Mixing
Constructing a polynomial given the Galois Group of it's splitting field
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Let $red_p : mathbb{Z}[x]tomathbb{Z}/(p)[x]$ be the canonical ring morphism sending a polynomial with integer coefficients to a polynomial with integer coefficients modulo $p$, with $p$ a prime, by taking modulo to each coefficient.
My objective is to find a polynomial $finmathbb{Z}[x]$ of degree $8$ such that $mathrm{Gal}(F/mathbb{Q})cong S_8$, where $F$ denotes the splitting field of $f$ over $mathbb{Q}$, and such that $red_7(f)$ is irreducible in $mathbb{Z}/(7)[x]$. I don't even know how to begin. Any help would be greatly appreciated.
I've made some attemps to construct an 8 degree polynomial such that $mathrm{Gal}(F/mathbb{Q})cong S_8$ but I can't seem to solve that problem either. I think that would be a great starting point.
polynomials ring-theory field-theory galois-theory automorphism-group
asked yesterday
Ray Bern
1109
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up vote
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Let $red_p : mathbb{Z}[x]tomathbb{Z}/(p)[x]$ be the canonical ring morphism sending a polynomial with integer coefficients to a polynomial with integer coefficients modulo $p$, with $p$ a prime, by taking modulo to each coefficient.
My objective is to find a polynomial $finmathbb{Z}[x]$ of degree $8$ such that $mathrm{Gal}(F/mathbb{Q})cong S_8$, where $F$ denotes the splitting field of $f$ over $mathbb{Q}$, and such that $red_7(f)$ is irreducible in $mathbb{Z}/(7)[x]$. I don't even know how to begin. Any help would be greatly appreciated.
I've made some attemps to construct an 8 degree polynomial such that $mathrm{Gal}(F/mathbb{Q})cong S_8$ but I can't seem to solve that problem either. I think that would be a great starting point.
polynomials ring-theory field-theory galois-theory automorphism-group
asked yesterday
Ray Bern
1109
Replace $S_8$ by $S_3$. Take $f in mathbb{Z}[x], deg(f) = 3$ with $f in mathbb{Z}/(p)[x] $ irreducible. Thus $f in mathbb{Z}[x]$ is irreducible and $R = mathbb{Z}[x]/(f(x))$ is an integral domain. Then let $g(y) = frac{f(y)}{y-x} in R[y]$ and assume $g(y) in R/(q)[y]$ is irreducible (note we can change $gbmod q$ without changing $f bmod p$), thus $g(y) in R[y]$ is irreducible and $R[y]/(g(y))$ is an integral domain. Whence $F = Frac(R[y]/(g(y)))$ is the splitting field of $f(x) in mathbb{Q}[x]$ and $[F:mathbb{Q}] = 6$ implies $Gal(F/mathbb{Q}) = S_3$.
– reuns
yesterday
2
Are you familiar with Dedekind's theorem on the relation between irreducible factors of $f$ mod primes and cycle types of the elements of $operatorname{Gal}(f)$? This would suggest to start with an irreducible polynomial in $Bbb{Z}/7Bbb{Z}[X]$ and lift it to $Bbb{Z}[X]$; this will give you a Galois group with a $7$-cycle. Then it does not take much to get all of $S_8$, e.g. a transposition will do.
– Servaes
yesterday
*In the above it should say $8$-cycle in stead of $7$-cycle.
– Servaes
yesterday
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $red_p : mathbb{Z}[x]tomathbb{Z}/(p)[x]$ be the canonical ring morphism sending a polynomial with integer coefficients to a polynomial with integer coefficients modulo $p$, with $p$ a prime, by taking modulo to each coefficient.
My objective is to find a polynomial $finmathbb{Z}[x]$ of degree $8$ such that $mathrm{Gal}(F/mathbb{Q})cong S_8$, where $F$ denotes the splitting field of $f$ over $mathbb{Q}$, and such that $red_7(f)$ is irreducible in $mathbb{Z}/(7)[x]$. I don't even know how to begin. Any help would be greatly appreciated.
I've made some attemps to construct an 8 degree polynomial such that $mathrm{Gal}(F/mathbb{Q})cong S_8$ but I can't seem to solve that problem either. I think that would be a great starting point.
polynomials ring-theory field-theory galois-theory automorphism-group
asked yesterday
Ray Bern
1109
Let $red_p : mathbb{Z}[x]tomathbb{Z}/(p)[x]$ be the canonical ring morphism sending a polynomial with integer coefficients to a polynomial with integer coefficients modulo $p$, with $p$ a prime, by taking modulo to each coefficient.
