Mixing
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Irreducibility of higher order polynomials over $mathbbQ$
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I wish to show that the polynomial $f(x)=x^9+3x^6+165x^3+1inmathbbQ[x]$ is irreducible over $mathbbQ$.
My guess; reducing $f(x)inmathbbF_p[x]$ for suitable prime $p$, it might be irreducible over $mathbbF_p$, where $mathbbF_p$ is finite field of order $p$.
But, how to find the suitable prime $p$, and to show irreducibility over $mathbbF_p$?
The order of $f(x)$ is too much high for me ;(
Give some hint or advice! Thank you!
abstract-algebra irreducible-polynomials
asked Jul 19 at 8:21
Primavera
1838
 |Â
show 2 more comments
up vote
5
down vote
favorite
I wish to show that the polynomial $f(x)=x^9+3x^6+165x^3+1inmathbbQ[x]$ is irreducible over $mathbbQ$.
My guess; reducing $f(x)inmathbbF_p[x]$ for suitable prime $p$, it might be irreducible over $mathbbF_p$, where $mathbbF_p$ is finite field of order $p$.
But, how to find the suitable prime $p$, and to show irreducibility over $mathbbF_p$?
The order of $f(x)$ is too much high for me ;(
Give some hint or advice! Thank you!
abstract-algebra irreducible-polynomials
asked Jul 19 at 8:21
Primavera
1838
Given that $3mid 3$ and $3, 5, 11mid 165$, I would at least try $p = 3, 5$ or $11$ first, see if anything immediate pops up.
– Arthur
Jul 19 at 8:24
1
It is enough to show that $x^3+3x^2+165x+1$ is irreducible, and it is possible to obtain the roots analytically.
– Yves Daoust
Jul 19 at 8:31
1
@Yves Daoust, why is that enough? I mean, could you please elaborate on "it is possible to obtain the roots analytically"?
– Ennar
Jul 19 at 8:42
If the roots of the reduced polynomial aren't rational, neither are those of the original. (In fact it suffices that the roots of the reduced aren't perfect cubes, but this is not needed here.) wolframalpha.com/input/?i=solve+x%5E3%2B3x%5E2%2B165x%2B1%3D0
– Yves Daoust
Jul 19 at 9:15
7
@YvesDaoust Why would that be enough? There is no general result stating that if $p(x)$ is irreducible, so would $p(x^3)$. Consider $p(x)=x^2+1$. It is irreducible, but $p(x^3)=x^6+1=(x^2+1)(x^4-x^2+1)$.
– Jyrki Lahtonen
Jul 19 at 10:33
 |Â
show 2 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I wish to show that the polynomial $f(x)=x^9+3x^6+165x^3+1inmathbbQ[x]$ is irreducible over $mathbbQ$.
My guess; reducing $f(x)inmathbbF_p[x]$ for suitable prime $p$, it might be irreducible over $mathbbF_p$, where $mathbbF_p$ is finite field of order $p$.
But, how to find the suitable prime $p$, and to show irreducibility over $mathbbF_p$?
The order of $f(x)$ is too much high for me ;(
Give some hint or advice! Thank you!
abstract-algebra irreducible-polynomials
asked Jul 19 at 8:21
Primavera
1838
I wish to show that the polynomial $f(x)=x^9+3x^6+165x^3+1inmathbbQ[x]$ is irreducible over $mathbbQ$.
My guess; reducing $f(x)inmathbbF_p[x]$ for suitable prime $p$, it might be irreducible over $mathbbF_p$, where $mathbbF_p$ is finite field of order $p$.
But, how to find the suitable prime $p$, and to show irreducibility over $mathbbF_p$?
The order of $f(x)$ is too much high for me ;(
Give some hint or advice! Thank you!
abstract-algebra irreducible-polynomials
asked Jul 19 at 8:21
Primavera
1838
asked Jul 19 at 8:21
Primavera
1838
asked Jul 19 at 8:21
Primavera
1838
asked Jul 19 at 8:21
Primavera
1838
1838
Given that $3mid 3$ and $3, 5, 11mid 165$, I would at least try $p = 3, 5$ or $11$ first, see if anything immediate pops up.
– Arthur
Jul 19 at 8:24
1
It is enough to show that $x^3+3x^2+165x+1$ is irreducible, and it is possible to obtain the roots analytically.
– Yves Daoust
Jul 19 at 8:31
1
@Yves Daoust, why is that enough? I mean, could you please elaborate on "it is possible to obtain the roots analytically"?
– Ennar
Jul 19 at 8:42
If the roots of the reduced polynomial aren't rational, neither are those of the original. (In fact it suffices that the roots of the reduced aren't perfect cubes, but this is not needed here.) wolframalpha.com/input/?i=solve+x%5E3%2B3x%5E2%2B165x%2B1%3D0
– Yves Daoust
Jul 19 at 9:15
7
@YvesDaoust Why would that be enough? There is no general result stating that if $p(x)$ is irreducible, so would $p(x^3)$. Consider $p(x)=x^2+1$. It is irreducible, but $p(x^3)=x^6+1=(x^2+1)(x^4-x^2+1)$.
