When we say two fields are isomorphic, does that just mean they are isomophic as rings? [closed]
If we say fields $A$ and $B$ are isomorphic, does that just mean they are isomorphic as rings, or is there something else?
abstract-algebra field-theory
asked Dec 11 '18 at 18:40
Ovi
12.4k1038111
closed as off-topic by Saad, Gibbs, Xander Henderson, amWhy, user 170039 Dec 14 '18 at 14:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Gibbs, Xander Henderson, amWhy, user 170039
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
If we say fields $A$ and $B$ are isomorphic, does that just mean they are isomorphic as rings, or is there something else?
abstract-algebra field-theory
asked Dec 11 '18 at 18:40
Ovi
12.4k1038111
closed as off-topic by Saad, Gibbs, Xander Henderson, amWhy, user 170039 Dec 14 '18 at 14:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Gibbs, Xander Henderson, amWhy, user 170039
If this question can be reworded to fit the rules in the help center, please edit the question.
8
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
Dec 11 '18 at 19:09
Please see my comment, which explains just what TonyK's answer means by "field structure".
– user21820
Dec 12 '18 at 12:46
add a comment |
If we say fields $A$ and $B$ are isomorphic, does that just mean they are isomorphic as rings, or is there something else?
abstract-algebra field-theory
asked Dec 11 '18 at 18:40
Ovi
12.4k1038111
If we say fields $A$ and $B$ are isomorphic, does that just mean they are isomorphic as rings, or is there something else?
abstract-algebra field-theory
abstract-algebra field-theory
asked Dec 11 '18 at 18:40
Ovi
12.4k1038111
asked Dec 11 '18 at 18:40
Ovi
12.4k1038111
asked Dec 11 '18 at 18:40
Ovi
12.4k1038111
asked Dec 11 '18 at 18:40
Ovi
12.4k1038111
asked Dec 11 '18 at 18:40
Ovi
12.4k1038111
12.4k1038111
closed as off-topic by Saad, Gibbs, Xander Henderson, amWhy, user 170039 Dec 14 '18 at 14:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Gibbs, Xander Henderson, amWhy, user 170039
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Gibbs, Xander Henderson, amWhy, user 170039 Dec 14 '18 at 14:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Gibbs, Xander Henderson, amWhy, user 170039
If this question can be reworded to fit the rules in the help center, please edit the question.
8
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
Dec 11 '18 at 19:09
Please see my comment, which explains just what TonyK's answer means by "field structure".
– user21820
Dec 12 '18 at 12:46
add a comment |
8
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
Dec 11 '18 at 19:09
Please see my comment, which explains just what TonyK's answer means by "field structure".
– user21820
Dec 12 '18 at 12:46
8
8
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
Dec 11 '18 at 19:09
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
Dec 11 '18 at 19:09
Please see my comment, which explains just what TonyK's answer means by "field structure".
– user21820
Dec 12 '18 at 12:46
Please see my comment, which explains just what TonyK's answer means by "field structure".
– user21820
Dec 12 '18 at 12:46
add a comment |
2 Answers
2
active
oldest
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In a sense, yes, that is what it means. But not really. When we say two structures $S$ and $T$ of a certain type are isomorphic, we mean that there is a bijection $varphi:Srightarrow T$ which preserves the structure. So, for instance, if $circ$ is a binary operation in the structure, then for $x,yin S$, we have $varphi(xcirc y)=varphi(x)circ varphi(y)$.
It turns out that preserving the ring structure is enough to preserve the field structure; a field is just a commutative ring with inverses, so the property of being a field is preserved if the operations $+$ and $times$ are preserved. Thus two fields are isomorphic if and only if they are isomorphic when considered as rings. But this is a contingent fact, and it's not really what we mean when we say that two fields are isomorphic.
I realise that this view verges on philosophy, and I wouldn't defend it to the death. I am just trying to give an idea of what mathematicians are thinking of when they say isomorphic.
answered Dec 11 '18 at 19:03
TonyK
41.5k353132
So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
– Daniel Schepler
Dec 11 '18 at 19:34
7
I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
– hunter
Dec 11 '18 at 22:59
8
@hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
– rschwieb
Dec 12 '18 at 1:48
1
@user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
– Joker_vD
Dec 12 '18 at 14:55
1
@Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
– user21820
Dec 12 '18 at 18:08
|
show 3 more comments
They are just isomorphic as rings.