My objective is to find a polynomial $finmathbb{Z}[x]$ of degree $8$ such that $mathrm{Gal}(F/mathbb{Q})cong S_8$, where $F$ denotes the splitting field of $f$ over $mathbb{Q}$, and such that $red_7(f)$ is irreducible in $mathbb{Z}/(7)[x]$. I don't even know how to begin. Any help would be greatly appreciated.
I've made some attemps to construct an 8 degree polynomial such that $mathrm{Gal}(F/mathbb{Q})cong S_8$ but I can't seem to solve that problem either. I think that would be a great starting point.
polynomials ring-theory field-theory galois-theory automorphism-group
polynomials ring-theory field-theory galois-theory automorphism-group
asked yesterday
Ray Bern
1109
asked yesterday
Ray Bern
1109
asked yesterday
Ray Bern
1109
asked yesterday
Ray Bern
1109
asked yesterday
Ray Bern
1109
1109
Replace $S_8$ by $S_3$. Take $f in mathbb{Z}[x], deg(f) = 3$ with $f in mathbb{Z}/(p)[x] $ irreducible. Thus $f in mathbb{Z}[x]$ is irreducible and $R = mathbb{Z}[x]/(f(x))$ is an integral domain. Then let $g(y) = frac{f(y)}{y-x} in R[y]$ and assume $g(y) in R/(q)[y]$ is irreducible (note we can change $gbmod q$ without changing $f bmod p$), thus $g(y) in R[y]$ is irreducible and $R[y]/(g(y))$ is an integral domain. Whence $F = Frac(R[y]/(g(y)))$ is the splitting field of $f(x) in mathbb{Q}[x]$ and $[F:mathbb{Q}] = 6$ implies $Gal(F/mathbb{Q}) = S_3$.
– reuns
yesterday
2
Are you familiar with Dedekind's theorem on the relation between irreducible factors of $f$ mod primes and cycle types of the elements of $operatorname{Gal}(f)$? This would suggest to start with an irreducible polynomial in $Bbb{Z}/7Bbb{Z}[X]$ and lift it to $Bbb{Z}[X]$; this will give you a Galois group with a $7$-cycle. Then it does not take much to get all of $S_8$, e.g. a transposition will do.
– Servaes
yesterday
*In the above it should say $8$-cycle in stead of $7$-cycle.
– Servaes
yesterday
add a comment |
Replace $S_8$ by $S_3$. Take $f in mathbb{Z}[x], deg(f) = 3$ with $f in mathbb{Z}/(p)[x] $ irreducible. Thus $f in mathbb{Z}[x]$ is irreducible and $R = mathbb{Z}[x]/(f(x))$ is an integral domain. Then let $g(y) = frac{f(y)}{y-x} in R[y]$ and assume $g(y) in R/(q)[y]$ is irreducible (note we can change $gbmod q$ without changing $f bmod p$), thus $g(y) in R[y]$ is irreducible and $R[y]/(g(y))$ is an integral domain. Whence $F = Frac(R[y]/(g(y)))$ is the splitting field of $f(x) in mathbb{Q}[x]$ and $[F:mathbb{Q}] = 6$ implies $Gal(F/mathbb{Q}) = S_3$.
– reuns
yesterday
2
Are you familiar with Dedekind's theorem on the relation between irreducible factors of $f$ mod primes and cycle types of the elements of $operatorname{Gal}(f)$? This would suggest to start with an irreducible polynomial in $Bbb{Z}/7Bbb{Z}[X]$ and lift it to $Bbb{Z}[X]$; this will give you a Galois group with a $7$-cycle. Then it does not take much to get all of $S_8$, e.g. a transposition will do.
– Servaes
yesterday
*In the above it should say $8$-cycle in stead of $7$-cycle.
– Servaes
yesterday
Replace $S_8$ by $S_3$. Take $f in mathbb{Z}[x], deg(f) = 3$ with $f in mathbb{Z}/(p)[x] $ irreducible. Thus $f in mathbb{Z}[x]$ is irreducible and $R = mathbb{Z}[x]/(f(x))$ is an integral domain. Then let $g(y) = frac{f(y)}{y-x} in R[y]$ and assume $g(y) in R/(q)[y]$ is irreducible (note we can change $gbmod q$ without changing $f bmod p$), thus $g(y) in R[y]$ is irreducible and $R[y]/(g(y))$ is an integral domain. Whence $F = Frac(R[y]/(g(y)))$ is the splitting field of $f(x) in mathbb{Q}[x]$ and $[F:mathbb{Q}] = 6$ implies $Gal(F/mathbb{Q}) = S_3$.
