Homomorphisms and automorphisms on polynomial rings
I am trying to prove a series of propositions:
Given any homomorphism p from $mathbb{R}$[X] to $mathbb{R}$[X], show that it is equal to $phi_g$ for a unique g in $mathbb{R}$[X], with $phi_g$(f) = f(g(X)). I've expanded the expression $p(f) = p(sum_{i=0}^n a_iX^i) =sum_{i=0}^n p(a_i)p(X)^i$ but I'm not sure how to show $p(f) = sum_{i=0}^n p(a_i)p(X)^i = sum_{i=0}^n a_ig(X)^i = phi_g(f)$.
Show that if h,g $in mathbb{R}$[X] are such that h(g(X)) = X, then g(X) = aX+b for a $in$ R$^x$ and b$in mathbb{R}$.
abstract-algebra ring-theory linear-transformations ring-homomorphism polynomial-rings
edited Nov 30 at 21:45
Servaes
22.3k33793
asked Nov 30 at 10:09
nlin08
91
add a comment |
I am trying to prove a series of propositions:
Given any homomorphism p from $mathbb{R}$[X] to $mathbb{R}$[X], show that it is equal to $phi_g$ for a unique g in $mathbb{R}$[X], with $phi_g$(f) = f(g(X)). I've expanded the expression $p(f) = p(sum_{i=0}^n a_iX^i) =sum_{i=0}^n p(a_i)p(X)^i$ but I'm not sure how to show $p(f) = sum_{i=0}^n p(a_i)p(X)^i = sum_{i=0}^n a_ig(X)^i = phi_g(f)$.
Show that if h,g $in mathbb{R}$[X] are such that h(g(X)) = X, then g(X) = aX+b for a $in$ R$^x$ and b$in mathbb{R}$.
abstract-algebra ring-theory linear-transformations ring-homomorphism polynomial-rings
edited Nov 30 at 21:45
Servaes
22.3k33793
asked Nov 30 at 10:09
nlin08
91
As it stands both statements you ask to show are false. What makes you believe they are true?
– Servaes
Nov 30 at 14:42
And Servaes meant the non-trivial automorphism of $mathbb{Q}(sqrt{3})$ extends to an automorphism of $mathbb{R}$ (that we can't define without things like the axiom of choice) and $mathbb{R}[X]$
– reuns
Nov 30 at 22:16
add a comment |
I am trying to prove a series of propositions:
Given any homomorphism p from $mathbb{R}$[X] to $mathbb{R}$[X], show that it is equal to $phi_g$ for a unique g in $mathbb{R}$[X], with $phi_g$(f) = f(g(X)). I've expanded the expression $p(f) = p(sum_{i=0}^n a_iX^i) =sum_{i=0}^n p(a_i)p(X)^i$ but I'm not sure how to show $p(f) = sum_{i=0}^n p(a_i)p(X)^i = sum_{i=0}^n a_ig(X)^i = phi_g(f)$.
Show that if h,g $in mathbb{R}$[X] are such that h(g(X)) = X, then g(X) = aX+b for a $in$ R$^x$ and b$in mathbb{R}$.
abstract-algebra ring-theory linear-transformations ring-homomorphism polynomial-rings
edited Nov 30 at 21:45
Servaes
22.3k33793
asked Nov 30 at 10:09
nlin08
91
I am trying to prove a series of propositions:
Given any homomorphism p from $mathbb{R}$[X] to $mathbb{R}$[X], show that it is equal to $phi_g$ for a unique g in $mathbb{R}$[X], with $phi_g$(f) = f(g(X)). I've expanded the expression $p(f) = p(sum_{i=0}^n a_iX^i) =sum_{i=0}^n p(a_i)p(X)^i$ but I'm not sure how to show $p(f) = sum_{i=0}^n p(a_i)p(X)^i = sum_{i=0}^n a_ig(X)^i = phi_g(f)$.
