Mixing
Show that it doesn't exist any homomorphism $phi: mathbb{C} to mathbb{R}^{mathbb{R}}$
$begingroup$
- Considering the ring of all of the real functions $ mathbb{R}^{mathbb{R}}$. Show that it doesn't exist any homomorphism $phi: mathbb{C} to mathbb{R}^{mathbb{R}}$
- Let $R$ be an integral domain and consider $f:Rto R,,,,$ s.t. $,,,, f(a)=a^2,,,,,forall a in R$. Show that $f$ is injective $iff$ $f$ is a homomorphism.
About 1. any hint on how to show that?
About the point 2.: I have been thinking about that $forall a in R, ,,, f(a) = a^2 = (-a)^2 = f(-a)$ and, assuming $f$ injective, it means $a=-a Rightarrow a + a = 2a =0_R,,,,$ Then: $,,,,,a^2+b^2 = a^2+ (2a)b +b^2=(a+b)^2=f((a+b))=f(a)+f(b)=a^2+b^2,,,,,$
The fact that $R$ is an I.D. gives the product because we can say that $(ab)^2=a^2b^2$, so $f$ is homomorphism. To show that homomorphism implies injectivity do I need only to retrace this demonstration?
abstract-algebra ring-theory
asked Jan 4 at 1:55
F.incF.inc
394110
$endgroup$
add a comment |
$begingroup$
- Considering the ring of all of the real functions $ mathbb{R}^{mathbb{R}}$. Show that it doesn't exist any homomorphism $phi: mathbb{C} to mathbb{R}^{mathbb{R}}$
- Let $R$ be an integral domain and consider $f:Rto R,,,,$ s.t. $,,,, f(a)=a^2,,,,,forall a in R$. Show that $f$ is injective $iff$ $f$ is a homomorphism.
About 1. any hint on how to show that?
About the point 2.: I have been thinking about that $forall a in R, ,,, f(a) = a^2 = (-a)^2 = f(-a)$ and, assuming $f$ injective, it means $a=-a Rightarrow a + a = 2a =0_R,,,,$ Then: $,,,,,a^2+b^2 = a^2+ (2a)b +b^2=(a+b)^2=f((a+b))=f(a)+f(b)=a^2+b^2,,,,,$
The fact that $R$ is an I.D. gives the product because we can say that $(ab)^2=a^2b^2$, so $f$ is homomorphism. To show that homomorphism implies injectivity do I need only to retrace this demonstration?
abstract-algebra ring-theory
asked Jan 4 at 1:55
F.incF.inc
394110
$endgroup$
1
$begingroup$
Do your homomorphisms preserve $1$s?
$endgroup$
– Randall
Jan 4 at 2:42
add a comment |
$begingroup$
- Considering the ring of all of the real functions $ mathbb{R}^{mathbb{R}}$. Show that it doesn't exist any homomorphism $phi: mathbb{C} to mathbb{R}^{mathbb{R}}$
- Let $R$ be an integral domain and consider $f:Rto R,,,,$ s.t. $,,,, f(a)=a^2,,,,,forall a in R$. Show that $f$ is injective $iff$ $f$ is a homomorphism.
About 1. any hint on how to show that?
About the point 2.: I have been thinking about that $forall a in R, ,,, f(a) = a^2 = (-a)^2 = f(-a)$ and, assuming $f$ injective, it means $a=-a Rightarrow a + a = 2a =0_R,,,,$ Then: $,,,,,a^2+b^2 = a^2+ (2a)b +b^2=(a+b)^2=f((a+b))=f(a)+f(b)=a^2+b^2,,,,,$
The fact that $R$ is an I.D. gives the product because we can say that $(ab)^2=a^2b^2$, so $f$ is homomorphism. To show that homomorphism implies injectivity do I need only to retrace this demonstration?
abstract-algebra ring-theory
asked Jan 4 at 1:55
F.incF.inc
394110
$endgroup$
- Considering the ring of all of the real functions $ mathbb{R}^{mathbb{R}}$. Show that it doesn't exist any homomorphism $phi: mathbb{C} to mathbb{R}^{mathbb{R}}$
- Let $R$ be an integral domain and consider $f:Rto R,,,,$ s.t. $,,,, f(a)=a^2,,,,,forall a in R$. Show that $f$ is injective $iff$ $f$ is a homomorphism.
About 1. any hint on how to show that?
