Mixing
If $q(X)$ polynomial over $Bbb Z/pBbb Z$, then $q(X)^p=q(X^p)$
$begingroup$
Theorem: Let $p$ be a prime. If $a_i in Bbb Z/pBbb Z$ for $i=0,1,2...,n$, then the polynomial $$left(sum_{i=0}^{n} a_iX^i right)^p=sum_{i=0}^{n}a_iX^{pi}$$
I am not sure how to approach this proof. My initial thought was to use binomial expansion but It doesn't seem to lead anywhere. I'd like to mention that I want to keep the level of techniques used at a minimum as the course only assumes bare minimum knowledge (for some reason the course focuses on particular topics in field theory namely, polynomial rings over finite fields without proper treatment of rings and fields).
abstract-algebra finite-fields
asked Jan 13 at 20:05
Sei SakataSei Sakata
10810
$endgroup$
add a comment |
$begingroup$
Theorem: Let $p$ be a prime. If $a_i in Bbb Z/pBbb Z$ for $i=0,1,2...,n$, then the polynomial $$left(sum_{i=0}^{n} a_iX^i right)^p=sum_{i=0}^{n}a_iX^{pi}$$
I am not sure how to approach this proof. My initial thought was to use binomial expansion but It doesn't seem to lead anywhere. I'd like to mention that I want to keep the level of techniques used at a minimum as the course only assumes bare minimum knowledge (for some reason the course focuses on particular topics in field theory namely, polynomial rings over finite fields without proper treatment of rings and fields).
abstract-algebra finite-fields
asked Jan 13 at 20:05
Sei SakataSei Sakata
10810
$endgroup$
$begingroup$
This might help.
$endgroup$
– J. W. Tanner
Jan 13 at 20:08
add a comment |
$begingroup$
Theorem: Let $p$ be a prime. If $a_i in Bbb Z/pBbb Z$ for $i=0,1,2...,n$, then the polynomial $$left(sum_{i=0}^{n} a_iX^i right)^p=sum_{i=0}^{n}a_iX^{pi}$$
I am not sure how to approach this proof. My initial thought was to use binomial expansion but It doesn't seem to lead anywhere. I'd like to mention that I want to keep the level of techniques used at a minimum as the course only assumes bare minimum knowledge (for some reason the course focuses on particular topics in field theory namely, polynomial rings over finite fields without proper treatment of rings and fields).
abstract-algebra finite-fields
asked Jan 13 at 20:05
Sei SakataSei Sakata
10810
$endgroup$
Theorem: Let $p$ be a prime. If $a_i in Bbb Z/pBbb Z$ for $i=0,1,2...,n$, then the polynomial $$left(sum_{i=0}^{n} a_iX^i right)^p=sum_{i=0}^{n}a_iX^{pi}$$
I am not sure how to approach this proof. My initial thought was to use binomial expansion but It doesn't seem to lead anywhere. I'd like to mention that I want to keep the level of techniques used at a minimum as the course only assumes bare minimum knowledge (for some reason the course focuses on particular topics in field theory namely, polynomial rings over finite fields without proper treatment of rings and fields).
abstract-algebra finite-fields
abstract-algebra finite-fields
asked Jan 13 at 20:05
Sei SakataSei Sakata
10810
asked Jan 13 at 20:05
Sei SakataSei Sakata
10810
asked Jan 13 at 20:05
Sei SakataSei Sakata
10810
asked Jan 13 at 20:05
Sei SakataSei Sakata
10810
asked Jan 13 at 20:05
Sei SakataSei Sakata
10810
10810
$begingroup$
This might help.
$endgroup$
– J. W. Tanner
Jan 13 at 20:08
add a comment |
$begingroup$
This might help.
$endgroup$
– J. W. Tanner
Jan 13 at 20:08
$begingroup$
This might help.
$endgroup$
– J. W. Tanner
Jan 13 at 20:08
$begingroup$
This might help.