Show that if h,g $in mathbb{R}$[X] are such that h(g(X)) = X, then g(X) = aX+b for a $in$ R$^x$ and b$in mathbb{R}$.
abstract-algebra ring-theory linear-transformations ring-homomorphism polynomial-rings
abstract-algebra ring-theory linear-transformations ring-homomorphism polynomial-rings
edited Nov 30 at 21:45
Servaes
22.3k33793
asked Nov 30 at 10:09
nlin08
91
edited Nov 30 at 21:45
Servaes
22.3k33793
asked Nov 30 at 10:09
nlin08
91
edited Nov 30 at 21:45
Servaes
22.3k33793
edited Nov 30 at 21:45
Servaes
22.3k33793
edited Nov 30 at 21:45
Servaes
22.3k33793
22.3k33793
asked Nov 30 at 10:09
nlin08
91
asked Nov 30 at 10:09
nlin08
91
asked Nov 30 at 10:09
nlin08
91
91
As it stands both statements you ask to show are false. What makes you believe they are true?
– Servaes
Nov 30 at 14:42
And Servaes meant the non-trivial automorphism of $mathbb{Q}(sqrt{3})$ extends to an automorphism of $mathbb{R}$ (that we can't define without things like the axiom of choice) and $mathbb{R}[X]$
– reuns
Nov 30 at 22:16
add a comment |
As it stands both statements you ask to show are false. What makes you believe they are true?
– Servaes
Nov 30 at 14:42
And Servaes meant the non-trivial automorphism of $mathbb{Q}(sqrt{3})$ extends to an automorphism of $mathbb{R}$ (that we can't define without things like the axiom of choice) and $mathbb{R}[X]$
– reuns
Nov 30 at 22:16
As it stands both statements you ask to show are false. What makes you believe they are true?
– Servaes
Nov 30 at 14:42
As it stands both statements you ask to show are false. What makes you believe they are true?
– Servaes
Nov 30 at 14:42
And Servaes meant the non-trivial automorphism of $mathbb{Q}(sqrt{3})$ extends to an automorphism of $mathbb{R}$ (that we can't define without things like the axiom of choice) and $mathbb{R}[X]$
– reuns
Nov 30 at 22:16
And Servaes meant the non-trivial automorphism of $mathbb{Q}(sqrt{3})$ extends to an automorphism of $mathbb{R}$ (that we can't define without things like the axiom of choice) and $mathbb{R}[X]$
– reuns
Nov 30 at 22:16
add a comment |
1 Answer
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For the claim to be true you need the additional hypothesis that $p$ is $Bbb{R}$-linear.
An $Bbb{R}$-linear ring homomorphism
$$p: Bbb{R}[X] longrightarrow Bbb{R}[X]$$
is determined by where it maps $X$. It follows immediately from the ring axioms that $p=phi_{p(X)}$. Indeed, if $p$ is $Bbb{R}$-linear then $p(r)=r$ for all $rinBbb{R}$, and your algebraic manipulations show that then
$$p(f)=f(p(X))=phi_{p(X)}(f),$$
for all $finBbb{R}[X]$. To see that the $Bbb{R}$-linear automorphisms are precisely the linear maps $p$ for which $p(X)$ is linear, note that $deg p(f)=deg p(X)cdotdeg f$ for all $finBbb{R}[X]$, so for $p$ to be surjective we must have $deg p(X)=1$. Check that $phi_g$ is invertible for all linear $g$.
answered Nov 30 at 14:29
Servaes
22.3k33793
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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votes
For the claim to be true you need the additional hypothesis that $p$ is $Bbb{R}$-linear.
An $Bbb{R}$-linear ring homomorphism
$$p: Bbb{R}[X] longrightarrow Bbb{R}[X]$$
is determined by where it maps $X$. It follows immediately from the ring axioms that $p=phi_{p(X)}$. Indeed, if $p$ is $Bbb{R}$-linear then $p(r)=r$ for all $rinBbb{R}$, and your algebraic manipulations show that then
$$p(f)=f(p(X))=phi_{p(X)}(f),$$
for all $finBbb{R}[X]$. To see that the $Bbb{R}$-linear automorphisms are precisely the linear maps $p$ for which $p(X)$ is linear, note that $deg p(f)=deg p(X)cdotdeg f$ for all $finBbb{R}[X]$, so for $p$ to be surjective we must have $deg p(X)=1$. Check that $phi_g$ is invertible for all linear $g$.
answered Nov 30 at 14:29
Servaes
22.3k33793
add a comment |
For the claim to be true you need the additional hypothesis that $p$ is $Bbb{R}$-linear.