– Jyrki Lahtonen
Jul 19 at 10:33
 |Â
show 2 more comments
Given that $3mid 3$ and $3, 5, 11mid 165$, I would at least try $p = 3, 5$ or $11$ first, see if anything immediate pops up.
– Arthur
Jul 19 at 8:24
1
It is enough to show that $x^3+3x^2+165x+1$ is irreducible, and it is possible to obtain the roots analytically.
– Yves Daoust
Jul 19 at 8:31
1
@Yves Daoust, why is that enough? I mean, could you please elaborate on "it is possible to obtain the roots analytically"?
– Ennar
Jul 19 at 8:42
If the roots of the reduced polynomial aren't rational, neither are those of the original. (In fact it suffices that the roots of the reduced aren't perfect cubes, but this is not needed here.) wolframalpha.com/input/?i=solve+x%5E3%2B3x%5E2%2B165x%2B1%3D0
– Yves Daoust
Jul 19 at 9:15
7
@YvesDaoust Why would that be enough? There is no general result stating that if $p(x)$ is irreducible, so would $p(x^3)$. Consider $p(x)=x^2+1$. It is irreducible, but $p(x^3)=x^6+1=(x^2+1)(x^4-x^2+1)$.
– Jyrki Lahtonen
Jul 19 at 10:33
Given that $3mid 3$ and $3, 5, 11mid 165$, I would at least try $p = 3, 5$ or $11$ first, see if anything immediate pops up.
– Arthur
Jul 19 at 8:24
Given that $3mid 3$ and $3, 5, 11mid 165$, I would at least try $p = 3, 5$ or $11$ first, see if anything immediate pops up.
– Arthur
Jul 19 at 8:24
1
1
It is enough to show that $x^3+3x^2+165x+1$ is irreducible, and it is possible to obtain the roots analytically.
– Yves Daoust
Jul 19 at 8:31
It is enough to show that $x^3+3x^2+165x+1$ is irreducible, and it is possible to obtain the roots analytically.
– Yves Daoust
Jul 19 at 8:31
1
1
@Yves Daoust, why is that enough? I mean, could you please elaborate on "it is possible to obtain the roots analytically"?
– Ennar
Jul 19 at 8:42
@Yves Daoust, why is that enough? I mean, could you please elaborate on "it is possible to obtain the roots analytically"?
– Ennar
Jul 19 at 8:42
If the roots of the reduced polynomial aren't rational, neither are those of the original. (In fact it suffices that the roots of the reduced aren't perfect cubes, but this is not needed here.) wolframalpha.com/input/?i=solve+x%5E3%2B3x%5E2%2B165x%2B1%3D0
– Yves Daoust
Jul 19 at 9:15
If the roots of the reduced polynomial aren't rational, neither are those of the original. (In fact it suffices that the roots of the reduced aren't perfect cubes, but this is not needed here.) wolframalpha.com/input/?i=solve+x%5E3%2B3x%5E2%2B165x%2B1%3D0
– Yves Daoust
Jul 19 at 9:15
7
7
@YvesDaoust Why would that be enough? There is no general result stating that if $p(x)$ is irreducible, so would $p(x^3)$. Consider $p(x)=x^2+1$. It is irreducible, but $p(x^3)=x^6+1=(x^2+1)(x^4-x^2+1)$.
– Jyrki Lahtonen
Jul 19 at 10:33
@YvesDaoust Why would that be enough? There is no general result stating that if $p(x)$ is irreducible, so would $p(x^3)$. Consider $p(x)=x^2+1$. It is irreducible, but $p(x^3)=x^6+1=(x^2+1)(x^4-x^2+1)$.
– Jyrki Lahtonen
Jul 19 at 10:33
 |Â
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
An argument somewhat inspired by pisco's nice answer. Trying to get away with less theory.
Consider the polynomial
$$g(x)=x^3+3x^2+165x+1.$$
Let's factor $g(x)equiv x^3+3x^2-4x+1$ modulo $p=13$. By regrouping we see that
$$
g(x)equiv x^3+3x^2-4x-12=(x^3-4x)+3(x^2-4)=(x-2)(x+2)(x+3).
$$
This means that modulo $13$
$$
f(x)equiv (x^3+2)(x^3-2)(x^3+3).qquad(*)
$$
Pisco already observed that $2$ is not a cubic residue modulo $13$. More generally, because $13-1=3cdot4$, the (non-zero) residue class of $a$ is a cubic residue modulo $13$ if and only if $a^4equiv1pmod13$. From this it follows easily that none of $pm2,-3$ is a cubic reside modulo $13$.
It follows that the three cubic factors in $(*)$ are all irreducible modulo $13$.
The rest is easy. If $f(x)$ factors over $BbbQ$, it must factor over $BbbZ$ (Gauss' Lemma), i.e. $f(x)=g(x)h(x)$ with $g(x),h(x)inBbbZ[x]$ both monic. Such a factorization survives reduction modulo $13$. Factorization over $BbbZ_13$ is unique, so we can conclude that
- $g(x)$ and $h(x)$ have degrees $3$ and $6$ (w.l.o.g. we can assume that $g(x)$ is a cubic), and
- $g(x)$ is congruent to one of the factors in $(*)$ modulo $13$.
