Mixing
What elements does a map of rings fix if both rings contain the same algebraically closed field as a subring?
$begingroup$
Let $k$ be an algebraically closed field. Let $R,S$ be rings such that $R,S$ both contain $k$ as a subring. Let $varphi:Rto S$ be a ring homomorphism. Then does $varphi(a)=a$ for all $ain k$?
So far I think: $varphi:Rto S$ is a ring homomorphism $implies varphi|_{k}:kto k$ is a field homomorphism $implies varphi(1_{k})=1_{k}$.
But I think this only shows that $varphi$ fixes elements of $k$ generated by $1_{k}$. So let's say $k$ is $mathbb{C}$ (I don't know any other algebraically closed field). Then I reckon that this would imply $varphi$ fixes $mathbb{Z}$ (and by the answer I got from this, $mathbb{Q}$ as well), but I don't know what I can say about irrational or imaginary numbers.
abstract-algebra ring-theory field-theory ring-homomorphism
asked Jan 29 at 22:07
anonanon444anonanon444
1756
$endgroup$
add a comment |
$begingroup$
Let $k$ be an algebraically closed field. Let $R,S$ be rings such that $R,S$ both contain $k$ as a subring. Let $varphi:Rto S$ be a ring homomorphism. Then does $varphi(a)=a$ for all $ain k$?
So far I think: $varphi:Rto S$ is a ring homomorphism $implies varphi|_{k}:kto k$ is a field homomorphism $implies varphi(1_{k})=1_{k}$.
But I think this only shows that $varphi$ fixes elements of $k$ generated by $1_{k}$. So let's say $k$ is $mathbb{C}$ (I don't know any other algebraically closed field). Then I reckon that this would imply $varphi$ fixes $mathbb{Z}$ (and by the answer I got from this, $mathbb{Q}$ as well), but I don't know what I can say about irrational or imaginary numbers.
abstract-algebra ring-theory field-theory ring-homomorphism
asked Jan 29 at 22:07
anonanon444anonanon444
1756
$endgroup$
3
$begingroup$
Choose $R=S=k=Bbb C$ and consider $varphi(z)=bar z$.
$endgroup$
– Berci
Jan 29 at 22:21
$begingroup$
Thanks for your comment, but a question, can we have a homomorphism (besides the one that maps $z$ to $z$) not defined this way? So can something that sends $z$ to something besides $overline{z}$ still be a homomorphism?
$endgroup$
– anonanon444
Jan 30 at 3:23
add a comment |
$begingroup$
Let $k$ be an algebraically closed field. Let $R,S$ be rings such that $R,S$ both contain $k$ as a subring. Let $varphi:Rto S$ be a ring homomorphism. Then does $varphi(a)=a$ for all $ain k$?
So far I think: $varphi:Rto S$ is a ring homomorphism $implies varphi|_{k}:kto k$ is a field homomorphism $implies varphi(1_{k})=1_{k}$.
But I think this only shows that $varphi$ fixes elements of $k$ generated by $1_{k}$. So let's say $k$ is $mathbb{C}$ (I don't know any other algebraically closed field). Then I reckon that this would imply $varphi$ fixes $mathbb{Z}$ (and by the answer I got from this, $mathbb{Q}$ as well), but I don't know what I can say about irrational or imaginary numbers.
abstract-algebra ring-theory field-theory ring-homomorphism
asked Jan 29 at 22:07
anonanon444anonanon444
1756
$endgroup$
Let $k$ be an algebraically closed field. Let $R,S$ be rings such that $R,S$ both contain $k$ as a subring. Let $varphi:Rto S$ be a ring homomorphism. Then does $varphi(a)=a$ for all $ain k$?
So far I think: $varphi:Rto S$ is a ring homomorphism $implies varphi|_{k}:kto k$ is a field homomorphism $implies varphi(1_{k})=1_{k}$.
But I think this only shows that $varphi$ fixes elements of $k$ generated by $1_{k}$. So let's say $k$ is $mathbb{C}$ (I don't know any other algebraically closed field). Then I reckon that this would imply $varphi$ fixes $mathbb{Z}$ (and by the answer I got from this, $mathbb{Q}$ as well), but I don't know what I can say about irrational or imaginary numbers.
abstract-algebra ring-theory field-theory ring-homomorphism
abstract-algebra ring-theory field-theory ring-homomorphism
asked Jan 29 at 22:07
anonanon444anonanon444
1756
asked Jan 29 at 22:07
anonanon444anonanon444
1756
asked Jan 29 at 22:07
anonanon444anonanon444
1756
asked Jan 29 at 22:07
anonanon444anonanon444
1756
asked Jan 29 at 22:07
anonanon444anonanon444
1756
1756
3
$begingroup$
Choose $R=S=k=Bbb C$ and consider $varphi(z)=bar z$.
$endgroup$
– Berci
Jan 29 at 22:21
$begingroup$
Thanks for your comment, but a question, can we have a homomorphism (besides the one that maps $z$ to $z$) not defined this way? So can something that sends $z$ to something besides $overline{z}$ still be a homomorphism?
$endgroup$
– anonanon444
Jan 30 at 3:23
add a comment |
3
$begingroup$
Choose $R=S=k=Bbb C$ and consider $varphi(z)=bar z$.
$endgroup$
– Berci
Jan 29 at 22:21
$begingroup$
Thanks for your comment, but a question, can we have a homomorphism (besides the one that maps $z$ to $z$) not defined this way? So can something that sends $z$ to something besides $overline{z}$ still be a homomorphism?
