Classifying homomorphisms on polynomial rings with real coefficients.
Show that every homomorphism $mathbb{R}$[X] $rightarrow$ $mathbb{R}$[X] can is equal to $φ_g$ for a unique g $in$ $mathbb{R}$[X], given by $φ_g(f)$ = $f(g(X))$
My guess for any homomorphism $h$, $g = h(X)$ but I'm not sure how to proceed from there.
abstract-algebra ring-homomorphism polynomial-rings
add a comment |
Show that every homomorphism $mathbb{R}$[X] $rightarrow$ $mathbb{R}$[X] can is equal to $φ_g$ for a unique g $in$ $mathbb{R}$[X], given by $φ_g(f)$ = $f(g(X))$
My guess for any homomorphism $h$, $g = h(X)$ but I'm not sure how to proceed from there.
abstract-algebra ring-homomorphism polynomial-rings
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 30 at 8:40
1
Your guess is right! Now you can directly calculate $h(f) = varphi_g(f)$ for any $f$ explicitly, using that $h$ is a homomorphism.
– Gnampfissimo
Nov 30 at 9:00
Every homomorphism of $mathbb R$-algebras, that is, fixing $mathbb R$.
– lhf
Nov 30 at 10:07
add a comment |
Show that every homomorphism $mathbb{R}$[X] $rightarrow$ $mathbb{R}$[X] can is equal to $φ_g$ for a unique g $in$ $mathbb{R}$[X], given by $φ_g(f)$ = $f(g(X))$
My guess for any homomorphism $h$, $g = h(X)$ but I'm not sure how to proceed from there.
abstract-algebra ring-homomorphism polynomial-rings
Show that every homomorphism $mathbb{R}$[X] $rightarrow$ $mathbb{R}$[X] can is equal to $φ_g$ for a unique g $in$ $mathbb{R}$[X], given by $φ_g(f)$ = $f(g(X))$
My guess for any homomorphism $h$, $g = h(X)$ but I'm not sure how to proceed from there.
abstract-algebra ring-homomorphism polynomial-rings
abstract-algebra ring-homomorphism polynomial-rings
asked Nov 30 at 8:40
user621286
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 30 at 8:40
1
Your guess is right! Now you can directly calculate $h(f) = varphi_g(f)$ for any $f$ explicitly, using that $h$ is a homomorphism.
– Gnampfissimo
Nov 30 at 9:00
Every homomorphism of $mathbb R$-algebras, that is, fixing $mathbb R$.
– lhf
Nov 30 at 10:07
add a comment |
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 30 at 8:40
1
Your guess is right! Now you can directly calculate $h(f) = varphi_g(f)$ for any $f$ explicitly, using that $h$ is a homomorphism.
– Gnampfissimo
Nov 30 at 9:00
Every homomorphism of $mathbb R$-algebras, that is, fixing $mathbb R$.
– lhf
Nov 30 at 10:07
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 30 at 8:40
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 30 at 8:40
1
1
Your guess is right! Now you can directly calculate $h(f) = varphi_g(f)$ for any $f$ explicitly, using that $h$ is a homomorphism.
– Gnampfissimo
Nov 30 at 9:00
Your guess is right! Now you can directly calculate $h(f) = varphi_g(f)$ for any $f$ explicitly, using that $h$ is a homomorphism.
– Gnampfissimo
Nov 30 at 9:00
Every homomorphism of $mathbb R$-algebras, that is, fixing $mathbb R$.
– lhf
Nov 30 at 10:07
Every homomorphism of $mathbb R$-algebras, that is, fixing $mathbb R$.
– lhf
Nov 30 at 10:07
add a comment |
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Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 30 at 8:40
1
Your guess is right! Now you can directly calculate $h(f) = varphi_g(f)$ for any $f$ explicitly, using that $h$ is a homomorphism.
– Gnampfissimo
Nov 30 at 9:00
Every homomorphism of $mathbb R$-algebras, that is, fixing $mathbb R$.
– lhf
Nov 30 at 10:07