When is $x^{2^n} + 1$ Reducible in $mathbf{F}_p$ For All Primes $p$
$begingroup$
Define $f_n = x^{2^n} + 1$.
Then we want to show that there is an integer $n$ such that $f_n$ is reducible in $mathbb{F}_p[x]$ for all primes, $p$.
However, I want to do this using the hint
The group of units modulo $2^r$, $(mathbb{Z}/2^rmathbb{Z})^*$, is not cyclic for $r ge 3 $.
Indeed, there are ways to show this without the hint - one such method forms the basis of this answer.
However, I just can't seem to figure out how to use the hint! Hopefully using the hint will also shine some light on the question of for which $n$ does the above hold?
galois-theory finite-fields cyclic-groups splitting-field
$endgroup$
add a comment |
$begingroup$
Define $f_n = x^{2^n} + 1$.
Then we want to show that there is an integer $n$ such that $f_n$ is reducible in $mathbb{F}_p[x]$ for all primes, $p$.
However, I want to do this using the hint
The group of units modulo $2^r$, $(mathbb{Z}/2^rmathbb{Z})^*$, is not cyclic for $r ge 3 $.
Indeed, there are ways to show this without the hint - one such method forms the basis of this answer.
However, I just can't seem to figure out how to use the hint! Hopefully using the hint will also shine some light on the question of for which $n$ does the above hold?
galois-theory finite-fields cyclic-groups splitting-field
$endgroup$
$begingroup$
I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:30
$begingroup$
Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:41
add a comment |
$begingroup$
Define $f_n = x^{2^n} + 1$.
Then we want to show that there is an integer $n$ such that $f_n$ is reducible in $mathbb{F}_p[x]$ for all primes, $p$.
However, I want to do this using the hint
The group of units modulo $2^r$, $(mathbb{Z}/2^rmathbb{Z})^*$, is not cyclic for $r ge 3 $.
Indeed, there are ways to show this without the hint - one such method forms the basis of this answer.
However, I just can't seem to figure out how to use the hint! Hopefully using the hint will also shine some light on the question of for which $n$ does the above hold?
galois-theory finite-fields cyclic-groups splitting-field
$endgroup$
Define $f_n = x^{2^n} + 1$.
Then we want to show that there is an integer $n$ such that $f_n$ is reducible in $mathbb{F}_p[x]$ for all primes, $p$.
However, I want to do this using the hint
The group of units modulo $2^r$, $(mathbb{Z}/2^rmathbb{Z})^*$, is not cyclic for $r ge 3 $.
Indeed, there are ways to show this without the hint - one such method forms the basis of this answer.
However, I just can't seem to figure out how to use the hint! Hopefully using the hint will also shine some light on the question of for which $n$ does the above hold?
galois-theory finite-fields cyclic-groups splitting-field
galois-theory finite-fields cyclic-groups splitting-field
asked Dec 15 '18 at 15:14
John DonJohn Don
337115
337115
$begingroup$
I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:30
$begingroup$
Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:41
add a comment |
$begingroup$
I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:30
$begingroup$
Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:41
$begingroup$
I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:30
$begingroup$
I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:30
$begingroup$
Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:41
$begingroup$
Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:41
add a comment |
2 Answers
2
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oldest
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$begingroup$
It's always reducible modulo $2$.
Modulo an odd prime $p$, its zeroes are the primitive $2^{n+1}$-th roots of unity.
Let $alpha$ be any of them. Then $Bbb F_p(alpha)=Bbb F_{p^k}$
where $k$ is the least positive integer with $alpha^{p^k}=alpha$.
That's equivalent to $p^kequiv1pmod{2^{m+1}}$. If $k<2^m$, then the
degree of $Bbb F_p(alpha)$ over $Bbb F_p$ is less than $2^m$, so that $x^{2^m}+1$
cannot be irreducible over $Bbb F_p$.
$endgroup$
add a comment |
$begingroup$
Assume $f_n$ is irreducible in some $K=mathbb{F}_p$, where $p$ is odd. Then let $omega in L=mathbb{F}_q$ a root of $f_n$ in some field extension.
Then, since $L$ is a Galois extension of $K$, the Galois group of $L/K$ (a generator of which is the Frobenius automorphism) has to act transitively on the roots of $f_n$. In other words, every root of $f_n$ is $omega^{p^k}$ for some $k$. Now, $f_n$ has $2^n$ distinct roots, which are the odd powers of $omega$. Thus, every odd integer mist be congruent mod $2^{n+1}$ to some $p^k$. Thus $p$ is a generator of your non-cyclic group.
When $p=2$, then $f_n=(X+1)^{2^n}$.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's always reducible modulo $2$.
Modulo an odd prime $p$, its zeroes are the primitive $2^{n+1}$-th roots of unity.
Let $alpha$ be any of them. Then $Bbb F_p(alpha)=Bbb F_{p^k}$
where $k$ is the least positive integer with $alpha^{p^k}=alpha$.
That's equivalent to $p^kequiv1pmod{2^{m+1}}$. If $k<2^m$, then the
degree of $Bbb F_p(alpha)$ over $Bbb F_p$ is less than $2^m$, so that $x^{2^m}+1$
cannot be irreducible over $Bbb F_p$.
$endgroup$
add a comment |
$begingroup$
It's always reducible modulo $2$.
Modulo an odd prime $p$, its zeroes are the primitive $2^{n+1}$-th roots of unity.
