A ring homomorphism from $mathbb{C}$ to $mathbb{C}$
$begingroup$
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
DannyDanny
1,151412
$endgroup$
add a comment |
$begingroup$
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
DannyDanny
1,151412
$endgroup$
add a comment |
$begingroup$
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
DannyDanny
1,151412
$endgroup$
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
DannyDanny
1,151412
asked Dec 11 '18 at 4:19
DannyDanny
1,151412
asked Dec 11 '18 at 4:19
DannyDanny
1,151412
asked Dec 11 '18 at 4:19
DannyDanny
1,151412
asked Dec 11 '18 at 4:19
DannyDanny
1,151412
1,151412
add a comment |
add a comment |
1 Answer
1
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$begingroup$
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 UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
$endgroup$
– 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
$begingroup$
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 UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
$endgroup$
– Danny
Dec 11 '18 at 4:31
add a comment |
$begingroup$
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 UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
$endgroup$
– Danny
Dec 11 '18 at 4:31
add a comment |
$begingroup$
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 UnknownLord Shark the Unknown
107k1162135
$endgroup$
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 UnknownLord Shark the Unknown
107k1162135
answered Dec 11 '18 at 4:29
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Dec 11 '18 at 4:29
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Dec 11 '18 at 4:29
Lord Shark the UnknownLord Shark the Unknown
107k1162135
107k1162135
$begingroup$
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
$endgroup$
– Danny
Dec 11 '18 at 4:31
add a comment |
$begingroup$
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
$endgroup$
– Danny
Dec 11 '18 at 4:31
$begingroup$
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
$endgroup$
– Danny
Dec 11 '18 at 4:31
$begingroup$
Is there a result of Hom($mathbb{C}$,$mathbb{C}$)$?
$endgroup$
– Danny
Dec 11 '18 at 4:31
add a comment |
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