$endgroup$
– anonanon444
Jan 30 at 3:23
3
3
$begingroup$
Choose $R=S=k=Bbb C$ and consider $varphi(z)=bar z$.
$endgroup$
– Berci
Jan 29 at 22:21
$begingroup$
Choose $R=S=k=Bbb C$ and consider $varphi(z)=bar z$.
$endgroup$
– Berci
Jan 29 at 22:21
$begingroup$
Thanks for your comment, but a question, can we have a homomorphism (besides the one that maps $z$ to $z$) not defined this way? So can something that sends $z$ to something besides $overline{z}$ still be a homomorphism?
$endgroup$
– anonanon444
Jan 30 at 3:23
$begingroup$
Thanks for your comment, but a question, can we have a homomorphism (besides the one that maps $z$ to $z$) not defined this way? So can something that sends $z$ to something besides $overline{z}$ still be a homomorphism?
$endgroup$
– anonanon444
Jan 30 at 3:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The case where all the rings are $mathbb C$ is already interesting enough! As Berci mentioned in the comments, conjugation gives an example of an automorphism that only fixes the reals. One can show that this is the only non-trivial automorphism fixing the reals.
To answer your second question in the comments:
Assuming choice, there are even more exotic counterexamples, so called wild automorphisms.
You might find the discussion on this MSE post interesting.
answered Jan 30 at 15:19
ffffforallffffforall
36028
$endgroup$
$begingroup$
I see, thanks for your answer. I didn't know that it would be so "wild" to have an automorphism besides the identity/conjugation, lol!
$endgroup$
– anonanon444
Jan 30 at 18:00
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
The case where all the rings are $mathbb C$ is already interesting enough! As Berci mentioned in the comments, conjugation gives an example of an automorphism that only fixes the reals. One can show that this is the only non-trivial automorphism fixing the reals.
To answer your second question in the comments:
Assuming choice, there are even more exotic counterexamples, so called wild automorphisms.
You might find the discussion on this MSE post interesting.
answered Jan 30 at 15:19
ffffforallffffforall
36028
$endgroup$
$begingroup$
I see, thanks for your answer. I didn't know that it would be so "wild" to have an automorphism besides the identity/conjugation, lol!
$endgroup$
– anonanon444
Jan 30 at 18:00
add a comment |
$begingroup$
The case where all the rings are $mathbb C$ is already interesting enough! As Berci mentioned in the comments, conjugation gives an example of an automorphism that only fixes the reals. One can show that this is the only non-trivial automorphism fixing the reals.
To answer your second question in the comments:
Assuming choice, there are even more exotic counterexamples, so called wild automorphisms.
You might find the discussion on this MSE post interesting.
answered Jan 30 at 15:19
ffffforallffffforall
36028
$endgroup$
$begingroup$
I see, thanks for your answer. I didn't know that it would be so "wild" to have an automorphism besides the identity/conjugation, lol!
$endgroup$
– anonanon444
Jan 30 at 18:00
add a comment |
$begingroup$
The case where all the rings are $mathbb C$ is already interesting enough! As Berci mentioned in the comments, conjugation gives an example of an automorphism that only fixes the reals. One can show that this is the only non-trivial automorphism fixing the reals.
To answer your second question in the comments:
Assuming choice, there are even more exotic counterexamples, so called wild automorphisms.
You might find the discussion on this MSE post interesting.
answered Jan 30 at 15:19
ffffforallffffforall
36028
$endgroup$
The case where all the rings are $mathbb C$ is already interesting enough! As Berci mentioned in the comments, conjugation gives an example of an automorphism that only fixes the reals. One can show that this is the only non-trivial automorphism fixing the reals.
To answer your second question in the comments:
Assuming choice, there are even more exotic counterexamples, so called wild automorphisms.
You might find the discussion on this MSE post interesting.
answered Jan 30 at 15:19
ffffforallffffforall
36028
answered Jan 30 at 15:19
ffffforallffffforall
36028
answered Jan 30 at 15:19
ffffforallffffforall
36028
answered Jan 30 at 15:19
ffffforallffffforall
36028
36028
$begingroup$
I see, thanks for your answer. I didn't know that it would be so "wild" to have an automorphism besides the identity/conjugation, lol!
$endgroup$
– anonanon444
Jan 30 at 18:00
add a comment |
$begingroup$
I see, thanks for your answer. I didn't know that it would be so "wild" to have an automorphism besides the identity/conjugation, lol!
$endgroup$
– anonanon444
Jan 30 at 18:00
$begingroup$
I see, thanks for your answer. I didn't know that it would be so "wild" to have an automorphism besides the identity/conjugation, lol!
$endgroup$
– anonanon444
Jan 30 at 18:00
$begingroup$
I see, thanks for your answer. I didn't know that it would be so "wild" to have an automorphism besides the identity/conjugation, lol!
$endgroup$
– anonanon444
Jan 30 at 18:00
add a comment |
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$begingroup$
Choose $R=S=k=Bbb C$ and consider $varphi(z)=bar z$.
$endgroup$
– Berci
Jan 29 at 22:21
$begingroup$
Thanks for your comment, but a question, can we have a homomorphism (besides the one that maps $z$ to $z$) not defined this way? So can something that sends $z$ to something besides $overline{z}$ still be a homomorphism?
$endgroup$
– anonanon444
Jan 30 at 3:23