A ring homomorphism from $mathbb{C}$ to $mathbb{C}$
For a nontrivial homomorphism from $mathbb{C}$ to $mathbb{C}$, I know it’s invective and maps $1$ to $1$ and i to +i or -i. Then is such a map onto? I can’t go further.
abstract-algebra field-theory
asked Dec 11 '18 at 4:19
Danny
1,062412
add a comment |
For a nontrivial homomorphism from $mathbb{C}$ to $mathbb{C}$, I know it’s invective and maps $1$ to $1$ and i to +i or -i. Then is such a map onto? I can’t go further.
abstract-algebra field-theory
asked Dec 11 '18 at 4:19
Danny
1,062412
add a comment |
For a nontrivial homomorphism from $mathbb{C}$ to $mathbb{C}$, I know it’s invective and maps $1$ to $1$ and i to +i or -i. Then is such a map onto? I can’t go further.
abstract-algebra field-theory
asked Dec 11 '18 at 4:19
Danny
1,062412
For a nontrivial homomorphism from $mathbb{C}$ to $mathbb{C}$, I know it’s invective and maps $1$ to $1$ and i to +i or -i. Then is such a map onto? I can’t go further.
abstract-algebra field-theory
abstract-algebra field-theory
asked Dec 11 '18 at 4:19
Danny
1,062412
asked Dec 11 '18 at 4:19
Danny
1,062412
asked Dec 11 '18 at 4:19
Danny
1,062412
asked Dec 11 '18 at 4:19
Danny
1,062412
asked Dec 11 '18 at 4:19
Danny
1,062412
1,062412
add a comment |
add a comment |
1 Answer
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If you believe Zorn's lemma, there are lots of field maps from $Bbb C$
to itself, not all surjective. Take a transcendence basis $A$ of $Bbb C$
over $Bbb Q$. Then $A$ has cardinality $|Bbb C|$. Take an injection $phi:Ato A$. Then, $phi$ extends to a field map $Phi$ from $Bbb C$ to itself (this requires
Zorn, as does the existence of $A$). If $phi$ is not surjective, neither is $Phi$.
answered Dec 11 '18 at 4:29
Lord Shark the Unknown
101k958132
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
– Danny
Dec 11 '18 at 4:31
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you believe Zorn's lemma, there are lots of field maps from $Bbb C$
to itself, not all surjective. Take a transcendence basis $A$ of $Bbb C$
over $Bbb Q$. Then $A$ has cardinality $|Bbb C|$. Take an injection $phi:Ato A$. Then, $phi$ extends to a field map $Phi$ from $Bbb C$ to itself (this requires
Zorn, as does the existence of $A$). If $phi$ is not surjective, neither is $Phi$.
answered Dec 11 '18 at 4:29
Lord Shark the Unknown
101k958132
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
– Danny
Dec 11 '18 at 4:31
add a comment |
If you believe Zorn's lemma, there are lots of field maps from $Bbb C$
to itself, not all surjective. Take a transcendence basis $A$ of $Bbb C$
over $Bbb Q$. Then $A$ has cardinality $|Bbb C|$. Take an injection $phi:Ato A$. Then, $phi$ extends to a field map $Phi$ from $Bbb C$ to itself (this requires
Zorn, as does the existence of $A$). If $phi$ is not surjective, neither is $Phi$.
answered Dec 11 '18 at 4:29
Lord Shark the Unknown
101k958132
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
– Danny
Dec 11 '18 at 4:31
add a comment |
If you believe Zorn's lemma, there are lots of field maps from $Bbb C$
to itself, not all surjective. Take a transcendence basis $A$ of $Bbb C$
over $Bbb Q$. Then $A$ has cardinality $|Bbb C|$. Take an injection $phi:Ato A$. Then, $phi$ extends to a field map $Phi$ from $Bbb C$ to itself (this requires
Zorn, as does the existence of $A$). If $phi$ is not surjective, neither is $Phi$.
answered Dec 11 '18 at 4:29
Lord Shark the Unknown
101k958132
If you believe Zorn's lemma, there are lots of field maps from $Bbb C$
to itself, not all surjective. Take a transcendence basis $A$ of $Bbb C$
over $Bbb Q$. Then $A$ has cardinality $|Bbb C|$. Take an injection $phi:Ato A$. Then, $phi$ extends to a field map $Phi$ from $Bbb C$ to itself (this requires
Zorn, as does the existence of $A$). If $phi$ is not surjective, neither is $Phi$.
answered Dec 11 '18 at 4:29
Lord Shark the Unknown
101k958132
answered Dec 11 '18 at 4:29
Lord Shark the Unknown
101k958132
answered Dec 11 '18 at 4:29
Lord Shark the Unknown
101k958132
answered Dec 11 '18 at 4:29
Lord Shark the Unknown
101k958132
101k958132
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
– Danny
Dec 11 '18 at 4:31
add a comment |
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
– Danny
Dec 11 '18 at 4:31
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
– Danny
Dec 11 '18 at 4:31
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
– Danny
Dec 11 '18 at 4:31
add a comment |
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