- It follows that the constant term of $g(x)$ is thus congruent to one of $-3,-2,2$ modulo $13$.
This is a contradiction, because the constant term of $g(x)$ must be a factor of the constant term of $f(x)$, i.e. we should have $g(0)=pm1$.
Therefore $f(x)$ is irreducible.
Addendum: In case it interests someone I did a bit of testing with Mathematica. I collected data on the factorization types of $f(x)$ modulo $1800$ smallest primes not dividing the discriminant (so leaving out $2,3$ and $5$, i.e. multiple factors).
- Modulo 804 tested primes we get a factorization as a product of three cubics.
- Modulo 909 tested primes we get a factorization with a single linear and four quadratic factors.
- Modulo 87 tested primes we get a full factorization with nine linear factors.
In other words, the frequencies follow approximately the $8:9:1$ ratio. Keep in mind that the relative frequency of full splitting is $1/|G|$ and $9mid |G|$ by irreducibility. Here $G$ is the Galois group of this polynomial.
In light of Dedekind's theorem and Chebotarev's density theorem this strongly (= I will bet a beer saying this is the case) suggests to me that the Galois group of this nonic has order $18$ only. When viewed as a group of permutations of the nine zeros the group has $8$ elements that are products of three disjoint $3$-cycles, and $9$ elements that are products of four disjoint $2$-cycles. Looks an awful lot like $(C_3times C_3)rtimes C_2$ with conjugation by the generator of the $C_2$ factor inverting every element of the Sylow $3$-subgroup.
answered Jul 19 at 16:51
Jyrki Lahtonen
105k12161355
Thanks for kind answer! I could understand without difficulty :)
– Primavera
Jul 20 at 4:57
add a comment |Â
up vote
4
down vote
Let $alphain mathbbC$ be any root of $f(x)=x^9+3x^6+165x^3+1$. It is easily seen that $x^3+3x^2+165x+1$ is irreducible. We first show that $[mathbbQ(alpha):mathbbQ] > 3$.
Let $beta=alpha^3$, consider the ring of integers $mathcalO$ of $mathbbQ(beta)$, the discriminant of $x^3+3x^2+165x+1$ is $-2^2times 3^11times 5^2$, hence $13$ is relatively prime to the conductor $[mathcalO:mathbbZ[beta]]$. Since $13mid f(2)$, there exists a homomorphism:
$$mathcalOto mathbbF_13: beta mapsto 2$$
If $alphain mathbbQ(beta)$, then $2$ will be a cubic residue modulo $13$, which is obviously false. Hence $alpha notin mathbbQ(beta)$.
Therefore $$[mathbbQ(alpha) : mathbbQ(beta)] > 1 implies [mathbbQ(alpha) : mathbbQ] > 3$$
Therefore, we conclude that if $f$ is reducible, then $f$ must be a product of a quartic and quintic. This is easily ruled out, as $[mathbbQ(alpha) : mathbbQ]$ is a multiple of $3$.
answered Jul 19 at 12:33
pisco
9,93421335
What a nice approach. Thank you!
– Primavera
Jul 20 at 4:58
add a comment |Â
up vote
4
down vote
Here is a way using Newton polygons for $f(x-1)$. Specifically $$f(x-1)=x^9-9x^8+36x^7-81x^6+108x^5-81x^4+189x^3-486x^2+486x-162,$$ and looking at the powers of $3$ in its coefficients (the polygon coordinates depends only on these powers, we can ignore signs and other primes), we can write the coefficients as:
$$3^0, 3^2, 3^2, 3^4, 3^3, 3^4, 3^3, 3^5, 3^5, 3^4.$$
The Newton polygon corresponding to $p=3$ is then lower convex hull of points
$$[0,4],[1,5],[2,5],[3,3],[4,4],[5,3],[6,4],[7,2],[8,2],[9,0],$$
which can be easily seen to be line segment between $[0,4]$ and $[9,0]$, because all of the points lie above.
$ $
We also have $(4,9)=1$ and so this line does not pass through any other integer points. By Dumas' theorem (see for example this Variants of Eisenstein irreducibility, or also Advanced explanation on Eisenstein's criterion), the polynomial $f(x-1)$ is irreducible over $mathbbQ$, and so especially $f(x)$ is irreducible over $mathbbQ$.
answered Jul 20 at 6:15
Sil
5,19021342
Very nice. Have some reading up to do... good!
– WimC
Jul 22 at 15:02
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
An argument somewhat inspired by pisco's nice answer. Trying to get away with less theory.
Consider the polynomial
$$g(x)=x^3+3x^2+165x+1.$$
Let's factor $g(x)equiv x^3+3x^2-4x+1$ modulo $p=13$. By regrouping we see that
$$
g(x)equiv x^3+3x^2-4x-12=(x^3-4x)+3(x^2-4)=(x-2)(x+2)(x+3).
$$
This means that modulo $13$
$$
f(x)equiv (x^3+2)(x^3-2)(x^3+3).qquad(*)
$$
Pisco already observed that $2$ is not a cubic residue modulo $13$. More generally, because $13-1=3cdot4$, the (non-zero) residue class of $a$ is a cubic residue modulo $13$ if and only if $a^4equiv1pmod13$. From this it follows easily that none of $pm2,-3$ is a cubic reside modulo $13$.
