Field homomorphism
$begingroup$
Theorem is
If φ is a field homomorphism from field F to K then φ is either identical zero or injective.
I know how this come but when we say φ is identical zero then what about multiplicative identity of F. Is it map to zero? If so then we know that multiplicative identity map to multiplicative identity then how is it possible.
field-theory
edited Nov 30 '18 at 8:38
Arthur
112k7107190
asked Nov 30 '18 at 8:37
user499117user499117
409
$endgroup$
add a comment |
$begingroup$
Theorem is
If φ is a field homomorphism from field F to K then φ is either identical zero or injective.
I know how this come but when we say φ is identical zero then what about multiplicative identity of F. Is it map to zero? If so then we know that multiplicative identity map to multiplicative identity then how is it possible.
field-theory
edited Nov 30 '18 at 8:38
Arthur
112k7107190
asked Nov 30 '18 at 8:37
user499117user499117
409
$endgroup$
$begingroup$
Zero is (trivially) a multiplicative identity in the zero ring, but you don't have $1neq 0$ which is often included in definitions to exclude the trivial case.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:46
$begingroup$
Sorry but I don't understand....Is it necessary multiplicative identity map to multiplicative identity if yes that means in theorem only said about injective not about zero map.
$endgroup$
– user499117
Nov 30 '18 at 8:51
$begingroup$
In the zero ring (if you allow it by not insisting $1neq 0$) there is just one possible multiplication: $0times 0=0$, and this means that $0$ satisfies the property of being a multiplicative identity and is also its own multiplicative inverse.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:53
add a comment |
$begingroup$
Theorem is
If φ is a field homomorphism from field F to K then φ is either identical zero or injective.
I know how this come but when we say φ is identical zero then what about multiplicative identity of F. Is it map to zero? If so then we know that multiplicative identity map to multiplicative identity then how is it possible.
field-theory
edited Nov 30 '18 at 8:38
Arthur
112k7107190
asked Nov 30 '18 at 8:37
user499117user499117
409
$endgroup$
Theorem is
If φ is a field homomorphism from field F to K then φ is either identical zero or injective.
I know how this come but when we say φ is identical zero then what about multiplicative identity of F. Is it map to zero? If so then we know that multiplicative identity map to multiplicative identity then how is it possible.
field-theory
field-theory
edited Nov 30 '18 at 8:38
Arthur
112k7107190
asked Nov 30 '18 at 8:37
user499117user499117
409
edited Nov 30 '18 at 8:38
Arthur
112k7107190
asked Nov 30 '18 at 8:37
user499117user499117
409
edited Nov 30 '18 at 8:38
Arthur
112k7107190
edited Nov 30 '18 at 8:38
Arthur
112k7107190
edited Nov 30 '18 at 8:38
Arthur
112k7107190
112k7107190
asked Nov 30 '18 at 8:37
user499117user499117
409
asked Nov 30 '18 at 8:37
user499117user499117
409
asked Nov 30 '18 at 8:37
user499117user499117
409
409
$begingroup$
Zero is (trivially) a multiplicative identity in the zero ring, but you don't have $1neq 0$ which is often included in definitions to exclude the trivial case.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:46
$begingroup$
Sorry but I don't understand....Is it necessary multiplicative identity map to multiplicative identity if yes that means in theorem only said about injective not about zero map.
$endgroup$
– user499117
Nov 30 '18 at 8:51
$begingroup$
In the zero ring (if you allow it by not insisting $1neq 0$) there is just one possible multiplication: $0times 0=0$, and this means that $0$ satisfies the property of being a multiplicative identity and is also its own multiplicative inverse.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:53
add a comment |
$begingroup$
Zero is (trivially) a multiplicative identity in the zero ring, but you don't have $1neq 0$ which is often included in definitions to exclude the trivial case.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:46
$begingroup$
Sorry but I don't understand....Is it necessary multiplicative identity map to multiplicative identity if yes that means in theorem only said about injective not about zero map.
$endgroup$
– user499117
Nov 30 '18 at 8:51
$begingroup$
In the zero ring (if you allow it by not insisting $1neq 0$) there is just one possible multiplication: $0times 0=0$, and this means that $0$ satisfies the property of being a multiplicative identity and is also its own multiplicative inverse.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:53
$begingroup$
Zero is (trivially) a multiplicative identity in the zero ring, but you don't have $1neq 0$ which is often included in definitions to exclude the trivial case.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:46
$begingroup$
Zero is (trivially) a multiplicative identity in the zero ring, but you don't have $1neq 0$ which is often included in definitions to exclude the trivial case.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:46
$begingroup$
Sorry but I don't understand....Is it necessary multiplicative identity map to multiplicative identity if yes that means in theorem only said about injective not about zero map.
$endgroup$
– user499117
Nov 30 '18 at 8:51
$begingroup$
Sorry but I don't understand....Is it necessary multiplicative identity map to multiplicative identity if yes that means in theorem only said about injective not about zero map.
