When we say two fields are isomorphic, does that just mean they are isomophic as rings?
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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 5 hours ago
Ovi
12.1k938108
add a comment |
up vote
4
down vote
favorite
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 5 hours ago
Ovi
12.1k938108
6
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
4 hours ago
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
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 5 hours ago
Ovi
12.1k938108
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 5 hours ago
Ovi
12.1k938108
asked 5 hours ago
Ovi
12.1k938108
asked 5 hours ago
Ovi
12.1k938108
asked 5 hours ago
Ovi
12.1k938108
asked 5 hours ago
Ovi
12.1k938108
12.1k938108
6
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
4 hours ago
add a comment |
6
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
4 hours ago
6
6
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
4 hours ago
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
9
down vote
accepted
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 4 hours ago
TonyK
40.8k352130
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
4 hours ago
@Daniel: all that follows from the elementary ring operations (don't forget that $varphi$ is a bijection).
– TonyK
2 hours ago
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
46 mins ago
add a comment |
up vote
8
down vote
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 5 hours ago
rschwieb
104k1299238
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
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 4 hours ago
TonyK
40.8k352130
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
4 hours ago
@Daniel: all that follows from the elementary ring operations (don't forget that $varphi$ is a bijection).
– TonyK
2 hours ago
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
46 mins ago
add a comment |
up vote
9
down vote
accepted
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 4 hours ago
TonyK
40.8k352130
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
4 hours ago
@Daniel: all that follows from the elementary ring operations (don't forget that $varphi$ is a bijection).
– TonyK
2 hours ago
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
46 mins ago
add a comment |
up vote
9
down vote
accepted
up vote
9
down vote
accepted
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 4 hours ago
TonyK
40.8k352130
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 4 hours ago
TonyK
40.8k352130
answered 4 hours ago
TonyK
40.8k352130
answered 4 hours ago
TonyK
40.8k352130
answered 4 hours ago
TonyK
40.8k352130
40.8k352130
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
4 hours ago
@Daniel: all that follows from the elementary ring operations (don't forget that $varphi$ is a bijection).
– TonyK
2 hours ago
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
46 mins ago
add a comment |
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
4 hours ago
@Daniel: all that follows from the elementary ring operations (don't forget that $varphi$ is a bijection).
– TonyK
2 hours ago
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
46 mins ago
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
4 hours ago
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
4 hours ago
@Daniel: all that follows from the elementary ring operations (don't forget that $varphi$ is a bijection).
– TonyK
2 hours ago
@Daniel: all that follows from the elementary ring operations (don't forget that $varphi$ is a bijection).
– TonyK
2 hours ago
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
46 mins ago
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
46 mins ago
add a comment |
up vote
8
down vote
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 5 hours ago
rschwieb
104k1299238
add a comment |
up vote
8
down vote
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 5 hours ago
rschwieb
104k1299238
add a comment |
up vote
8
down vote
up vote
8
down vote
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 5 hours ago
rschwieb
104k1299238
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 5 hours ago
rschwieb
104k1299238
answered 5 hours ago
rschwieb
104k1299238
answered 5 hours ago
rschwieb
104k1299238
answered 5 hours ago
rschwieb
104k1299238
104k1299238
add a comment |
add a comment |
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6
A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
4 hours ago