$endgroup$
– J. W. Tanner
Jan 13 at 20:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Given the polynomial
$q(X) = displaystyle sum_0^n a_k X^k in Bbb Z/pBbb Z[X], tag 1$
where $p in Bbb P$ is prime, so that
$a_k in Bbb Z/pBbb Z, ; 0 le k le n, tag 2$
we have
$(q(X))^p = left ( displaystyle sum_0^n a_k X^k right )^p, tag 3$
and we find the expansion of the right-hand side via a simple induction on $n$:
first, for $n = 1$ we have
$q(X) = displaystyle sum_0^1 a_k X^k = a_0 X^0 + a_1 X = a_0 + a_1 X; tag 4$
thus,
$(q(X))^p = left ( displaystyle sum_0^1 a_k X^k right )^p = (a_0 + a_1 X)^p; tag 5$
we may use the plain-and-ordinary, vanilla-flavored binomial theorem to write
$(a_0 + a_1 X)^p = displaystyle sum_0^p dfrac{p!}{i!(p - i)!} a_0^{p - i}a_1^iX^i; tag 6$
the binomial is applicable in this situation since it holds over any unital, commutative ring; inspecting the coefficients in (6), we see that, as integers,
$p mid dfrac{p!}{i!(p - i)!}, ; 1 le i le p - 1, tag 7$
whenever $p in Bbb P$, since neither $i!$ nor $(p - i)!$ contain $p$ as a factor; $p$ can't can't cancel out of the numerator of the binomial coefficient $p!/(i!(p - i)!)$. Thus, (6) reduces to
$(a_0 + a_1 X)^p = a_0^p a_1^0 X^0 + a_0^0 a_1^p X^p = a_0^p + a_1^p X^p; tag 8$
now since $a_0, a_1 in Bbb Z / pBbb Z$, we have by Fermat's Little Theorem that
$a_0^p = a_0, ; a_1^p = a_1, tag 9$
whence (8) becomes
$(a_0 + a_1 X)^p = a_0 + a_1 X^p, tag{10}$
and the case $n = 1$ is thus established. The next step is to assume that
$(q(X))^p = left ( displaystyle sum_0^m a_k X^k right )^p = displaystyle sum_0^m a_k X^{kp} = q(X^p) tag{11}$
binds for polynomials $q(X)$ of degree $m$; now with
$q(X) = displaystyle sum_0^{m + 1} a_k X^k tag{12}$
we may write
$(q(X))^p = left ( displaystyle sum_0^{m + 1} a_k X^k right )^p = left (left ( displaystyle sum_0^m a_k X^k right ) + a_{m + 1}X^{m + 1} right )^p$
$= displaystyle sum_0^p dfrac{p!}{j!(p - j)!}left ( displaystyle sum_0^m a_k X^k right )^{p - j} (a_{m + 1}X^{m + 1} )^j, tag{13}$
again by the binomial theorem, and again by virtue of (7) this is transformed into
$(q(X))^p = left ( displaystyle sum_0^m a_k X^k right )^p + (a_{m + 1}X^{m + 1} )^p = left ( displaystyle sum_0^m a_k X^k right )^p + a_{m + 1}^p X^{p(m + 1)}; tag{14}$
we now substitue in the inductive assumption (11), and again apply Fermat's Little to
replace $a_{m + 1}^p$ with $a_{m + 1}$:
$(q(X))^p = displaystyle sum_0^m a_k X^{kp} + (a_{m + 1}X^{m + 1} )^p$
$= displaystyle sum_0^m a_k X^{kp} + a_{m + 1} X^{(m + 1)p} = sum_0^{m + 1} a_k X^{kp} = q(X^p); tag{15}$
thus our induction is complete and we have established
$(q(X))^p = q(X^p) tag{16}$
for all polynomials $q(Z) in Bbb Z / pBbb Z[X]$. $OEDelta$
answered Jan 13 at 23:52
Robert LewisRobert Lewis
46.5k23067
$endgroup$
$begingroup$
Very nice and detailed answer. Thanks, it adds completeness to the question.
$endgroup$
– Sei Sakata
Jan 14 at 13:31
add a comment |
$begingroup$
Hint$_1$: $binom{p}{k} = frac{p(p-1)!}{k!(p-k)!}$, where $k, p-k < p$ when $0 < k < p$. Does the factor $p$ change?
Hint$_2$: $(a+b)^p equiv a^p + b^p$ is a consequence from above. What happens if you plug more numbers, i.e., $a, b, c, dots$? Induction is a good try.
Hint$_3$: Fermat's little theorem states that $a^p equiv a$.
answered Jan 13 at 20:21
Lucas HenriqueLucas Henrique
1,059414
$endgroup$
1
$begingroup$
Thanks that makes the problem so much easier.