An $Bbb{R}$-linear ring homomorphism
$$p: Bbb{R}[X] longrightarrow Bbb{R}[X]$$
is determined by where it maps $X$. It follows immediately from the ring axioms that $p=phi_{p(X)}$. Indeed, if $p$ is $Bbb{R}$-linear then $p(r)=r$ for all $rinBbb{R}$, and your algebraic manipulations show that then
$$p(f)=f(p(X))=phi_{p(X)}(f),$$
for all $finBbb{R}[X]$. To see that the $Bbb{R}$-linear automorphisms are precisely the linear maps $p$ for which $p(X)$ is linear, note that $deg p(f)=deg p(X)cdotdeg f$ for all $finBbb{R}[X]$, so for $p$ to be surjective we must have $deg p(X)=1$. Check that $phi_g$ is invertible for all linear $g$.
answered Nov 30 at 14:29
Servaes
22.3k33793
add a comment |
For the claim to be true you need the additional hypothesis that $p$ is $Bbb{R}$-linear.
An $Bbb{R}$-linear ring homomorphism
$$p: Bbb{R}[X] longrightarrow Bbb{R}[X]$$
is determined by where it maps $X$. It follows immediately from the ring axioms that $p=phi_{p(X)}$. Indeed, if $p$ is $Bbb{R}$-linear then $p(r)=r$ for all $rinBbb{R}$, and your algebraic manipulations show that then
$$p(f)=f(p(X))=phi_{p(X)}(f),$$
for all $finBbb{R}[X]$. To see that the $Bbb{R}$-linear automorphisms are precisely the linear maps $p$ for which $p(X)$ is linear, note that $deg p(f)=deg p(X)cdotdeg f$ for all $finBbb{R}[X]$, so for $p$ to be surjective we must have $deg p(X)=1$. Check that $phi_g$ is invertible for all linear $g$.
answered Nov 30 at 14:29
Servaes
22.3k33793
For the claim to be true you need the additional hypothesis that $p$ is $Bbb{R}$-linear.
An $Bbb{R}$-linear ring homomorphism
$$p: Bbb{R}[X] longrightarrow Bbb{R}[X]$$
is determined by where it maps $X$. It follows immediately from the ring axioms that $p=phi_{p(X)}$. Indeed, if $p$ is $Bbb{R}$-linear then $p(r)=r$ for all $rinBbb{R}$, and your algebraic manipulations show that then
$$p(f)=f(p(X))=phi_{p(X)}(f),$$
for all $finBbb{R}[X]$. To see that the $Bbb{R}$-linear automorphisms are precisely the linear maps $p$ for which $p(X)$ is linear, note that $deg p(f)=deg p(X)cdotdeg f$ for all $finBbb{R}[X]$, so for $p$ to be surjective we must have $deg p(X)=1$. Check that $phi_g$ is invertible for all linear $g$.
answered Nov 30 at 14:29
Servaes
22.3k33793
edited Nov 30 at 21:43
answered Nov 30 at 14:29
Servaes
22.3k33793
answered Nov 30 at 14:29
Servaes
22.3k33793
answered Nov 30 at 14:29
Servaes
22.3k33793
22.3k33793
add a comment |
add a comment |
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As it stands both statements you ask to show are false. What makes you believe they are true?
– Servaes
Nov 30 at 14:42
And Servaes meant the non-trivial automorphism of $mathbb{Q}(sqrt{3})$ extends to an automorphism of $mathbb{R}$ (that we can't define without things like the axiom of choice) and $mathbb{R}[X]$
– reuns
Nov 30 at 22:16