– reuns
yesterday
Replace $S_8$ by $S_3$. Take $f in mathbb{Z}[x], deg(f) = 3$ with $f in mathbb{Z}/(p)[x] $ irreducible. Thus $f in mathbb{Z}[x]$ is irreducible and $R = mathbb{Z}[x]/(f(x))$ is an integral domain. Then let $g(y) = frac{f(y)}{y-x} in R[y]$ and assume $g(y) in R/(q)[y]$ is irreducible (note we can change $gbmod q$ without changing $f bmod p$), thus $g(y) in R[y]$ is irreducible and $R[y]/(g(y))$ is an integral domain. Whence $F = Frac(R[y]/(g(y)))$ is the splitting field of $f(x) in mathbb{Q}[x]$ and $[F:mathbb{Q}] = 6$ implies $Gal(F/mathbb{Q}) = S_3$.
– reuns
yesterday
2
2
Are you familiar with Dedekind's theorem on the relation between irreducible factors of $f$ mod primes and cycle types of the elements of $operatorname{Gal}(f)$? This would suggest to start with an irreducible polynomial in $Bbb{Z}/7Bbb{Z}[X]$ and lift it to $Bbb{Z}[X]$; this will give you a Galois group with a $7$-cycle. Then it does not take much to get all of $S_8$, e.g. a transposition will do.
– Servaes
yesterday
Are you familiar with Dedekind's theorem on the relation between irreducible factors of $f$ mod primes and cycle types of the elements of $operatorname{Gal}(f)$? This would suggest to start with an irreducible polynomial in $Bbb{Z}/7Bbb{Z}[X]$ and lift it to $Bbb{Z}[X]$; this will give you a Galois group with a $7$-cycle. Then it does not take much to get all of $S_8$, e.g. a transposition will do.
– Servaes
yesterday
*In the above it should say $8$-cycle in stead of $7$-cycle.
– Servaes
yesterday
*In the above it should say $8$-cycle in stead of $7$-cycle.
– Servaes
yesterday
add a comment |
2 Answers
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up vote
3
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This is not a complete answer, just a start as you indicate that you don't even know where to begin.
Let $finBbb{Z}[x]$ monic. If its image $f_p$ in $Bbb{F}_p[x]$ is separable and factors as $f_p=prod_{i=1}^kg_k$, then $operatorname{Gal}(f)$ contains an element of cycle type $(deg g_1,ldots,deg g_k)$. So it makes sense to start from an irreducibe polynomial $hinBbb{F}_7[x]$ as then any lift $tilde{h}inBbb{Z}[x]$ already has an element of order $8$ in $operatorname{Gal}(tilde{h})$. An easy first candidate is
$$h_7=x^8+x+3inBbb{F}_7[x].$$
To make sure we also have a transpotion in $operatorname{Gal}(f)$, we choose a lift $tilde{h}inBbb{Z}[x]$ that factors into one quadratic and six linear factors mod $2$. A bit of fiddling around yields, for example
$$tilde{h}=x^8+x+3+7(x^7+x^6+x+1)inBbb{Z}[x],$$
so that
$$h_2=x^8+x^7+x^6=x^6(x^2+x+1)inBbb{F}_2[x].$$
which shows that $operatorname{Gal}(tilde{h})$ contains a transposition. This doesn't quite give you that $operatorname{Gal}(tilde{h})cong S_8$, but gets you on the right track.
EDIT: As pointed out in the comments below $h_2$ is not separable, so this choice of $tilde{h}$ doesn't quite work. I have no doubt the argument can be salvaged, but the result will likely not be as pretty. I'll give it some thought tomorrow.
UPDATE: One way to salvage the argument is to take a larger prime $p$, so that a lift $tilde{h}$ splits into six distinct linear factors and one irreducible quadratic factor. As $degtilde{h}=8$ this requires $pgeq6$, hence $p=11$ is the smallest prime that might work. And indeed, surprisingly little fiddling around shows that the lift
$$tilde{h}=x^8+x+3+7(x+8)big((x+3)(x+4)+6(x+1)(x+2)(x+5)(x+10)big)inBbb{Z}[x],$$
satisfies
$$h_p=(x+1)(x+2)(x+3)(x+5)(x+8)(x+10)(x^2+4x+5)inBbb{F}_p[x],$$
which shows that $operatorname{Gal}(tilde{h})$ contains a transposition.
answered yesterday
Servaes
20.6k33789
How did you find $h_7$?
– Ray Bern
yesterday
An educated guess; there is quite often a $cinBbb{F}_p$ such that $x^n+x+c$ is irreducible.
– Servaes
yesterday
A good technique. But I'm afraid I need to complain about $h_2(x)$ not being separable in $Bbb{F}_2[x]$.