$endgroup$
– Sei Sakata
Jan 13 at 20:26
$begingroup$
If you've managed to solve the question with the tips, please check this as the answer. :)
$endgroup$
– Lucas Henrique
Jan 13 at 22:02
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Given the polynomial
$q(X) = displaystyle sum_0^n a_k X^k in Bbb Z/pBbb Z[X], tag 1$
where $p in Bbb P$ is prime, so that
$a_k in Bbb Z/pBbb Z, ; 0 le k le n, tag 2$
we have
$(q(X))^p = left ( displaystyle sum_0^n a_k X^k right )^p, tag 3$
and we find the expansion of the right-hand side via a simple induction on $n$:
first, for $n = 1$ we have
$q(X) = displaystyle sum_0^1 a_k X^k = a_0 X^0 + a_1 X = a_0 + a_1 X; tag 4$
thus,
$(q(X))^p = left ( displaystyle sum_0^1 a_k X^k right )^p = (a_0 + a_1 X)^p; tag 5$
we may use the plain-and-ordinary, vanilla-flavored binomial theorem to write
$(a_0 + a_1 X)^p = displaystyle sum_0^p dfrac{p!}{i!(p - i)!} a_0^{p - i}a_1^iX^i; tag 6$
the binomial is applicable in this situation since it holds over any unital, commutative ring; inspecting the coefficients in (6), we see that, as integers,
$p mid dfrac{p!}{i!(p - i)!}, ; 1 le i le p - 1, tag 7$
whenever $p in Bbb P$, since neither $i!$ nor $(p - i)!$ contain $p$ as a factor; $p$ can't can't cancel out of the numerator of the binomial coefficient $p!/(i!(p - i)!)$. Thus, (6) reduces to
$(a_0 + a_1 X)^p = a_0^p a_1^0 X^0 + a_0^0 a_1^p X^p = a_0^p + a_1^p X^p; tag 8$
now since $a_0, a_1 in Bbb Z / pBbb Z$, we have by Fermat's Little Theorem that
$a_0^p = a_0, ; a_1^p = a_1, tag 9$
whence (8) becomes
$(a_0 + a_1 X)^p = a_0 + a_1 X^p, tag{10}$
and the case $n = 1$ is thus established. The next step is to assume that
$(q(X))^p = left ( displaystyle sum_0^m a_k X^k right )^p = displaystyle sum_0^m a_k X^{kp} = q(X^p) tag{11}$
binds for polynomials $q(X)$ of degree $m$; now with
$q(X) = displaystyle sum_0^{m + 1} a_k X^k tag{12}$
we may write
$(q(X))^p = left ( displaystyle sum_0^{m + 1} a_k X^k right )^p = left (left ( displaystyle sum_0^m a_k X^k right ) + a_{m + 1}X^{m + 1} right )^p$
$= displaystyle sum_0^p dfrac{p!}{j!(p - j)!}left ( displaystyle sum_0^m a_k X^k right )^{p - j} (a_{m + 1}X^{m + 1} )^j, tag{13}$
again by the binomial theorem, and again by virtue of (7) this is transformed into
$(q(X))^p = left ( displaystyle sum_0^m a_k X^k right )^p + (a_{m + 1}X^{m + 1} )^p = left ( displaystyle sum_0^m a_k X^k right )^p + a_{m + 1}^p X^{p(m + 1)}; tag{14}$
we now substitue in the inductive assumption (11), and again apply Fermat's Little to
replace $a_{m + 1}^p$ with $a_{m + 1}$:
$(q(X))^p = displaystyle sum_0^m a_k X^{kp} + (a_{m + 1}X^{m + 1} )^p$
$= displaystyle sum_0^m a_k X^{kp} + a_{m + 1} X^{(m + 1)p} = sum_0^{m + 1} a_k X^{kp} = q(X^p); tag{15}$
thus our induction is complete and we have established
$(q(X))^p = q(X^p) tag{16}$
for all polynomials $q(Z) in Bbb Z / pBbb Z[X]$. $OEDelta$
answered Jan 13 at 23:52
Robert LewisRobert Lewis
46.5k23067
$endgroup$
$begingroup$
Very nice and detailed answer. Thanks, it adds completeness to the question.
