Mixing
The kernel of $D$ if $char(F)=0$
I am attempting the part (b) and (c) of the following question from Judson:
I wrote this: $ker(D) = left { f(x) in F[x] : f'(x) = 0right } =left {f'(x) = a_{1} + 2a_{2}x + cdots + na_{n}x^{n-1} : a_{1} = 0, 2a_{2} = 0, ..., na_{n} = 0 right }$
For part (b), I followed this answer. I am not sure why $char(F)=0$ implies $k$ of $ka_{n}$ is invertible for all $1leq k leq n$.
For part(c), I followed this answer. I understood it until they said "polynomial in $x^{p}$". What does that mean?
abstract-algebra field-theory
asked Nov 21 '18 at 18:48
numericalorangenumericalorange
1,726311
|
show 3 more comments
I am attempting the part (b) and (c) of the following question from Judson:
I wrote this: $ker(D) = left { f(x) in F[x] : f'(x) = 0right } =left {f'(x) = a_{1} + 2a_{2}x + cdots + na_{n}x^{n-1} : a_{1} = 0, 2a_{2} = 0, ..., na_{n} = 0 right }$
For part (b), I followed this answer. I am not sure why $char(F)=0$ implies $k$ of $ka_{n}$ is invertible for all $1leq k leq n$.
For part(c), I followed this answer. I understood it until they said "polynomial in $x^{p}$". What does that mean?
abstract-algebra field-theory
asked Nov 21 '18 at 18:48
numericalorangenumericalorange
1,726311
Characteristic of any field is either $0$ or a prime number. Hence $k$ is invertible.
– Yadati Kiran
Nov 21 '18 at 19:14
"$f(x)$ is polynomial in $x^p$" means that there is a polynomial $g in F[x]$ such that $f(x) = g(x^p)$. More concretely, it means that every monomial in $f$ has a power that is divisible by $p$. And as the comment above says, since $k$ and $a_n$ are nonzero, then $k a_n$ is invertible.
– André 3000
Nov 21 '18 at 21:18
@André3000 Does that mean the kernel is the coefficients of $f'(x)$ when $char(F)=0$ and the kernel is the monomials in $f$ when $char(F) = p$? I am a little confused...
– numericalorange
Nov 22 '18 at 18:19
@YadatiKiran I am a little confused about how characteristic relates to invertibility? If there does not exist a positive integer $n$ such that $nr=0$ for all $rin F$, why does that show that $r$ is invertible?
– numericalorange
Nov 22 '18 at 18:20
@numericalorange: What are the proper ideals of ring with characteristic $0$?
– Yadati Kiran
Nov 22 '18 at 18:28
|
show 3 more comments
I am attempting the part (b) and (c) of the following question from Judson:
I wrote this: $ker(D) = left { f(x) in F[x] : f'(x) = 0right } =left {f'(x) = a_{1} + 2a_{2}x + cdots + na_{n}x^{n-1} : a_{1} = 0, 2a_{2} = 0, ..., na_{n} = 0 right }$
For part (b), I followed this answer. I am not sure why $char(F)=0$ implies $k$ of $ka_{n}$ is invertible for all $1leq k leq n$.
For part(c), I followed this answer. I understood it until they said "polynomial in $x^{p}$". What does that mean?
abstract-algebra field-theory
asked Nov 21 '18 at 18:48
numericalorangenumericalorange
1,726311
I am attempting the part (b) and (c) of the following question from Judson:
I wrote this: $ker(D) = left { f(x) in F[x] : f'(x) = 0right } =left {f'(x) = a_{1} + 2a_{2}x + cdots + na_{n}x^{n-1} : a_{1} = 0, 2a_{2} = 0, ..., na_{n} = 0 right }$
For part (b), I followed this answer. I am not sure why $char(F)=0$ implies $k$ of $ka_{n}$ is invertible for all $1leq k leq n$.
For part(c), I followed this answer. I understood it until they said "polynomial in $x^{p}$". What does that mean?
abstract-algebra field-theory
abstract-algebra field-theory
asked Nov 21 '18 at 18:48
numericalorangenumericalorange
1,726311
asked Nov 21 '18 at 18:48
numericalorangenumericalorange
1,726311
asked Nov 21 '18 at 18:48
numericalorangenumericalorange
1,726311
asked Nov 21 '18 at 18:48
numericalorangenumericalorange
1,726311
asked Nov 21 '18 at 18:48
numericalorangenumericalorange
1,726311
1,726311
Characteristic of any field is either $0$ or a prime number. Hence $k$ is invertible.
– Yadati Kiran
Nov 21 '18 at 19:14
"$f(x)$ is polynomial in $x^p$" means that there is a polynomial $g in F[x]$ such that $f(x) = g(x^p)$. More concretely, it means that every monomial in $f$ has a power that is divisible by $p$. And as the comment above says, since $k$ and $a_n$ are nonzero, then $k a_n$ is invertible.
– André 3000
Nov 21 '18 at 21:18
@André3000 Does that mean the kernel is the coefficients of $f'(x)$ when $char(F)=0$ and the kernel is the monomials in $f$ when $char(F) = p$? I am a little confused...