$endgroup$
– user499117
Nov 30 '18 at 8:51
$begingroup$
In the zero ring (if you allow it by not insisting $1neq 0$) there is just one possible multiplication: $0times 0=0$, and this means that $0$ satisfies the property of being a multiplicative identity and is also its own multiplicative inverse.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:53
$begingroup$
In the zero ring (if you allow it by not insisting $1neq 0$) there is just one possible multiplication: $0times 0=0$, and this means that $0$ satisfies the property of being a multiplicative identity and is also its own multiplicative inverse.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It depends if you postulate field morphisms to conserve the multiplicative unit. If you do, all field morphisms are injective. If you don't, the zero morphism is permitted too.
answered Nov 30 '18 at 8:39
GnampfissimoGnampfissimo
18011
$endgroup$
add a comment |
$begingroup$
There are different conventions about ring (and field) homomorphisms. Some people require that the multiplicative identity is mapped to the multiplicative identity, and some don't. In this case you've clearly come across someone who doesn't recuire such a ting.
And yes, when they say that $varphi$ is identically zero, that means, among other things, that $varphi(1_F) = 0_K$.
answered Nov 30 '18 at 8:40
ArthurArthur
112k7107190
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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$begingroup$
It depends if you postulate field morphisms to conserve the multiplicative unit. If you do, all field morphisms are injective. If you don't, the zero morphism is permitted too.
answered Nov 30 '18 at 8:39
GnampfissimoGnampfissimo
18011
$endgroup$
add a comment |
$begingroup$
It depends if you postulate field morphisms to conserve the multiplicative unit. If you do, all field morphisms are injective. If you don't, the zero morphism is permitted too.
answered Nov 30 '18 at 8:39
GnampfissimoGnampfissimo
18011
$endgroup$
add a comment |
$begingroup$
It depends if you postulate field morphisms to conserve the multiplicative unit. If you do, all field morphisms are injective. If you don't, the zero morphism is permitted too.
answered Nov 30 '18 at 8:39
GnampfissimoGnampfissimo
18011
$endgroup$
It depends if you postulate field morphisms to conserve the multiplicative unit. If you do, all field morphisms are injective. If you don't, the zero morphism is permitted too.
answered Nov 30 '18 at 8:39
GnampfissimoGnampfissimo
18011
answered Nov 30 '18 at 8:39
GnampfissimoGnampfissimo
18011
answered Nov 30 '18 at 8:39
GnampfissimoGnampfissimo
18011
answered Nov 30 '18 at 8:39
GnampfissimoGnampfissimo
18011
18011
add a comment |
add a comment |
$begingroup$
There are different conventions about ring (and field) homomorphisms. Some people require that the multiplicative identity is mapped to the multiplicative identity, and some don't. In this case you've clearly come across someone who doesn't recuire such a ting.
And yes, when they say that $varphi$ is identically zero, that means, among other things, that $varphi(1_F) = 0_K$.
answered Nov 30 '18 at 8:40
ArthurArthur
112k7107190
$endgroup$
add a comment |
$begingroup$
There are different conventions about ring (and field) homomorphisms. Some people require that the multiplicative identity is mapped to the multiplicative identity, and some don't. In this case you've clearly come across someone who doesn't recuire such a ting.
And yes, when they say that $varphi$ is identically zero, that means, among other things, that $varphi(1_F) = 0_K$.
answered Nov 30 '18 at 8:40
ArthurArthur
112k7107190
$endgroup$
add a comment |
$begingroup$
There are different conventions about ring (and field) homomorphisms. Some people require that the multiplicative identity is mapped to the multiplicative identity, and some don't. In this case you've clearly come across someone who doesn't recuire such a ting.
And yes, when they say that $varphi$ is identically zero, that means, among other things, that $varphi(1_F) = 0_K$.
answered Nov 30 '18 at 8:40
ArthurArthur
112k7107190
$endgroup$
There are different conventions about ring (and field) homomorphisms. Some people require that the multiplicative identity is mapped to the multiplicative identity, and some don't. In this case you've clearly come across someone who doesn't recuire such a ting.
And yes, when they say that $varphi$ is identically zero, that means, among other things, that $varphi(1_F) = 0_K$.
answered Nov 30 '18 at 8:40
ArthurArthur
112k7107190
answered Nov 30 '18 at 8:40
ArthurArthur
112k7107190
answered Nov 30 '18 at 8:40
ArthurArthur
112k7107190
answered Nov 30 '18 at 8:40
ArthurArthur
112k7107190
112k7107190
add a comment |
add a comment |
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$begingroup$
Zero is (trivially) a multiplicative identity in the zero ring, but you don't have $1neq 0$ which is often included in definitions to exclude the trivial case.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:46
$begingroup$
Sorry but I don't understand....Is it necessary multiplicative identity map to multiplicative identity if yes that means in theorem only said about injective not about zero map.
$endgroup$
– user499117
Nov 30 '18 at 8:51
$begingroup$
In the zero ring (if you allow it by not insisting $1neq 0$) there is just one possible multiplication: $0times 0=0$, and this means that $0$ satisfies the property of being a multiplicative identity and is also its own multiplicative inverse.
$endgroup$
– Mark Bennet
Nov 30 '18 at 8:53