Homomorphism in Rings
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I have the following true/false statement from a worked example:
If $F_1, F_2$ are fields and $phi: F_1 rightarrow F_2$ is a homomorphism, then $phi$ is either the zero map or an isomorphism.
The solution says that this is False and gives the identity map $mathbbR rightarrow mathbbC$ as a counterexample.
I am a bit confused by this. Obviously the identity map is not the zero map. When I try to check the conditions for isomoporphism I get confused on the domainrange over which the identity map is defined.
Can someone explain why this example disproves the statement?
Thank you in advance for any answers/comments.
abstract-algebra ring-theory ring-isomorphism
asked Aug 20 at 9:18
Nick
474
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up vote
2
down vote
favorite
I have the following true/false statement from a worked example:
If $F_1, F_2$ are fields and $phi: F_1 rightarrow F_2$ is a homomorphism, then $phi$ is either the zero map or an isomorphism.
The solution says that this is False and gives the identity map $mathbbR rightarrow mathbbC$ as a counterexample.
I am a bit confused by this. Obviously the identity map is not the zero map. When I try to check the conditions for isomoporphism I get confused on the domainrange over which the identity map is defined.
Can someone explain why this example disproves the statement?
Thank you in advance for any answers/comments.
abstract-algebra ring-theory ring-isomorphism
asked Aug 20 at 9:18
Nick
474
Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
– Bernard
Aug 20 at 9:27
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the following true/false statement from a worked example:
If $F_1, F_2$ are fields and $phi: F_1 rightarrow F_2$ is a homomorphism, then $phi$ is either the zero map or an isomorphism.
The solution says that this is False and gives the identity map $mathbbR rightarrow mathbbC$ as a counterexample.
I am a bit confused by this. Obviously the identity map is not the zero map. When I try to check the conditions for isomoporphism I get confused on the domainrange over which the identity map is defined.
Can someone explain why this example disproves the statement?
Thank you in advance for any answers/comments.
abstract-algebra ring-theory ring-isomorphism
asked Aug 20 at 9:18
Nick
474
I have the following true/false statement from a worked example:
If $F_1, F_2$ are fields and $phi: F_1 rightarrow F_2$ is a homomorphism, then $phi$ is either the zero map or an isomorphism.
The solution says that this is False and gives the identity map $mathbbR rightarrow mathbbC$ as a counterexample.
I am a bit confused by this. Obviously the identity map is not the zero map. When I try to check the conditions for isomoporphism I get confused on the domainrange over which the identity map is defined.
Can someone explain why this example disproves the statement?
Thank you in advance for any answers/comments.
abstract-algebra ring-theory ring-isomorphism
asked Aug 20 at 9:18
Nick
474
asked Aug 20 at 9:18
Nick
474
asked Aug 20 at 9:18
Nick
474
asked Aug 20 at 9:18
Nick
474
474
Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
– Bernard
Aug 20 at 9:27
add a comment |Â
Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
– Bernard
Aug 20 at 9:27
Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
– Bernard
Aug 20 at 9:27
Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
– Bernard
Aug 20 at 9:27
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
So $mathbbR$ is a subset of $mathbbC$ so the identity map is clearly a homomorphism from $mathbbR$ to $mathbbC.$ As you said, it is clearly not the zero map but the map is not surjective! What maps to $i?$ Hence, it cannot be an isomorphism as it is not bijective.
answered Aug 20 at 9:22
伽罗瓦
934615
add a comment |Â
up vote
2
down vote
The identity map, by definition is a function $i : mathbbC to mathbbC$. With a little abuse of language we refer to the identity map of $mathbbR$ in $mathbbC$ as the restriction of the above function to $mathbbR$, hence the morphism you are actually considering is
$$
i_ : mathbbR to mathbbC
$$
which is both non zero and non surjective
edited Aug 20 at 9:26
Bernard
111k635103
answered Aug 20 at 9:25
JayTuma
1,333118
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
So $mathbbR$ is a subset of $mathbbC$ so the identity map is clearly a homomorphism from $mathbbR$ to $mathbbC.$ As you said, it is clearly not the zero map but the map is not surjective! What maps to $i?$ Hence, it cannot be an isomorphism as it is not bijective.
