgroup$
– 6005
Jan 17 at 10:57
10
10
$begingroup$
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
Jan 16 at 17:27
$begingroup$
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
Jan 16 at 17:27
$begingroup$
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
Jan 16 at 23:22
$begingroup$
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
Jan 16 at 23:22
$begingroup$
@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♦
Jan 16 at 23:39
$begingroup$
@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♦
Jan 16 at 23:39
1
1
$begingroup$
@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
Jan 16 at 23:44
$begingroup$
@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
Jan 16 at 23:44
1
1
$begingroup$
@hunter Agree with that distinction. $mathbb{R} hookrightarrow mathbb{C}$ is canonical, though.
$endgroup$
– 6005
Jan 17 at 10:57
$begingroup$
@hunter Agree with that distinction. $mathbb{R} hookrightarrow mathbb{C}$ is canonical, though.
$endgroup$
– 6005
Jan 17 at 10:57
|
show 1 more 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 Jan 16 at 9:42
Emilio NovatiEmilio Novati
52k43474
$endgroup$
$begingroup$
I've heard it called a "non-commutative field" in many places
$endgroup$
– Pacopaco
Jan 17 at 5:14
$begingroup$
@Pacopaco The source I'm now looking at calls the quaternions a skew-field.
$endgroup$
– Steven Hatton
Jan 29 at 3:29
$begingroup$
Skew field, Non-commutative field, division ring, are synonymous. (en.wikipedia.org/wiki/Division_ring)
$endgroup$
– Emilio Novati
Jan 29 at 8:22
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 Jan 16 at 9:42
Emilio NovatiEmilio Novati
52k43474
$endgroup$
$begingroup$
I've heard it called a "non-commutative field" in many places
$endgroup$
– Pacopaco
Jan 17 at 5:14
$begingroup$
@Pacopaco The source I'm now looking at calls the quaternions a skew-field.
$endgroup$
– Steven Hatton
Jan 29 at 3:29
$begingroup$
Skew field, Non-commutative field, division ring, are synonymous. (en.wikipedia.org/wiki/Division_ring)
$endgroup$
– Emilio Novati
Jan 29 at 8:22
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 Jan 16 at 9:42
Emilio NovatiEmilio Novati
52k43474
$endgroup$
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 Jan 16 at 9:42
Emilio NovatiEmilio Novati
52k43474
answered Jan 16 at 9:42
Emilio NovatiEmilio Novati
52k43474
answered Jan 16 at 9:42
Emilio NovatiEmilio Novati
52k43474
answered Jan 16 at 9:42
Emilio NovatiEmilio Novati
52k43474
52k43474
$begingroup$
I've heard it called a "non-commutative field" in many places
$endgroup$
– Pacopaco
Jan 17 at 5:14
$begingroup$
@Pacopaco The source I'm now looking at calls the quaternions a skew-field.
$endgroup$
– Steven Hatton
Jan 29 at 3:29
$begingroup$
Skew field, Non-commutative field, division ring, are synonymous. (en.wikipedia.org/wiki/Division_ring)
$endgroup$
– Emilio Novati
Jan 29 at 8:22
add a comment |
$begingroup$
I've heard it called a "non-commutative field" in many places
$endgroup$
– Pacopaco
Jan 17 at 5:14
$begingroup$
@Pacopaco The source I'm now looking at calls the quaternions a skew-field.
$endgroup$
– Steven Hatton
Jan 29 at 3:29
$begingroup$
Skew field, Non-commutative field, division ring, are synonymous. (en.wikipedia.org/wiki/Division_ring)
$endgroup$
– Emilio Novati
Jan 29 at 8:22
$begingroup$
I've heard it called a "non-commutative field" in many places
$endgroup$
– Pacopaco
Jan 17 at 5:14
$begingroup$
I've heard it called a "non-commutative field" in many places
$endgroup$
– Pacopaco
Jan 17 at 5:14
$begingroup$
@Pacopaco The source I'm now looking at calls the quaternions a skew-field.
$endgroup$
– Steven Hatton
Jan 29 at 3:29
$begingroup$
@Pacopaco The source I'm now looking at calls the quaternions a skew-field.
$endgroup$
– Steven Hatton
Jan 29 at 3:29
$begingroup$
Skew field, Non-commutative field, division ring, are synonymous. (en.wikipedia.org/wiki/Division_ring)
$endgroup$
– Emilio Novati
Jan 29 at 8:22
$begingroup$
Skew field, Non-commutative field, division ring, are synonymous. (en.wikipedia.org/wiki/Division_ring)
$endgroup$
– Emilio Novati
Jan 29 at 8:22
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 Jan 16 at 10:47
WuestenfuxWuestenfux
4,7311513
$endgroup$
1
$begingroup$
For the sake of completeness, I guess we should also demonstrate commutativity of multiplication for $(z,0),$ which seems obvious.
$endgroup$
– Steven Hatton
Jan 29 at 3:48
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 Jan 16 at 10:47
WuestenfuxWuestenfux
4,7311513
$endgroup$
1
$begingroup$
For the sake of completeness, I guess we should also demonstrate commutativity of multiplication for $(z,0),$ which seems obvious.
$endgroup$
– Steven Hatton
Jan 29 at 3:48
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 Jan 16 at 10:47
WuestenfuxWuestenfux
4,7311513
$endgroup$
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 Jan 16 at 10:47
WuestenfuxWuestenfux
4,7311513
edited Jan 16 at 14:35
answered Jan 16 at 10:47
WuestenfuxWuestenfux
4,7311513
answered Jan 16 at 10:47
WuestenfuxWuestenfux
4,7311513
answered Jan 16 at 10:47
WuestenfuxWuestenfux
4,7311513
4,7311513
1
$begingroup$
For the sake of completeness, I guess we should also demonstrate commutativity of multiplication for $(z,0),$ which seems obvious.
$endgroup$
– Steven Hatton
Jan 29 at 3:48
add a comment |
1
$begingroup$
For the sake of completeness, I guess we should also demonstrate commutativity of multiplication for $(z,0),$ which seems obvious.
$endgroup$
– Steven Hatton
Jan 29 at 3:48
1
1
$begingroup$
For the sake of completeness, I guess we should also demonstrate commutativity of multiplication for $(z,0),$ which seems obvious.
$endgroup$
– Steven Hatton
Jan 29 at 3:48
$begingroup$
For the sake of completeness, I guess we should also demonstrate commutativity of multiplication for $(z,0),$ which seems obvious.
$endgroup$
– Steven Hatton
Jan 29 at 3:48
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 Jan 16 at 23:18
Hurwitz_showed_the_wayHurwitz_showed_the_way
111
$endgroup$
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 Jan 16 at 23:18
Hurwitz_showed_the_wayHurwitz_showed_the_way
111
$endgroup$
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 Jan 16 at 23:18
Hurwitz_showed_the_wayHurwitz_showed_the_way
111
$endgroup$
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 Jan 16 at 23:18
Hurwitz_showed_the_wayHurwitz_showed_the_way
111
answered Jan 16 at 23:18
Hurwitz_showed_the_wayHurwitz_showed_the_way
111
answered Jan 16 at 23:18
Hurwitz_showed_the_wayHurwitz_showed_the_way
111
answered Jan 16 at 23:18
Hurwitz_showed_the_wayHurwitz_showed_the_way
111
111
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
Jan 16 at 9:43
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
Jan 16 at 9:48
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♦
Jan 16 at 12:58
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♦
Jan 16 at 13:02
$begingroup$
@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
Jan 16 at 13:44