About the point 2.: I have been thinking about that $forall a in R, ,,, f(a) = a^2 = (-a)^2 = f(-a)$ and, assuming $f$ injective, it means $a=-a Rightarrow a + a = 2a =0_R,,,,$ Then: $,,,,,a^2+b^2 = a^2+ (2a)b +b^2=(a+b)^2=f((a+b))=f(a)+f(b)=a^2+b^2,,,,,$
The fact that $R$ is an I.D. gives the product because we can say that $(ab)^2=a^2b^2$, so $f$ is homomorphism. To show that homomorphism implies injectivity do I need only to retrace this demonstration?
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Jan 4 at 1:55
F.incF.inc
394110
asked Jan 4 at 1:55
F.incF.inc
394110
asked Jan 4 at 1:55
F.incF.inc
394110
asked Jan 4 at 1:55
F.incF.inc
394110
asked Jan 4 at 1:55
F.incF.inc
394110
394110
1
$begingroup$
Do your homomorphisms preserve $1$s?
$endgroup$
– Randall
Jan 4 at 2:42
add a comment |
1
$begingroup$
Do your homomorphisms preserve $1$s?
$endgroup$
– Randall
Jan 4 at 2:42
1
1
$begingroup$
Do your homomorphisms preserve $1$s?
$endgroup$
– Randall
Jan 4 at 2:42
$begingroup$
Do your homomorphisms preserve $1$s?
$endgroup$
– Randall
Jan 4 at 2:42
add a comment |
2 Answers
2
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$begingroup$
- You must have $phi(i)^2=-1$ and it is impossible.
answered Jan 4 at 2:00
Tsemo AristideTsemo Aristide
57k11444
$endgroup$
add a comment |
$begingroup$
For your other question: if $-1 neq 1$, then $f $ can't be injective since $f(-1) = f(1) = 1$. And if $-1 = 1$, then $2=0$ and so $f$ respects addition as per your calculation.
answered Jan 4 at 2:47
hunterhunter
14.5k22438
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
- You must have $phi(i)^2=-1$ and it is impossible.
answered Jan 4 at 2:00
Tsemo AristideTsemo Aristide
57k11444
$endgroup$
add a comment |
$begingroup$
- You must have $phi(i)^2=-1$ and it is impossible.
answered Jan 4 at 2:00
Tsemo AristideTsemo Aristide
57k11444
$endgroup$
add a comment |
$begingroup$
- You must have $phi(i)^2=-1$ and it is impossible.
answered Jan 4 at 2:00
Tsemo AristideTsemo Aristide
57k11444
$endgroup$
- You must have $phi(i)^2=-1$ and it is impossible.
answered Jan 4 at 2:00
Tsemo AristideTsemo Aristide
57k11444
answered Jan 4 at 2:00
Tsemo AristideTsemo Aristide
57k11444
answered Jan 4 at 2:00
Tsemo AristideTsemo Aristide
57k11444
answered Jan 4 at 2:00
Tsemo AristideTsemo Aristide
57k11444
57k11444
add a comment |
add a comment |
$begingroup$
For your other question: if $-1 neq 1$, then $f $ can't be injective since $f(-1) = f(1) = 1$. And if $-1 = 1$, then $2=0$ and so $f$ respects addition as per your calculation.
answered Jan 4 at 2:47
hunterhunter
14.5k22438
$endgroup$
add a comment |
$begingroup$
For your other question: if $-1 neq 1$, then $f $ can't be injective since $f(-1) = f(1) = 1$. And if $-1 = 1$, then $2=0$ and so $f$ respects addition as per your calculation.
answered Jan 4 at 2:47
hunterhunter
14.5k22438
$endgroup$
add a comment |
$begingroup$
For your other question: if $-1 neq 1$, then $f $ can't be injective since $f(-1) = f(1) = 1$. And if $-1 = 1$, then $2=0$ and so $f$ respects addition as per your calculation.
answered Jan 4 at 2:47
hunterhunter
14.5k22438
$endgroup$
For your other question: if $-1 neq 1$, then $f $ can't be injective since $f(-1) = f(1) = 1$. And if $-1 = 1$, then $2=0$ and so $f$ respects addition as per your calculation.
answered Jan 4 at 2:47
hunterhunter
14.5k22438
answered Jan 4 at 2:47
hunterhunter
14.5k22438
answered Jan 4 at 2:47
hunterhunter
14.5k22438
answered Jan 4 at 2:47
hunterhunter
14.5k22438
14.5k22438
add a comment |
add a comment |
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$begingroup$
Do your homomorphisms preserve $1$s?
$endgroup$
– Randall
Jan 4 at 2:42