answered Aug 20 at 9:22
伽罗瓦
934615
add a comment |Â
up vote
3
down vote
accepted
So $mathbbR$ is a subset of $mathbbC$ so the identity map is clearly a homomorphism from $mathbbR$ to $mathbbC.$ As you said, it is clearly not the zero map but the map is not surjective! What maps to $i?$ Hence, it cannot be an isomorphism as it is not bijective.
answered Aug 20 at 9:22
伽罗瓦
934615
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
So $mathbbR$ is a subset of $mathbbC$ so the identity map is clearly a homomorphism from $mathbbR$ to $mathbbC.$ As you said, it is clearly not the zero map but the map is not surjective! What maps to $i?$ Hence, it cannot be an isomorphism as it is not bijective.
answered Aug 20 at 9:22
伽罗瓦
934615
So $mathbbR$ is a subset of $mathbbC$ so the identity map is clearly a homomorphism from $mathbbR$ to $mathbbC.$ As you said, it is clearly not the zero map but the map is not surjective! What maps to $i?$ Hence, it cannot be an isomorphism as it is not bijective.
answered Aug 20 at 9:22
伽罗瓦
934615
answered Aug 20 at 9:22
伽罗瓦
934615
answered Aug 20 at 9:22
伽罗瓦
934615
answered Aug 20 at 9:22
伽罗瓦
934615
934615
add a comment |Â
add a comment |Â
up vote
2
down vote
The identity map, by definition is a function $i : mathbbC to mathbbC$. With a little abuse of language we refer to the identity map of $mathbbR$ in $mathbbC$ as the restriction of the above function to $mathbbR$, hence the morphism you are actually considering is
$$
i_ : mathbbR to mathbbC
$$
which is both non zero and non surjective
edited Aug 20 at 9:26
Bernard
111k635103
answered Aug 20 at 9:25
JayTuma
1,333118
add a comment |Â
up vote
2
down vote
The identity map, by definition is a function $i : mathbbC to mathbbC$. With a little abuse of language we refer to the identity map of $mathbbR$ in $mathbbC$ as the restriction of the above function to $mathbbR$, hence the morphism you are actually considering is
$$
i_ : mathbbR to mathbbC
$$
which is both non zero and non surjective
edited Aug 20 at 9:26
Bernard
111k635103
answered Aug 20 at 9:25
JayTuma
1,333118
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The identity map, by definition is a function $i : mathbbC to mathbbC$. With a little abuse of language we refer to the identity map of $mathbbR$ in $mathbbC$ as the restriction of the above function to $mathbbR$, hence the morphism you are actually considering is
$$
i_ : mathbbR to mathbbC
$$
which is both non zero and non surjective
edited Aug 20 at 9:26
Bernard
111k635103
answered Aug 20 at 9:25
JayTuma
1,333118
The identity map, by definition is a function $i : mathbbC to mathbbC$. With a little abuse of language we refer to the identity map of $mathbbR$ in $mathbbC$ as the restriction of the above function to $mathbbR$, hence the morphism you are actually considering is
$$
i_ : mathbbR to mathbbC
$$
which is both non zero and non surjective
edited Aug 20 at 9:26
Bernard
111k635103
answered Aug 20 at 9:25
JayTuma
1,333118
edited Aug 20 at 9:26
Bernard
111k635103
edited Aug 20 at 9:26
Bernard
111k635103
edited Aug 20 at 9:26
Bernard
111k635103
111k635103
answered Aug 20 at 9:25
JayTuma
1,333118
answered Aug 20 at 9:25
JayTuma
1,333118
answered Aug 20 at 9:25
JayTuma
1,333118
1,333118
add a comment |Â
add a comment |Â
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Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
– Bernard
Aug 20 at 9:27