A ring isomorphism already preserves both operations of the field, and it's trivial to prove that a ring isomorphism "preserves inverses," so there's nothing else you could ask of an isomorphism between fields that isn't already there.
answered Dec 11 '18 at 18:44
rschwieb
105k1299244
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
In a sense, yes, that is what it means. But not really. When we say two structures $S$ and $T$ of a certain type are isomorphic, we mean that there is a bijection $varphi:Srightarrow T$ which preserves the structure. So, for instance, if $circ$ is a binary operation in the structure, then for $x,yin S$, we have $varphi(xcirc y)=varphi(x)circ varphi(y)$.
It turns out that preserving the ring structure is enough to preserve the field structure; a field is just a commutative ring with inverses, so the property of being a field is preserved if the operations $+$ and $times$ are preserved. Thus two fields are isomorphic if and only if they are isomorphic when considered as rings. But this is a contingent fact, and it's not really what we mean when we say that two fields are isomorphic.
I realise that this view verges on philosophy, and I wouldn't defend it to the death. I am just trying to give an idea of what mathematicians are thinking of when they say isomorphic.
answered Dec 11 '18 at 19:03
TonyK
41.5k353132
So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
– Daniel Schepler
Dec 11 '18 at 19:34
7
I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
– hunter
Dec 11 '18 at 22:59
8
@hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
– rschwieb
Dec 12 '18 at 1:48
1
@user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
– Joker_vD
Dec 12 '18 at 14:55
1
@Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
– user21820
Dec 12 '18 at 18:08
|
show 3 more comments
In a sense, yes, that is what it means. But not really. When we say two structures $S$ and $T$ of a certain type are isomorphic, we mean that there is a bijection $varphi:Srightarrow T$ which preserves the structure. So, for instance, if $circ$ is a binary operation in the structure, then for $x,yin S$, we have $varphi(xcirc y)=varphi(x)circ varphi(y)$.
It turns out that preserving the ring structure is enough to preserve the field structure; a field is just a commutative ring with inverses, so the property of being a field is preserved if the operations $+$ and $times$ are preserved. Thus two fields are isomorphic if and only if they are isomorphic when considered as rings. But this is a contingent fact, and it's not really what we mean when we say that two fields are isomorphic.
I realise that this view verges on philosophy, and I wouldn't defend it to the death. I am just trying to give an idea of what mathematicians are thinking of when they say isomorphic.
answered Dec 11 '18 at 19:03
TonyK
41.5k353132
So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
– Daniel Schepler
Dec 11 '18 at 19:34
7
I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
– hunter
Dec 11 '18 at 22:59
8
@hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
– rschwieb
Dec 12 '18 at 1:48
1
@user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
– Joker_vD
Dec 12 '18 at 14:55
1
@Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
– user21820
Dec 12 '18 at 18:08
|
show 3 more comments
In a sense, yes, that is what it means. But not really. When we say two structures $S$ and $T$ of a certain type are isomorphic, we mean that there is a bijection $varphi:Srightarrow T$ which preserves the structure. So, for instance, if $circ$ is a binary operation in the structure, then for $x,yin S$, we have $varphi(xcirc y)=varphi(x)circ varphi(y)$.
It turns out that preserving the ring structure is enough to preserve the field structure; a field is just a commutative ring with inverses, so the property of being a field is preserved if the operations $+$ and $times$ are preserved. Thus two fields are isomorphic if and only if they are isomorphic when considered as rings. But this is a contingent fact, and it's not really what we mean when we say that two fields are isomorphic.
I realise that this view verges on philosophy, and I wouldn't defend it to the death. I am just trying to give an idea of what mathematicians are thinking of when they say isomorphic.
answered Dec 11 '18 at 19:03
TonyK
41.5k353132
In a sense, yes, that is what it means. But not really. When we say two structures $S$ and $T$ of a certain type are isomorphic, we mean that there is a bijection $varphi:Srightarrow T$ which preserves the structure. So, for instance, if $circ$ is a binary operation in the structure, then for $x,yin S$, we have $varphi(xcirc y)=varphi(x)circ varphi(y)$.
