Prove that the application $psi_a:F[X] rightarrow K := psi_a(f) = f(a)$ is an homomorphism of an Integral...
$begingroup$
I'm trying to prove that
the application $psi_a:F[X] rightarrow K := psi_a(f) = f(a)$ is an
homomorphism of an Integral Domain.
Def: I know that an Integral Domain is an abelian ring $A$ with $(a,b) in A rightarrow ab=0 iff a=0$ or $b=0$.
edit: K is an extention of F
So I take $b=0 $ and define another application $psi_b:F[X] rightarrow K := psi_b(g) = 0$.
$psi_b(g)psi_a(f)= (g)(0)(f)(a)$ because there is an application in this product that gives $0$, whatever value the other application could get.
You could also rearrange the $0$ and the $a$ to pass the value to the other function, and the claim still holds.
I appreciate any edit/explanation/correction/suggestion because this is one of my first approach to abstract algebra.
Thanks
abstract-algebra
asked Dec 12 '18 at 12:35
AlessarAlessar
308115
$endgroup$
|
show 6 more comments
$begingroup$
I'm trying to prove that
the application $psi_a:F[X] rightarrow K := psi_a(f) = f(a)$ is an
homomorphism of an Integral Domain.
Def: I know that an Integral Domain is an abelian ring $A$ with $(a,b) in A rightarrow ab=0 iff a=0$ or $b=0$.
edit: K is an extention of F
So I take $b=0 $ and define another application $psi_b:F[X] rightarrow K := psi_b(g) = 0$.
$psi_b(g)psi_a(f)= (g)(0)(f)(a)$ because there is an application in this product that gives $0$, whatever value the other application could get.
You could also rearrange the $0$ and the $a$ to pass the value to the other function, and the claim still holds.
I appreciate any edit/explanation/correction/suggestion because this is one of my first approach to abstract algebra.
Thanks
abstract-algebra
asked Dec 12 '18 at 12:35
AlessarAlessar
308115
$endgroup$
1
$begingroup$
For fixed $ain F$, $psi_a(fg)=f(a)g(a)$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 12:38
1
$begingroup$
Is $F=K$ or at least $F subseteq K$ ?
$endgroup$
– lhf
Dec 12 '18 at 12:47
1
$begingroup$
Could you elaborate what your understanding of a homomorphism of integral domains is? In particular why you are choosing some $b$?
$endgroup$
– kesa
Dec 12 '18 at 13:59
1
$begingroup$
Do you agree that a homomorphism of integral domains is just a usual ring homomorphism where both, source and target are integral domains? This might cause some confusion..
$endgroup$
– kesa
Dec 12 '18 at 14:11
1
$begingroup$
Ok, then instead of chosing an element $b$ (probably from $K$), shouldn't you rather take two elements, $f,g$ from the target $F[X]$ and check, that $psi_a(f+g) = psi_a(f) + psi_a(g)$ and similarly for multiplication (As Yadati Kiran suggested already)?
$endgroup$
– kesa
Dec 12 '18 at 14:16
|
show 6 more comments
$begingroup$
I'm trying to prove that
the application $psi_a:F[X] rightarrow K := psi_a(f) = f(a)$ is an
homomorphism of an Integral Domain.
Def: I know that an Integral Domain is an abelian ring $A$ with $(a,b) in A rightarrow ab=0 iff a=0$ or $b=0$.
edit: K is an extention of F
So I take $b=0 $ and define another application $psi_b:F[X] rightarrow K := psi_b(g) = 0$.
$psi_b(g)psi_a(f)= (g)(0)(f)(a)$ because there is an application in this product that gives $0$, whatever value the other application could get.
You could also rearrange the $0$ and the $a$ to pass the value to the other function, and the claim still holds.
I appreciate any edit/explanation/correction/suggestion because this is one of my first approach to abstract algebra.
Thanks
abstract-algebra
asked Dec 12 '18 at 12:35
AlessarAlessar
308115
$endgroup$
I'm trying to prove that
the application $psi_a:F[X] rightarrow K := psi_a(f) = f(a)$ is an
homomorphism of an Integral Domain.
Def: I know that an Integral Domain is an abelian ring $A$ with $(a,b) in A rightarrow ab=0 iff a=0$ or $b=0$.
edit: K is an extention of F
So I take $b=0 $ and define another application $psi_b:F[X] rightarrow K := psi_b(g) = 0$.
$psi_b(g)psi_a(f)= (g)(0)(f)(a)$ because there is an application in this product that gives $0$, whatever value the other application could get.
You could also rearrange the $0$ and the $a$ to pass the value to the other function, and the claim still holds.
I appreciate any edit/explanation/correction/suggestion because this is one of my first approach to abstract algebra.
Thanks
abstract-algebra
abstract-algebra
asked Dec 12 '18 at 12:35
AlessarAlessar
308115
asked Dec 12 '18 at 12:35
AlessarAlessar
308115
edited Dec 12 '18 at 13:07
Alessar
asked Dec 12 '18 at 12:35
AlessarAlessar
308115
asked Dec 12 '18 at 12:35
AlessarAlessar
308115
asked Dec 12 '18 at 12:35
AlessarAlessar
308115
308115
1
$begingroup$
For fixed $ain F$, $psi_a(fg)=f(a)g(a)$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 12:38
1
$begingroup$
Is $F=K$ or at least $F subseteq K$ ?