It follows that the three cubic factors in $(*)$ are all irreducible modulo $13$.
The rest is easy. If $f(x)$ factors over $BbbQ$, it must factor over $BbbZ$ (Gauss' Lemma), i.e. $f(x)=g(x)h(x)$ with $g(x),h(x)inBbbZ[x]$ both monic. Such a factorization survives reduction modulo $13$. Factorization over $BbbZ_13$ is unique, so we can conclude that
- $g(x)$ and $h(x)$ have degrees $3$ and $6$ (w.l.o.g. we can assume that $g(x)$ is a cubic), and
- $g(x)$ is congruent to one of the factors in $(*)$ modulo $13$.
- It follows that the constant term of $g(x)$ is thus congruent to one of $-3,-2,2$ modulo $13$.
This is a contradiction, because the constant term of $g(x)$ must be a factor of the constant term of $f(x)$, i.e. we should have $g(0)=pm1$.
Therefore $f(x)$ is irreducible.
Addendum: In case it interests someone I did a bit of testing with Mathematica. I collected data on the factorization types of $f(x)$ modulo $1800$ smallest primes not dividing the discriminant (so leaving out $2,3$ and $5$, i.e. multiple factors).
- Modulo 804 tested primes we get a factorization as a product of three cubics.
- Modulo 909 tested primes we get a factorization with a single linear and four quadratic factors.
- Modulo 87 tested primes we get a full factorization with nine linear factors.
In other words, the frequencies follow approximately the $8:9:1$ ratio. Keep in mind that the relative frequency of full splitting is $1/|G|$ and $9mid |G|$ by irreducibility. Here $G$ is the Galois group of this polynomial.
In light of Dedekind's theorem and Chebotarev's density theorem this strongly (= I will bet a beer saying this is the case) suggests to me that the Galois group of this nonic has order $18$ only. When viewed as a group of permutations of the nine zeros the group has $8$ elements that are products of three disjoint $3$-cycles, and $9$ elements that are products of four disjoint $2$-cycles. Looks an awful lot like $(C_3times C_3)rtimes C_2$ with conjugation by the generator of the $C_2$ factor inverting every element of the Sylow $3$-subgroup.
answered Jul 19 at 16:51
Jyrki Lahtonen
105k12161355
Thanks for kind answer! I could understand without difficulty :)
– Primavera
Jul 20 at 4:57
add a comment |Â
up vote
5
down vote
accepted
An argument somewhat inspired by pisco's nice answer. Trying to get away with less theory.
Consider the polynomial
$$g(x)=x^3+3x^2+165x+1.$$
Let's factor $g(x)equiv x^3+3x^2-4x+1$ modulo $p=13$. By regrouping we see that
$$
g(x)equiv x^3+3x^2-4x-12=(x^3-4x)+3(x^2-4)=(x-2)(x+2)(x+3).
$$
This means that modulo $13$
$$
f(x)equiv (x^3+2)(x^3-2)(x^3+3).qquad(*)
$$
Pisco already observed that $2$ is not a cubic residue modulo $13$. More generally, because $13-1=3cdot4$, the (non-zero) residue class of $a$ is a cubic residue modulo $13$ if and only if $a^4equiv1pmod13$. From this it follows easily that none of $pm2,-3$ is a cubic reside modulo $13$.
It follows that the three cubic factors in $(*)$ are all irreducible modulo $13$.
The rest is easy. If $f(x)$ factors over $BbbQ$, it must factor over $BbbZ$ (Gauss' Lemma), i.e. $f(x)=g(x)h(x)$ with $g(x),h(x)inBbbZ[x]$ both monic. Such a factorization survives reduction modulo $13$. Factorization over $BbbZ_13$ is unique, so we can conclude that
- $g(x)$ and $h(x)$ have degrees $3$ and $6$ (w.l.o.g. we can assume that $g(x)$ is a cubic), and
- $g(x)$ is congruent to one of the factors in $(*)$ modulo $13$.
- It follows that the constant term of $g(x)$ is thus congruent to one of $-3,-2,2$ modulo $13$.
This is a contradiction, because the constant term of $g(x)$ must be a factor of the constant term of $f(x)$, i.e. we should have $g(0)=pm1$.
Therefore $f(x)$ is irreducible.
Addendum: In case it interests someone I did a bit of testing with Mathematica. I collected data on the factorization types of $f(x)$ modulo $1800$ smallest primes not dividing the discriminant (so leaving out $2,3$ and $5$, i.e. multiple factors).
- Modulo 804 tested primes we get a factorization as a product of three cubics.
- Modulo 909 tested primes we get a factorization with a single linear and four quadratic factors.
- Modulo 87 tested primes we get a full factorization with nine linear factors.
In other words, the frequencies follow approximately the $8:9:1$ ratio. Keep in mind that the relative frequency of full splitting is $1/|G|$ and $9mid |G|$ by irreducibility. Here $G$ is the Galois group of this polynomial.