$endgroup$
– Sei Sakata
Jan 14 at 13:31
add a comment |
$begingroup$
Given the polynomial
$q(X) = displaystyle sum_0^n a_k X^k in Bbb Z/pBbb Z[X], tag 1$
where $p in Bbb P$ is prime, so that
$a_k in Bbb Z/pBbb Z, ; 0 le k le n, tag 2$
we have
$(q(X))^p = left ( displaystyle sum_0^n a_k X^k right )^p, tag 3$
and we find the expansion of the right-hand side via a simple induction on $n$:
first, for $n = 1$ we have
$q(X) = displaystyle sum_0^1 a_k X^k = a_0 X^0 + a_1 X = a_0 + a_1 X; tag 4$
thus,
$(q(X))^p = left ( displaystyle sum_0^1 a_k X^k right )^p = (a_0 + a_1 X)^p; tag 5$
we may use the plain-and-ordinary, vanilla-flavored binomial theorem to write
$(a_0 + a_1 X)^p = displaystyle sum_0^p dfrac{p!}{i!(p - i)!} a_0^{p - i}a_1^iX^i; tag 6$
the binomial is applicable in this situation since it holds over any unital, commutative ring; inspecting the coefficients in (6), we see that, as integers,
$p mid dfrac{p!}{i!(p - i)!}, ; 1 le i le p - 1, tag 7$
whenever $p in Bbb P$, since neither $i!$ nor $(p - i)!$ contain $p$ as a factor; $p$ can't can't cancel out of the numerator of the binomial coefficient $p!/(i!(p - i)!)$. Thus, (6) reduces to
$(a_0 + a_1 X)^p = a_0^p a_1^0 X^0 + a_0^0 a_1^p X^p = a_0^p + a_1^p X^p; tag 8$
now since $a_0, a_1 in Bbb Z / pBbb Z$, we have by Fermat's Little Theorem that
$a_0^p = a_0, ; a_1^p = a_1, tag 9$
whence (8) becomes
$(a_0 + a_1 X)^p = a_0 + a_1 X^p, tag{10}$
and the case $n = 1$ is thus established. The next step is to assume that
$(q(X))^p = left ( displaystyle sum_0^m a_k X^k right )^p = displaystyle sum_0^m a_k X^{kp} = q(X^p) tag{11}$
binds for polynomials $q(X)$ of degree $m$; now with
$q(X) = displaystyle sum_0^{m + 1} a_k X^k tag{12}$
we may write
$(q(X))^p = left ( displaystyle sum_0^{m + 1} a_k X^k right )^p = left (left ( displaystyle sum_0^m a_k X^k right ) + a_{m + 1}X^{m + 1} right )^p$
$= displaystyle sum_0^p dfrac{p!}{j!(p - j)!}left ( displaystyle sum_0^m a_k X^k right )^{p - j} (a_{m + 1}X^{m + 1} )^j, tag{13}$
again by the binomial theorem, and again by virtue of (7) this is transformed into
$(q(X))^p = left ( displaystyle sum_0^m a_k X^k right )^p + (a_{m + 1}X^{m + 1} )^p = left ( displaystyle sum_0^m a_k X^k right )^p + a_{m + 1}^p X^{p(m + 1)}; tag{14}$
we now substitue in the inductive assumption (11), and again apply Fermat's Little to
replace $a_{m + 1}^p$ with $a_{m + 1}$:
$(q(X))^p = displaystyle sum_0^m a_k X^{kp} + (a_{m + 1}X^{m + 1} )^p$
$= displaystyle sum_0^m a_k X^{kp} + a_{m + 1} X^{(m + 1)p} = sum_0^{m + 1} a_k X^{kp} = q(X^p); tag{15}$
thus our induction is complete and we have established
$(q(X))^p = q(X^p) tag{16}$
for all polynomials $q(Z) in Bbb Z / pBbb Z[X]$. $OEDelta$
answered Jan 13 at 23:52
Robert LewisRobert Lewis
46.5k23067
$endgroup$
$begingroup$
Very nice and detailed answer. Thanks, it adds completeness to the question.