It turns out that preserving the ring structure is enough to preserve the field structure; a field is just a commutative ring with inverses, so the property of being a field is preserved if the operations $+$ and $times$ are preserved. Thus two fields are isomorphic if and only if they are isomorphic when considered as rings. But this is a contingent fact, and it's not really what we mean when we say that two fields are isomorphic.
I realise that this view verges on philosophy, and I wouldn't defend it to the death. I am just trying to give an idea of what mathematicians are thinking of when they say isomorphic.
answered Dec 11 '18 at 19:03
TonyK
41.5k353132
answered Dec 11 '18 at 19:03
TonyK
41.5k353132
answered Dec 11 '18 at 19:03
TonyK
41.5k353132
answered Dec 11 '18 at 19:03
TonyK
41.5k353132
41.5k353132
So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
– Daniel Schepler
Dec 11 '18 at 19:34
7
I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
– hunter
Dec 11 '18 at 22:59
8
@hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
– rschwieb
Dec 12 '18 at 1:48
1
@user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
– Joker_vD
Dec 12 '18 at 14:55
1
@Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
– user21820
Dec 12 '18 at 18:08
|
show 3 more comments
So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
– Daniel Schepler
Dec 11 '18 at 19:34
7
I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
– hunter
Dec 11 '18 at 22:59
8
@hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
– rschwieb
Dec 12 '18 at 1:48
1
@user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
– Joker_vD
Dec 12 '18 at 14:55
1
@Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
– user21820
Dec 12 '18 at 18:08
So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
– Daniel Schepler
Dec 11 '18 at 19:34
So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
– Daniel Schepler
Dec 11 '18 at 19:34
7
7
I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
– hunter
Dec 11 '18 at 22:59
I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
– hunter
Dec 11 '18 at 22:59
8
8
@hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
– rschwieb
Dec 12 '18 at 1:48
@hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
– rschwieb
Dec 12 '18 at 1:48
1
1
@user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
– Joker_vD
Dec 12 '18 at 14:55
@user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
– Joker_vD
Dec 12 '18 at 14:55
1
1
@Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
– user21820
Dec 12 '18 at 18:08
@Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
– user21820
Dec 12 '18 at 18:08
|
show 3 more comments
They are just isomorphic as rings.
A ring isomorphism already preserves both operations of the field, and it's trivial to prove that a ring isomorphism "preserves inverses," so there's nothing else you could ask of an isomorphism between fields that isn't already there.
answered Dec 11 '18 at 18:44
rschwieb
105k1299244
add a comment |
They are just isomorphic as rings.
A ring isomorphism already preserves both operations of the field, and it's trivial to prove that a ring isomorphism "preserves inverses," so there's nothing else you could ask of an isomorphism between fields that isn't already there.
answered Dec 11 '18 at 18:44
rschwieb
105k1299244
add a comment |
They are just isomorphic as rings.
A ring isomorphism already preserves both operations of the field, and it's trivial to prove that a ring isomorphism "preserves inverses," so there's nothing else you could ask of an isomorphism between fields that isn't already there.
answered Dec 11 '18 at 18:44
rschwieb
105k1299244
They are just isomorphic as rings.
A ring isomorphism already preserves both operations of the field, and it's trivial to prove that a ring isomorphism "preserves inverses," so there's nothing else you could ask of an isomorphism between fields that isn't already there.
answered Dec 11 '18 at 18:44
rschwieb
105k1299244
answered Dec 11 '18 at 18:44
rschwieb
105k1299244
answered Dec 11 '18 at 18:44
rschwieb
105k1299244
answered Dec 11 '18 at 18:44
rschwieb
105k1299244
105k1299244
add a comment |
add a comment |
8
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
Dec 11 '18 at 19:09
Please see my comment, which explains just what TonyK's answer means by "field structure".
– user21820
Dec 12 '18 at 12:46