In light of Dedekind's theorem and Chebotarev's density theorem this strongly (= I will bet a beer saying this is the case) suggests to me that the Galois group of this nonic has order $18$ only. When viewed as a group of permutations of the nine zeros the group has $8$ elements that are products of three disjoint $3$-cycles, and $9$ elements that are products of four disjoint $2$-cycles. Looks an awful lot like $(C_3times C_3)rtimes C_2$ with conjugation by the generator of the $C_2$ factor inverting every element of the Sylow $3$-subgroup.
answered Jul 19 at 16:51
Jyrki Lahtonen
105k12161355
Thanks for kind answer! I could understand without difficulty :)
– Primavera
Jul 20 at 4:57
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
An argument somewhat inspired by pisco's nice answer. Trying to get away with less theory.
Consider the polynomial
$$g(x)=x^3+3x^2+165x+1.$$
Let's factor $g(x)equiv x^3+3x^2-4x+1$ modulo $p=13$. By regrouping we see that
$$
g(x)equiv x^3+3x^2-4x-12=(x^3-4x)+3(x^2-4)=(x-2)(x+2)(x+3).
$$
This means that modulo $13$
$$
f(x)equiv (x^3+2)(x^3-2)(x^3+3).qquad(*)
$$
Pisco already observed that $2$ is not a cubic residue modulo $13$. More generally, because $13-1=3cdot4$, the (non-zero) residue class of $a$ is a cubic residue modulo $13$ if and only if $a^4equiv1pmod13$. From this it follows easily that none of $pm2,-3$ is a cubic reside modulo $13$.
It follows that the three cubic factors in $(*)$ are all irreducible modulo $13$.
The rest is easy. If $f(x)$ factors over $BbbQ$, it must factor over $BbbZ$ (Gauss' Lemma), i.e. $f(x)=g(x)h(x)$ with $g(x),h(x)inBbbZ[x]$ both monic. Such a factorization survives reduction modulo $13$. Factorization over $BbbZ_13$ is unique, so we can conclude that
- $g(x)$ and $h(x)$ have degrees $3$ and $6$ (w.l.o.g. we can assume that $g(x)$ is a cubic), and
- $g(x)$ is congruent to one of the factors in $(*)$ modulo $13$.
- It follows that the constant term of $g(x)$ is thus congruent to one of $-3,-2,2$ modulo $13$.
This is a contradiction, because the constant term of $g(x)$ must be a factor of the constant term of $f(x)$, i.e. we should have $g(0)=pm1$.
Therefore $f(x)$ is irreducible.
Addendum: In case it interests someone I did a bit of testing with Mathematica. I collected data on the factorization types of $f(x)$ modulo $1800$ smallest primes not dividing the discriminant (so leaving out $2,3$ and $5$, i.e. multiple factors).
- Modulo 804 tested primes we get a factorization as a product of three cubics.
- Modulo 909 tested primes we get a factorization with a single linear and four quadratic factors.
- Modulo 87 tested primes we get a full factorization with nine linear factors.
In other words, the frequencies follow approximately the $8:9:1$ ratio. Keep in mind that the relative frequency of full splitting is $1/|G|$ and $9mid |G|$ by irreducibility. Here $G$ is the Galois group of this polynomial.
In light of Dedekind's theorem and Chebotarev's density theorem this strongly (= I will bet a beer saying this is the case) suggests to me that the Galois group of this nonic has order $18$ only. When viewed as a group of permutations of the nine zeros the group has $8$ elements that are products of three disjoint $3$-cycles, and $9$ elements that are products of four disjoint $2$-cycles. Looks an awful lot like $(C_3times C_3)rtimes C_2$ with conjugation by the generator of the $C_2$ factor inverting every element of the Sylow $3$-subgroup.
answered Jul 19 at 16:51
Jyrki Lahtonen
105k12161355
An argument somewhat inspired by pisco's nice answer. Trying to get away with less theory.
Consider the polynomial
$$g(x)=x^3+3x^2+165x+1.$$
Let's factor $g(x)equiv x^3+3x^2-4x+1$ modulo $p=13$. By regrouping we see that
$$
g(x)equiv x^3+3x^2-4x-12=(x^3-4x)+3(x^2-4)=(x-2)(x+2)(x+3).
$$
This means that modulo $13$
$$
f(x)equiv (x^3+2)(x^3-2)(x^3+3).qquad(*)
$$
Pisco already observed that $2$ is not a cubic residue modulo $13$. More generally, because $13-1=3cdot4$, the (non-zero) residue class of $a$ is a cubic residue modulo $13$ if and only if $a^4equiv1pmod13$. From this it follows easily that none of $pm2,-3$ is a cubic reside modulo $13$.
It follows that the three cubic factors in $(*)$ are all irreducible modulo $13$.