$endgroup$
– Sei Sakata
Jan 14 at 13:31
add a comment |
$begingroup$
Given the polynomial
$q(X) = displaystyle sum_0^n a_k X^k in Bbb Z/pBbb Z[X], tag 1$
where $p in Bbb P$ is prime, so that
$a_k in Bbb Z/pBbb Z, ; 0 le k le n, tag 2$
we have
$(q(X))^p = left ( displaystyle sum_0^n a_k X^k right )^p, tag 3$
and we find the expansion of the right-hand side via a simple induction on $n$:
first, for $n = 1$ we have
$q(X) = displaystyle sum_0^1 a_k X^k = a_0 X^0 + a_1 X = a_0 + a_1 X; tag 4$
thus,
$(q(X))^p = left ( displaystyle sum_0^1 a_k X^k right )^p = (a_0 + a_1 X)^p; tag 5$
we may use the plain-and-ordinary, vanilla-flavored binomial theorem to write
$(a_0 + a_1 X)^p = displaystyle sum_0^p dfrac{p!}{i!(p - i)!} a_0^{p - i}a_1^iX^i; tag 6$
the binomial is applicable in this situation since it holds over any unital, commutative ring; inspecting the coefficients in (6), we see that, as integers,
$p mid dfrac{p!}{i!(p - i)!}, ; 1 le i le p - 1, tag 7$
whenever $p in Bbb P$, since neither $i!$ nor $(p - i)!$ contain $p$ as a factor; $p$ can't can't cancel out of the numerator of the binomial coefficient $p!/(i!(p - i)!)$. Thus, (6) reduces to
$(a_0 + a_1 X)^p = a_0^p a_1^0 X^0 + a_0^0 a_1^p X^p = a_0^p + a_1^p X^p; tag 8$
now since $a_0, a_1 in Bbb Z / pBbb Z$, we have by Fermat's Little Theorem that
$a_0^p = a_0, ; a_1^p = a_1, tag 9$
whence (8) becomes
$(a_0 + a_1 X)^p = a_0 + a_1 X^p, tag{10}$
and the case $n = 1$ is thus established. The next step is to assume that
$(q(X))^p = left ( displaystyle sum_0^m a_k X^k right )^p = displaystyle sum_0^m a_k X^{kp} = q(X^p) tag{11}$
binds for polynomials $q(X)$ of degree $m$; now with
$q(X) = displaystyle sum_0^{m + 1} a_k X^k tag{12}$
we may write
$(q(X))^p = left ( displaystyle sum_0^{m + 1} a_k X^k right )^p = left (left ( displaystyle sum_0^m a_k X^k right ) + a_{m + 1}X^{m + 1} right )^p$
$= displaystyle sum_0^p dfrac{p!}{j!(p - j)!}left ( displaystyle sum_0^m a_k X^k right )^{p - j} (a_{m + 1}X^{m + 1} )^j, tag{13}$
again by the binomial theorem, and again by virtue of (7) this is transformed into
$(q(X))^p = left ( displaystyle sum_0^m a_k X^k right )^p + (a_{m + 1}X^{m + 1} )^p = left ( displaystyle sum_0^m a_k X^k right )^p + a_{m + 1}^p X^{p(m + 1)}; tag{14}$
we now substitue in the inductive assumption (11), and again apply Fermat's Little to
replace $a_{m + 1}^p$ with $a_{m + 1}$:
$(q(X))^p = displaystyle sum_0^m a_k X^{kp} + (a_{m + 1}X^{m + 1} )^p$
$= displaystyle sum_0^m a_k X^{kp} + a_{m + 1} X^{(m + 1)p} = sum_0^{m + 1} a_k X^{kp} = q(X^p); tag{15}$
thus our induction is complete and we have established
$(q(X))^p = q(X^p) tag{16}$
for all polynomials $q(Z) in Bbb Z / pBbb Z[X]$. $OEDelta$
answered Jan 13 at 23:52
Robert LewisRobert Lewis
46.5k23067
$endgroup$
Given the polynomial
$q(X) = displaystyle sum_0^n a_k X^k in Bbb Z/pBbb Z[X], tag 1$
where $p in Bbb P$ is prime, so that
$a_k in Bbb Z/pBbb Z, ; 0 le k le n, tag 2$
we have
$(q(X))^p = left ( displaystyle sum_0^n a_k X^k right )^p, tag 3$
and we find the expansion of the right-hand side via a simple induction on $n$:
first, for $n = 1$ we have
$q(X) = displaystyle sum_0^1 a_k X^k = a_0 X^0 + a_1 X = a_0 + a_1 X; tag 4$
thus,
$(q(X))^p = left ( displaystyle sum_0^1 a_k X^k right )^p = (a_0 + a_1 X)^p; tag 5$
we may use the plain-and-ordinary, vanilla-flavored binomial theorem to write
$(a_0 + a_1 X)^p = displaystyle sum_0^p dfrac{p!}{i!(p - i)!} a_0^{p - i}a_1^iX^i; tag 6$
the binomial is applicable in this situation since it holds over any unital, commutative ring; inspecting the coefficients in (6), we see that, as integers,
$p mid dfrac{p!}{i!(p - i)!}, ; 1 le i le p - 1, tag 7$
whenever $p in Bbb P$, since neither $i!$ nor $(p - i)!$ contain $p$ as a factor; $p$ can't can't cancel out of the numerator of the binomial coefficient $p!/(i!(p - i)!)$. Thus, (6) reduces to
$(a_0 + a_1 X)^p = a_0^p a_1^0 X^0 + a_0^0 a_1^p X^p = a_0^p + a_1^p X^p; tag 8$
now since $a_0, a_1 in Bbb Z / pBbb Z$, we have by Fermat's Little Theorem that
$a_0^p = a_0, ; a_1^p = a_1, tag 9$
whence (8) becomes
$(a_0 + a_1 X)^p = a_0 + a_1 X^p, tag{10}$
and the case $n = 1$ is thus established. The next step is to assume that
$(q(X))^p = left ( displaystyle sum_0^m a_k X^k right )^p = displaystyle sum_0^m a_k X^{kp} = q(X^p) tag{11}$
binds for polynomials $q(X)$ of degree $m$; now with
$q(X) = displaystyle sum_0^{m + 1} a_k X^k tag{12}$
we may write
$(q(X))^p = left ( displaystyle sum_0^{m + 1} a_k X^k right )^p = left (left ( displaystyle sum_0^m a_k X^k right ) + a_{m + 1}X^{m + 1} right )^p$
$= displaystyle sum_0^p dfrac{p!}{j!(p - j)!}left ( displaystyle sum_0^m a_k X^k right )^{p - j} (a_{m + 1}X^{m + 1} )^j, tag{13}$
again by the binomial theorem, and again by virtue of (7) this is transformed into
$(q(X))^p = left ( displaystyle sum_0^m a_k X^k right )^p + (a_{m + 1}X^{m + 1} )^p = left ( displaystyle sum_0^m a_k X^k right )^p + a_{m + 1}^p X^{p(m + 1)}; tag{14}$
we now substitue in the inductive assumption (11), and again apply Fermat's Little to
replace $a_{m + 1}^p$ with $a_{m + 1}$:
$(q(X))^p = displaystyle sum_0^m a_k X^{kp} + (a_{m + 1}X^{m + 1} )^p$
$= displaystyle sum_0^m a_k X^{kp} + a_{m + 1} X^{(m + 1)p} = sum_0^{m + 1} a_k X^{kp} = q(X^p); tag{15}$
thus our induction is complete and we have established
$(q(X))^p = q(X^p) tag{16}$
for all polynomials $q(Z) in Bbb Z / pBbb Z[X]$. $OEDelta$
answered Jan 13 at 23:52
Robert LewisRobert Lewis
46.5k23067
edited Jan 14 at 0:10
answered Jan 13 at 23:52
Robert LewisRobert Lewis
46.5k23067
answered Jan 13 at 23:52
Robert LewisRobert Lewis
46.5k23067
answered Jan 13 at 23:52
Robert LewisRobert Lewis
46.5k23067
46.5k23067
$begingroup$
Very nice and detailed answer. Thanks, it adds completeness to the question.
$endgroup$
– Sei Sakata
Jan 14 at 13:31
add a comment |
$begingroup$
Very nice and detailed answer. Thanks, it adds completeness to the question.
$endgroup$
– Sei Sakata
Jan 14 at 13:31
$begingroup$
Very nice and detailed answer. Thanks, it adds completeness to the question.
$endgroup$
– Sei Sakata
Jan 14 at 13:31
$begingroup$
Very nice and detailed answer. Thanks, it adds completeness to the question.