Let $alpha$ be any of them. Then $Bbb F_p(alpha)=Bbb F_{p^k}$
where $k$ is the least positive integer with $alpha^{p^k}=alpha$.
That's equivalent to $p^kequiv1pmod{2^{m+1}}$. If $k<2^m$, then the
degree of $Bbb F_p(alpha)$ over $Bbb F_p$ is less than $2^m$, so that $x^{2^m}+1$
cannot be irreducible over $Bbb F_p$.
$endgroup$
add a comment |
$begingroup$
It's always reducible modulo $2$.
Modulo an odd prime $p$, its zeroes are the primitive $2^{n+1}$-th roots of unity.
Let $alpha$ be any of them. Then $Bbb F_p(alpha)=Bbb F_{p^k}$
where $k$ is the least positive integer with $alpha^{p^k}=alpha$.
That's equivalent to $p^kequiv1pmod{2^{m+1}}$. If $k<2^m$, then the
degree of $Bbb F_p(alpha)$ over $Bbb F_p$ is less than $2^m$, so that $x^{2^m}+1$
cannot be irreducible over $Bbb F_p$.
$endgroup$
It's always reducible modulo $2$.
Modulo an odd prime $p$, its zeroes are the primitive $2^{n+1}$-th roots of unity.
Let $alpha$ be any of them. Then $Bbb F_p(alpha)=Bbb F_{p^k}$
where $k$ is the least positive integer with $alpha^{p^k}=alpha$.
That's equivalent to $p^kequiv1pmod{2^{m+1}}$. If $k<2^m$, then the
degree of $Bbb F_p(alpha)$ over $Bbb F_p$ is less than $2^m$, so that $x^{2^m}+1$
cannot be irreducible over $Bbb F_p$.
answered Dec 15 '18 at 15:28
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
add a comment |
add a comment |
$begingroup$
Assume $f_n$ is irreducible in some $K=mathbb{F}_p$, where $p$ is odd. Then let $omega in L=mathbb{F}_q$ a root of $f_n$ in some field extension.
Then, since $L$ is a Galois extension of $K$, the Galois group of $L/K$ (a generator of which is the Frobenius automorphism) has to act transitively on the roots of $f_n$. In other words, every root of $f_n$ is $omega^{p^k}$ for some $k$. Now, $f_n$ has $2^n$ distinct roots, which are the odd powers of $omega$. Thus, every odd integer mist be congruent mod $2^{n+1}$ to some $p^k$. Thus $p$ is a generator of your non-cyclic group.
When $p=2$, then $f_n=(X+1)^{2^n}$.
$endgroup$
add a comment |
$begingroup$
Assume $f_n$ is irreducible in some $K=mathbb{F}_p$, where $p$ is odd. Then let $omega in L=mathbb{F}_q$ a root of $f_n$ in some field extension.
Then, since $L$ is a Galois extension of $K$, the Galois group of $L/K$ (a generator of which is the Frobenius automorphism) has to act transitively on the roots of $f_n$. In other words, every root of $f_n$ is $omega^{p^k}$ for some $k$. Now, $f_n$ has $2^n$ distinct roots, which are the odd powers of $omega$. Thus, every odd integer mist be congruent mod $2^{n+1}$ to some $p^k$. Thus $p$ is a generator of your non-cyclic group.
When $p=2$, then $f_n=(X+1)^{2^n}$.
$endgroup$
add a comment |
$begingroup$
Assume $f_n$ is irreducible in some $K=mathbb{F}_p$, where $p$ is odd. Then let $omega in L=mathbb{F}_q$ a root of $f_n$ in some field extension.
Then, since $L$ is a Galois extension of $K$, the Galois group of $L/K$ (a generator of which is the Frobenius automorphism) has to act transitively on the roots of $f_n$. In other words, every root of $f_n$ is $omega^{p^k}$ for some $k$. Now, $f_n$ has $2^n$ distinct roots, which are the odd powers of $omega$. Thus, every odd integer mist be congruent mod $2^{n+1}$ to some $p^k$. Thus $p$ is a generator of your non-cyclic group.
When $p=2$, then $f_n=(X+1)^{2^n}$.
$endgroup$
Assume $f_n$ is irreducible in some $K=mathbb{F}_p$, where $p$ is odd. Then let $omega in L=mathbb{F}_q$ a root of $f_n$ in some field extension.
Then, since $L$ is a Galois extension of $K$, the Galois group of $L/K$ (a generator of which is the Frobenius automorphism) has to act transitively on the roots of $f_n$. In other words, every root of $f_n$ is $omega^{p^k}$ for some $k$. Now, $f_n$ has $2^n$ distinct roots, which are the odd powers of $omega$. Thus, every odd integer mist be congruent mod $2^{n+1}$ to some $p^k$. Thus $p$ is a generator of your non-cyclic group.
When $p=2$, then $f_n=(X+1)^{2^n}$.
answered Dec 15 '18 at 15:31
MindlackMindlack
3,86018
3,86018
add a comment |
add a comment |
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$begingroup$
I hate to blow my own trumpet, but see this. If $nge2$ $f_n$ is reducible in $Bbb{F}_p[x]$ for all primes $p$. Do read Qiaochu Yuan's answer in the linked thread for a nice piece of Galois theory.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:30
$begingroup$
Also observe Robert Israel's comment under my answer for a simpler argument. See this for the case $n=2$.
$endgroup$
– Jyrki Lahtonen
Dec 16 '18 at 6:41