The rest is easy. If $f(x)$ factors over $BbbQ$, it must factor over $BbbZ$ (Gauss' Lemma), i.e. $f(x)=g(x)h(x)$ with $g(x),h(x)inBbbZ[x]$ both monic. Such a factorization survives reduction modulo $13$. Factorization over $BbbZ_13$ is unique, so we can conclude that
- $g(x)$ and $h(x)$ have degrees $3$ and $6$ (w.l.o.g. we can assume that $g(x)$ is a cubic), and
- $g(x)$ is congruent to one of the factors in $(*)$ modulo $13$.
- It follows that the constant term of $g(x)$ is thus congruent to one of $-3,-2,2$ modulo $13$.
This is a contradiction, because the constant term of $g(x)$ must be a factor of the constant term of $f(x)$, i.e. we should have $g(0)=pm1$.
Therefore $f(x)$ is irreducible.
Addendum: In case it interests someone I did a bit of testing with Mathematica. I collected data on the factorization types of $f(x)$ modulo $1800$ smallest primes not dividing the discriminant (so leaving out $2,3$ and $5$, i.e. multiple factors).
- Modulo 804 tested primes we get a factorization as a product of three cubics.
- Modulo 909 tested primes we get a factorization with a single linear and four quadratic factors.
- Modulo 87 tested primes we get a full factorization with nine linear factors.
In other words, the frequencies follow approximately the $8:9:1$ ratio. Keep in mind that the relative frequency of full splitting is $1/|G|$ and $9mid |G|$ by irreducibility. Here $G$ is the Galois group of this polynomial.
In light of Dedekind's theorem and Chebotarev's density theorem this strongly (= I will bet a beer saying this is the case) suggests to me that the Galois group of this nonic has order $18$ only. When viewed as a group of permutations of the nine zeros the group has $8$ elements that are products of three disjoint $3$-cycles, and $9$ elements that are products of four disjoint $2$-cycles. Looks an awful lot like $(C_3times C_3)rtimes C_2$ with conjugation by the generator of the $C_2$ factor inverting every element of the Sylow $3$-subgroup.
answered Jul 19 at 16:51
Jyrki Lahtonen
105k12161355
edited Jul 20 at 5:31
answered Jul 19 at 16:51
Jyrki Lahtonen
105k12161355
answered Jul 19 at 16:51
Jyrki Lahtonen
105k12161355
answered Jul 19 at 16:51
Jyrki Lahtonen
105k12161355
105k12161355
Thanks for kind answer! I could understand without difficulty :)
– Primavera
Jul 20 at 4:57
add a comment |Â
Thanks for kind answer! I could understand without difficulty :)
– Primavera
Jul 20 at 4:57
Thanks for kind answer! I could understand without difficulty :)
– Primavera
Jul 20 at 4:57
Thanks for kind answer! I could understand without difficulty :)
– Primavera
Jul 20 at 4:57
add a comment |Â
up vote
4
down vote
Let $alphain mathbbC$ be any root of $f(x)=x^9+3x^6+165x^3+1$. It is easily seen that $x^3+3x^2+165x+1$ is irreducible. We first show that $[mathbbQ(alpha):mathbbQ] > 3$.
Let $beta=alpha^3$, consider the ring of integers $mathcalO$ of $mathbbQ(beta)$, the discriminant of $x^3+3x^2+165x+1$ is $-2^2times 3^11times 5^2$, hence $13$ is relatively prime to the conductor $[mathcalO:mathbbZ[beta]]$. Since $13mid f(2)$, there exists a homomorphism:
$$mathcalOto mathbbF_13: beta mapsto 2$$
If $alphain mathbbQ(beta)$, then $2$ will be a cubic residue modulo $13$, which is obviously false. Hence $alpha notin mathbbQ(beta)$.
Therefore $$[mathbbQ(alpha) : mathbbQ(beta)] > 1 implies [mathbbQ(alpha) : mathbbQ] > 3$$
Therefore, we conclude that if $f$ is reducible, then $f$ must be a product of a quartic and quintic. This is easily ruled out, as $[mathbbQ(alpha) : mathbbQ]$ is a multiple of $3$.
answered Jul 19 at 12:33
pisco
9,93421335
What a nice approach. Thank you!
– Primavera
Jul 20 at 4:58
add a comment |Â
up vote
4
down vote
Let $alphain mathbbC$ be any root of $f(x)=x^9+3x^6+165x^3+1$. It is easily seen that $x^3+3x^2+165x+1$ is irreducible. We first show that $[mathbbQ(alpha):mathbbQ] > 3$.
Let $beta=alpha^3$, consider the ring of integers $mathcalO$ of $mathbbQ(beta)$, the discriminant of $x^3+3x^2+165x+1$ is $-2^2times 3^11times 5^2$, hence $13$ is relatively prime to the conductor $[mathcalO:mathbbZ[beta]]$. Since $13mid f(2)$, there exists a homomorphism:
$$mathcalOto mathbbF_13: beta mapsto 2$$
If $alphain mathbbQ(beta)$, then $2$ will be a cubic residue modulo $13$, which is obviously false. Hence $alpha notin mathbbQ(beta)$.
Therefore $$[mathbbQ(alpha) : mathbbQ(beta)] > 1 implies [mathbbQ(alpha) : mathbbQ] > 3$$
Therefore, we conclude that if $f$ is reducible, then $f$ must be a product of a quartic and quintic. This is easily ruled out, as $[mathbbQ(alpha) : mathbbQ]$ is a multiple of $3$.
answered Jul 19 at 12:33
pisco
9,93421335
What a nice approach. Thank you!