– numericalorange
Nov 22 '18 at 18:19
@YadatiKiran I am a little confused about how characteristic relates to invertibility? If there does not exist a positive integer $n$ such that $nr=0$ for all $rin F$, why does that show that $r$ is invertible?
– numericalorange
Nov 22 '18 at 18:20
@numericalorange: What are the proper ideals of ring with characteristic $0$?
– Yadati Kiran
Nov 22 '18 at 18:28
|
show 3 more comments
Characteristic of any field is either $0$ or a prime number. Hence $k$ is invertible.
– Yadati Kiran
Nov 21 '18 at 19:14
"$f(x)$ is polynomial in $x^p$" means that there is a polynomial $g in F[x]$ such that $f(x) = g(x^p)$. More concretely, it means that every monomial in $f$ has a power that is divisible by $p$. And as the comment above says, since $k$ and $a_n$ are nonzero, then $k a_n$ is invertible.
– André 3000
Nov 21 '18 at 21:18
@André3000 Does that mean the kernel is the coefficients of $f'(x)$ when $char(F)=0$ and the kernel is the monomials in $f$ when $char(F) = p$? I am a little confused...
– numericalorange
Nov 22 '18 at 18:19
@YadatiKiran I am a little confused about how characteristic relates to invertibility? If there does not exist a positive integer $n$ such that $nr=0$ for all $rin F$, why does that show that $r$ is invertible?
– numericalorange
Nov 22 '18 at 18:20
@numericalorange: What are the proper ideals of ring with characteristic $0$?
– Yadati Kiran
Nov 22 '18 at 18:28
Characteristic of any field is either $0$ or a prime number. Hence $k$ is invertible.
– Yadati Kiran
Nov 21 '18 at 19:14
Characteristic of any field is either $0$ or a prime number. Hence $k$ is invertible.
– Yadati Kiran
Nov 21 '18 at 19:14
"$f(x)$ is polynomial in $x^p$" means that there is a polynomial $g in F[x]$ such that $f(x) = g(x^p)$. More concretely, it means that every monomial in $f$ has a power that is divisible by $p$. And as the comment above says, since $k$ and $a_n$ are nonzero, then $k a_n$ is invertible.
– André 3000
Nov 21 '18 at 21:18
"$f(x)$ is polynomial in $x^p$" means that there is a polynomial $g in F[x]$ such that $f(x) = g(x^p)$. More concretely, it means that every monomial in $f$ has a power that is divisible by $p$. And as the comment above says, since $k$ and $a_n$ are nonzero, then $k a_n$ is invertible.
– André 3000
Nov 21 '18 at 21:18
@André3000 Does that mean the kernel is the coefficients of $f'(x)$ when $char(F)=0$ and the kernel is the monomials in $f$ when $char(F) = p$? I am a little confused...
– numericalorange
Nov 22 '18 at 18:19
@André3000 Does that mean the kernel is the coefficients of $f'(x)$ when $char(F)=0$ and the kernel is the monomials in $f$ when $char(F) = p$? I am a little confused...
– numericalorange
Nov 22 '18 at 18:19
@YadatiKiran I am a little confused about how characteristic relates to invertibility? If there does not exist a positive integer $n$ such that $nr=0$ for all $rin F$, why does that show that $r$ is invertible?
– numericalorange
Nov 22 '18 at 18:20
@YadatiKiran I am a little confused about how characteristic relates to invertibility? If there does not exist a positive integer $n$ such that $nr=0$ for all $rin F$, why does that show that $r$ is invertible?
– numericalorange
Nov 22 '18 at 18:20
@numericalorange: What are the proper ideals of ring with characteristic $0$?
– Yadati Kiran
Nov 22 '18 at 18:28
@numericalorange: What are the proper ideals of ring with characteristic $0$?
– Yadati Kiran
Nov 22 '18 at 18:28
|
show 3 more comments
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Characteristic of any field is either $0$ or a prime number. Hence $k$ is invertible.
– Yadati Kiran
Nov 21 '18 at 19:14
"$f(x)$ is polynomial in $x^p$" means that there is a polynomial $g in F[x]$ such that $f(x) = g(x^p)$. More concretely, it means that every monomial in $f$ has a power that is divisible by $p$. And as the comment above says, since $k$ and $a_n$ are nonzero, then $k a_n$ is invertible.
– André 3000
Nov 21 '18 at 21:18
@André3000 Does that mean the kernel is the coefficients of $f'(x)$ when $char(F)=0$ and the kernel is the monomials in $f$ when $char(F) = p$? I am a little confused...
– numericalorange
Nov 22 '18 at 18:19
@YadatiKiran I am a little confused about how characteristic relates to invertibility? If there does not exist a positive integer $n$ such that $nr=0$ for all $rin F$, why does that show that $r$ is invertible?
– numericalorange
Nov 22 '18 at 18:20
@numericalorange: What are the proper ideals of ring with characteristic $0$?
– Yadati Kiran
Nov 22 '18 at 18:28