$endgroup$
– Sei Sakata
Jan 14 at 13:31
add a comment |
$begingroup$
Hint$_1$: $binom{p}{k} = frac{p(p-1)!}{k!(p-k)!}$, where $k, p-k < p$ when $0 < k < p$. Does the factor $p$ change?
Hint$_2$: $(a+b)^p equiv a^p + b^p$ is a consequence from above. What happens if you plug more numbers, i.e., $a, b, c, dots$? Induction is a good try.
Hint$_3$: Fermat's little theorem states that $a^p equiv a$.
answered Jan 13 at 20:21
Lucas HenriqueLucas Henrique
1,059414
$endgroup$
1
$begingroup$
Thanks that makes the problem so much easier.
$endgroup$
– Sei Sakata
Jan 13 at 20:26
$begingroup$
If you've managed to solve the question with the tips, please check this as the answer. :)
$endgroup$
– Lucas Henrique
Jan 13 at 22:02
add a comment |
$begingroup$
Hint$_1$: $binom{p}{k} = frac{p(p-1)!}{k!(p-k)!}$, where $k, p-k < p$ when $0 < k < p$. Does the factor $p$ change?
Hint$_2$: $(a+b)^p equiv a^p + b^p$ is a consequence from above. What happens if you plug more numbers, i.e., $a, b, c, dots$? Induction is a good try.
Hint$_3$: Fermat's little theorem states that $a^p equiv a$.
answered Jan 13 at 20:21
Lucas HenriqueLucas Henrique
1,059414
$endgroup$
1
$begingroup$
Thanks that makes the problem so much easier.
$endgroup$
– Sei Sakata
Jan 13 at 20:26
$begingroup$
If you've managed to solve the question with the tips, please check this as the answer. :)
$endgroup$
– Lucas Henrique
Jan 13 at 22:02
add a comment |
$begingroup$
Hint$_1$: $binom{p}{k} = frac{p(p-1)!}{k!(p-k)!}$, where $k, p-k < p$ when $0 < k < p$. Does the factor $p$ change?
Hint$_2$: $(a+b)^p equiv a^p + b^p$ is a consequence from above. What happens if you plug more numbers, i.e., $a, b, c, dots$? Induction is a good try.
Hint$_3$: Fermat's little theorem states that $a^p equiv a$.
answered Jan 13 at 20:21
Lucas HenriqueLucas Henrique
1,059414
$endgroup$
Hint$_1$: $binom{p}{k} = frac{p(p-1)!}{k!(p-k)!}$, where $k, p-k < p$ when $0 < k < p$. Does the factor $p$ change?
Hint$_2$: $(a+b)^p equiv a^p + b^p$ is a consequence from above. What happens if you plug more numbers, i.e., $a, b, c, dots$? Induction is a good try.
Hint$_3$: Fermat's little theorem states that $a^p equiv a$.
answered Jan 13 at 20:21
Lucas HenriqueLucas Henrique
1,059414
answered Jan 13 at 20:21
Lucas HenriqueLucas Henrique
1,059414
answered Jan 13 at 20:21
Lucas HenriqueLucas Henrique
1,059414
answered Jan 13 at 20:21
Lucas HenriqueLucas Henrique
1,059414
1,059414
1
$begingroup$
Thanks that makes the problem so much easier.
$endgroup$
– Sei Sakata
Jan 13 at 20:26
$begingroup$
If you've managed to solve the question with the tips, please check this as the answer. :)
$endgroup$
– Lucas Henrique
Jan 13 at 22:02
add a comment |
1
$begingroup$
Thanks that makes the problem so much easier.
$endgroup$
– Sei Sakata
Jan 13 at 20:26
$begingroup$
If you've managed to solve the question with the tips, please check this as the answer. :)
$endgroup$
– Lucas Henrique
Jan 13 at 22:02
1
1
$begingroup$
Thanks that makes the problem so much easier.
$endgroup$
– Sei Sakata
Jan 13 at 20:26
$begingroup$
Thanks that makes the problem so much easier.
$endgroup$
– Sei Sakata
Jan 13 at 20:26
$begingroup$
If you've managed to solve the question with the tips, please check this as the answer. :)
$endgroup$
– Lucas Henrique
Jan 13 at 22:02
$begingroup$
If you've managed to solve the question with the tips, please check this as the answer. :)
$endgroup$
– Lucas Henrique
Jan 13 at 22:02
add a comment |
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