– Primavera
Jul 20 at 4:58
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Let $alphain mathbbC$ be any root of $f(x)=x^9+3x^6+165x^3+1$. It is easily seen that $x^3+3x^2+165x+1$ is irreducible. We first show that $[mathbbQ(alpha):mathbbQ] > 3$.
Let $beta=alpha^3$, consider the ring of integers $mathcalO$ of $mathbbQ(beta)$, the discriminant of $x^3+3x^2+165x+1$ is $-2^2times 3^11times 5^2$, hence $13$ is relatively prime to the conductor $[mathcalO:mathbbZ[beta]]$. Since $13mid f(2)$, there exists a homomorphism:
$$mathcalOto mathbbF_13: beta mapsto 2$$
If $alphain mathbbQ(beta)$, then $2$ will be a cubic residue modulo $13$, which is obviously false. Hence $alpha notin mathbbQ(beta)$.
Therefore $$[mathbbQ(alpha) : mathbbQ(beta)] > 1 implies [mathbbQ(alpha) : mathbbQ] > 3$$
Therefore, we conclude that if $f$ is reducible, then $f$ must be a product of a quartic and quintic. This is easily ruled out, as $[mathbbQ(alpha) : mathbbQ]$ is a multiple of $3$.
answered Jul 19 at 12:33
pisco
9,93421335
Let $alphain mathbbC$ be any root of $f(x)=x^9+3x^6+165x^3+1$. It is easily seen that $x^3+3x^2+165x+1$ is irreducible. We first show that $[mathbbQ(alpha):mathbbQ] > 3$.
Let $beta=alpha^3$, consider the ring of integers $mathcalO$ of $mathbbQ(beta)$, the discriminant of $x^3+3x^2+165x+1$ is $-2^2times 3^11times 5^2$, hence $13$ is relatively prime to the conductor $[mathcalO:mathbbZ[beta]]$. Since $13mid f(2)$, there exists a homomorphism:
$$mathcalOto mathbbF_13: beta mapsto 2$$
If $alphain mathbbQ(beta)$, then $2$ will be a cubic residue modulo $13$, which is obviously false. Hence $alpha notin mathbbQ(beta)$.
Therefore $$[mathbbQ(alpha) : mathbbQ(beta)] > 1 implies [mathbbQ(alpha) : mathbbQ] > 3$$
Therefore, we conclude that if $f$ is reducible, then $f$ must be a product of a quartic and quintic. This is easily ruled out, as $[mathbbQ(alpha) : mathbbQ]$ is a multiple of $3$.
answered Jul 19 at 12:33
pisco
9,93421335
edited Jul 19 at 15:58
answered Jul 19 at 12:33
pisco
9,93421335
answered Jul 19 at 12:33
pisco
9,93421335
answered Jul 19 at 12:33
pisco
9,93421335
9,93421335
What a nice approach. Thank you!
– Primavera
Jul 20 at 4:58
add a comment |Â
What a nice approach. Thank you!
– Primavera
Jul 20 at 4:58
What a nice approach. Thank you!
– Primavera
Jul 20 at 4:58
What a nice approach. Thank you!
– Primavera
Jul 20 at 4:58
add a comment |Â
up vote
4
down vote
Here is a way using Newton polygons for $f(x-1)$. Specifically $$f(x-1)=x^9-9x^8+36x^7-81x^6+108x^5-81x^4+189x^3-486x^2+486x-162,$$ and looking at the powers of $3$ in its coefficients (the polygon coordinates depends only on these powers, we can ignore signs and other primes), we can write the coefficients as:
$$3^0, 3^2, 3^2, 3^4, 3^3, 3^4, 3^3, 3^5, 3^5, 3^4.$$
The Newton polygon corresponding to $p=3$ is then lower convex hull of points
$$[0,4],[1,5],[2,5],[3,3],[4,4],[5,3],[6,4],[7,2],[8,2],[9,0],$$
which can be easily seen to be line segment between $[0,4]$ and $[9,0]$, because all of the points lie above.
$ $
We also have $(4,9)=1$ and so this line does not pass through any other integer points. By Dumas' theorem (see for example this Variants of Eisenstein irreducibility, or also Advanced explanation on Eisenstein's criterion), the polynomial $f(x-1)$ is irreducible over $mathbbQ$, and so especially $f(x)$ is irreducible over $mathbbQ$.
answered Jul 20 at 6:15
Sil
5,19021342
Very nice. Have some reading up to do... good!
– WimC
Jul 22 at 15:02
add a comment |Â
up vote
4
down vote
Here is a way using Newton polygons for $f(x-1)$. Specifically $$f(x-1)=x^9-9x^8+36x^7-81x^6+108x^5-81x^4+189x^3-486x^2+486x-162,$$ and looking at the powers of $3$ in its coefficients (the polygon coordinates depends only on these powers, we can ignore signs and other primes), we can write the coefficients as:
$$3^0, 3^2, 3^2, 3^4, 3^3, 3^4, 3^3, 3^5, 3^5, 3^4.$$
The Newton polygon corresponding to $p=3$ is then lower convex hull of points
$$[0,4],[1,5],[2,5],[3,3],[4,4],[5,3],[6,4],[7,2],[8,2],[9,0],$$
which can be easily seen to be line segment between $[0,4]$ and $[9,0]$, because all of the points lie above.