$endgroup$
– lhf
Dec 12 '18 at 12:47
1
$begingroup$
Could you elaborate what your understanding of a homomorphism of integral domains is? In particular why you are choosing some $b$?
$endgroup$
– kesa
Dec 12 '18 at 13:59
1
$begingroup$
Do you agree that a homomorphism of integral domains is just a usual ring homomorphism where both, source and target are integral domains? This might cause some confusion..
$endgroup$
– kesa
Dec 12 '18 at 14:11
1
$begingroup$
Ok, then instead of chosing an element $b$ (probably from $K$), shouldn't you rather take two elements, $f,g$ from the target $F[X]$ and check, that $psi_a(f+g) = psi_a(f) + psi_a(g)$ and similarly for multiplication (As Yadati Kiran suggested already)?
$endgroup$
– kesa
Dec 12 '18 at 14:16
|
show 6 more comments
1
$begingroup$
For fixed $ain F$, $psi_a(fg)=f(a)g(a)$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 12:38
1
$begingroup$
Is $F=K$ or at least $F subseteq K$ ?
$endgroup$
– lhf
Dec 12 '18 at 12:47
1
$begingroup$
Could you elaborate what your understanding of a homomorphism of integral domains is? In particular why you are choosing some $b$?
$endgroup$
– kesa
Dec 12 '18 at 13:59
1
$begingroup$
Do you agree that a homomorphism of integral domains is just a usual ring homomorphism where both, source and target are integral domains? This might cause some confusion..
$endgroup$
– kesa
Dec 12 '18 at 14:11
1
$begingroup$
Ok, then instead of chosing an element $b$ (probably from $K$), shouldn't you rather take two elements, $f,g$ from the target $F[X]$ and check, that $psi_a(f+g) = psi_a(f) + psi_a(g)$ and similarly for multiplication (As Yadati Kiran suggested already)?
$endgroup$
– kesa
Dec 12 '18 at 14:16
1
1
$begingroup$
For fixed $ain F$, $psi_a(fg)=f(a)g(a)$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 12:38
$begingroup$
For fixed $ain F$, $psi_a(fg)=f(a)g(a)$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 12:38
1
1
$begingroup$
Is $F=K$ or at least $F subseteq K$ ?
$endgroup$
– lhf
Dec 12 '18 at 12:47
$begingroup$
Is $F=K$ or at least $F subseteq K$ ?
$endgroup$
– lhf
Dec 12 '18 at 12:47
1
1
$begingroup$
Could you elaborate what your understanding of a homomorphism of integral domains is? In particular why you are choosing some $b$?
$endgroup$
– kesa
Dec 12 '18 at 13:59
$begingroup$
Could you elaborate what your understanding of a homomorphism of integral domains is? In particular why you are choosing some $b$?
$endgroup$
– kesa
Dec 12 '18 at 13:59
1
1
$begingroup$
Do you agree that a homomorphism of integral domains is just a usual ring homomorphism where both, source and target are integral domains? This might cause some confusion..
$endgroup$
– kesa
Dec 12 '18 at 14:11
$begingroup$
Do you agree that a homomorphism of integral domains is just a usual ring homomorphism where both, source and target are integral domains? This might cause some confusion..
$endgroup$
– kesa
Dec 12 '18 at 14:11
1
1
$begingroup$
Ok, then instead of chosing an element $b$ (probably from $K$), shouldn't you rather take two elements, $f,g$ from the target $F[X]$ and check, that $psi_a(f+g) = psi_a(f) + psi_a(g)$ and similarly for multiplication (As Yadati Kiran suggested already)?
$endgroup$
– kesa
Dec 12 '18 at 14:16
$begingroup$
Ok, then instead of chosing an element $b$ (probably from $K$), shouldn't you rather take two elements, $f,g$ from the target $F[X]$ and check, that $psi_a(f+g) = psi_a(f) + psi_a(g)$ and similarly for multiplication (As Yadati Kiran suggested already)?
$endgroup$
– kesa
Dec 12 '18 at 14:16
|
show 6 more comments
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1
$begingroup$
For fixed $ain F$, $psi_a(fg)=f(a)g(a)$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 12:38
1
$begingroup$
Is $F=K$ or at least $F subseteq K$ ?
$endgroup$
– lhf
Dec 12 '18 at 12:47
1
$begingroup$
Could you elaborate what your understanding of a homomorphism of integral domains is? In particular why you are choosing some $b$?
$endgroup$
– kesa
Dec 12 '18 at 13:59
1
$begingroup$
Do you agree that a homomorphism of integral domains is just a usual ring homomorphism where both, source and target are integral domains? This might cause some confusion..
$endgroup$
– kesa
Dec 12 '18 at 14:11
1
$begingroup$
Ok, then instead of chosing an element $b$ (probably from $K$), shouldn't you rather take two elements, $f,g$ from the target $F[X]$ and check, that $psi_a(f+g) = psi_a(f) + psi_a(g)$ and similarly for multiplication (As Yadati Kiran suggested already)?
$endgroup$
– kesa
Dec 12 '18 at 14:16