$ $
We also have $(4,9)=1$ and so this line does not pass through any other integer points. By Dumas' theorem (see for example this Variants of Eisenstein irreducibility, or also Advanced explanation on Eisenstein's criterion), the polynomial $f(x-1)$ is irreducible over $mathbbQ$, and so especially $f(x)$ is irreducible over $mathbbQ$.
answered Jul 20 at 6:15
Sil
5,19021342
Very nice. Have some reading up to do... good!
– WimC
Jul 22 at 15:02
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Here is a way using Newton polygons for $f(x-1)$. Specifically $$f(x-1)=x^9-9x^8+36x^7-81x^6+108x^5-81x^4+189x^3-486x^2+486x-162,$$ and looking at the powers of $3$ in its coefficients (the polygon coordinates depends only on these powers, we can ignore signs and other primes), we can write the coefficients as:
$$3^0, 3^2, 3^2, 3^4, 3^3, 3^4, 3^3, 3^5, 3^5, 3^4.$$
The Newton polygon corresponding to $p=3$ is then lower convex hull of points
$$[0,4],[1,5],[2,5],[3,3],[4,4],[5,3],[6,4],[7,2],[8,2],[9,0],$$
which can be easily seen to be line segment between $[0,4]$ and $[9,0]$, because all of the points lie above.
$ $
We also have $(4,9)=1$ and so this line does not pass through any other integer points. By Dumas' theorem (see for example this Variants of Eisenstein irreducibility, or also Advanced explanation on Eisenstein's criterion), the polynomial $f(x-1)$ is irreducible over $mathbbQ$, and so especially $f(x)$ is irreducible over $mathbbQ$.
answered Jul 20 at 6:15
Sil
5,19021342
Here is a way using Newton polygons for $f(x-1)$. Specifically $$f(x-1)=x^9-9x^8+36x^7-81x^6+108x^5-81x^4+189x^3-486x^2+486x-162,$$ and looking at the powers of $3$ in its coefficients (the polygon coordinates depends only on these powers, we can ignore signs and other primes), we can write the coefficients as:
$$3^0, 3^2, 3^2, 3^4, 3^3, 3^4, 3^3, 3^5, 3^5, 3^4.$$
The Newton polygon corresponding to $p=3$ is then lower convex hull of points
$$[0,4],[1,5],[2,5],[3,3],[4,4],[5,3],[6,4],[7,2],[8,2],[9,0],$$
which can be easily seen to be line segment between $[0,4]$ and $[9,0]$, because all of the points lie above.
$ $
We also have $(4,9)=1$ and so this line does not pass through any other integer points. By Dumas' theorem (see for example this Variants of Eisenstein irreducibility, or also Advanced explanation on Eisenstein's criterion), the polynomial $f(x-1)$ is irreducible over $mathbbQ$, and so especially $f(x)$ is irreducible over $mathbbQ$.
answered Jul 20 at 6:15
Sil
5,19021342
edited Jul 21 at 12:14
answered Jul 20 at 6:15
Sil
5,19021342
answered Jul 20 at 6:15
Sil
5,19021342
answered Jul 20 at 6:15
Sil
5,19021342
5,19021342
Very nice. Have some reading up to do... good!
– WimC
Jul 22 at 15:02
add a comment |Â
Very nice. Have some reading up to do... good!
– WimC
Jul 22 at 15:02
Very nice. Have some reading up to do... good!
– WimC
Jul 22 at 15:02
Very nice. Have some reading up to do... good!
– WimC
Jul 22 at 15:02
add a comment |Â
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Given that $3mid 3$ and $3, 5, 11mid 165$, I would at least try $p = 3, 5$ or $11$ first, see if anything immediate pops up.
– Arthur
Jul 19 at 8:24
1
It is enough to show that $x^3+3x^2+165x+1$ is irreducible, and it is possible to obtain the roots analytically.
– Yves Daoust
Jul 19 at 8:31
1
@Yves Daoust, why is that enough? I mean, could you please elaborate on "it is possible to obtain the roots analytically"?
– Ennar
Jul 19 at 8:42
If the roots of the reduced polynomial aren't rational, neither are those of the original. (In fact it suffices that the roots of the reduced aren't perfect cubes, but this is not needed here.) wolframalpha.com/input/?i=solve+x%5E3%2B3x%5E2%2B165x%2B1%3D0
– Yves Daoust
Jul 19 at 9:15
7
@YvesDaoust Why would that be enough? There is no general result stating that if $p(x)$ is irreducible, so would $p(x^3)$. Consider $p(x)=x^2+1$. It is irreducible, but $p(x^3)=x^6+1=(x^2+1)(x^4-x^2+1)$.
– Jyrki Lahtonen
Jul 19 at 10:33