– Jyrki Lahtonen
yesterday
@Jyrki You're absolutely right. It seems the argument needs a bigger prime, the computations will likely not be so clean. I'll give it some thought tomorrow.
– Servaes
yesterday
Many ways to use this idea I think. One way is to specify the factors modulo any finite collection of primes, then do a Chinese remainder theorem combination of the products. And only then lift from $Bbb{Z}_{p_1p_2cdots p_k}$ to $Bbb{Z}$. As you pointed out the end result may not look nice :-)
– Jyrki Lahtonen
yesterday
add a comment |
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0
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This example was produced in the first place by randomly generating polynomials and checking the factorizations with computer help (it only took a modest number of polynomials before finding a satisfactory one), so one might (reasonably) object that this answer isn't actually a "construction". But any method is going to have to show somehow that $operatorname{red}_7 f$ is irreducible, and doing this naively is labor-intensive, as here are $588 + 112 + 24 + 7 = 631$ irreducible polynomials over $Bbb F_7$ of degree $1 leq d leq 4$. If one has a faster way of generating irreducible polynomials over $Bbb F_7$ of degree $8$ that we can then factor over $Bbb F_2, Bbb F_3$ (which is much faster to do manually, see below), one might be able to optimize considerably here.
Like Servaes' approach, the method here uses Dedekind's Theorem to show that $operatorname{Gal}(F / Bbb Q)$ contains certain cycle types. In particular, we'll use that a transitive subgroup of $S_n$ that contains an $(n - 1)$-cycle and a transposition is $S_n$ itself.
Take
$$f(x) := x^8 + 4 x^7 + 3 x^6 + 3 x^5 + 3 x^4 + 5 x^3 + x^2 + 4 x + 5 .$$
Factoring $operatorname{red}_p f$ over $Bbb F_p$ for the below $p$ gives:
$operatorname{red}_7 f$ is irreducible over $f$, so $f$ satisfies the given hypothesis and by Dedekind's Theorem $operatorname{Gal}(F / Bbb Q)$ acts transitively on the roots of $f$.
$operatorname{red}_{2} f = p_3 hat p_3 p_2$ for irreducible (and distinct) polynomials of respective degrees $3, 3, 2$. Again by Dedekind's Theorem $operatorname{Gal}(F / Bbb Q)$ contains a product $sigma$ of cycle type $(3, 3, 2)$, so $sigma^3 in operatorname{Gal}(F / Bbb Q)$ is a transposition.
$operatorname{red}_{3} f = q_7 q_1$ for irreducible polynomials $q_d$, so Dedekind's Theorem this time gives us that $operatorname{Gal}(F / Bbb Q)$ contains a $7$-cycle.
After checking the irreducibility of $operatorname{red}_7 f$ as discussed above, the most intensive verification is checking that $q_7$ is irreducible over $Bbb F_3$, but there are only $8 + 3 + 3 = 14$ irreducible polynomials of degree $1 leq d leq 3$ irreducible over $Bbb F_3$.
answered 14 hours ago
Travis
58.7k765142
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
This is not a complete answer, just a start as you indicate that you don't even know where to begin.
Let $finBbb{Z}[x]$ monic. If its image $f_p$ in $Bbb{F}_p[x]$ is separable and factors as $f_p=prod_{i=1}^kg_k$, then $operatorname{Gal}(f)$ contains an element of cycle type $(deg g_1,ldots,deg g_k)$. So it makes sense to start from an irreducibe polynomial $hinBbb{F}_7[x]$ as then any lift $tilde{h}inBbb{Z}[x]$ already has an element of order $8$ in $operatorname{Gal}(tilde{h})$. An easy first candidate is
$$h_7=x^8+x+3inBbb{F}_7[x].$$
To make sure we also have a transpotion in $operatorname{Gal}(f)$, we choose a lift $tilde{h}inBbb{Z}[x]$ that factors into one quadratic and six linear factors mod $2$. A bit of fiddling around yields, for example
$$tilde{h}=x^8+x+3+7(x^7+x^6+x+1)inBbb{Z}[x],$$
so that
$$h_2=x^8+x^7+x^6=x^6(x^2+x+1)inBbb{F}_2[x].$$
which shows that $operatorname{Gal}(tilde{h})$ contains a transposition. This doesn't quite give you that $operatorname{Gal}(tilde{h})cong S_8$, but gets you on the right track.
EDIT: As pointed out in the comments below $h_2$ is not separable, so this choice of $tilde{h}$ doesn't quite work. I have no doubt the argument can be salvaged, but the result will likely not be as pretty. I'll give it some thought tomorrow.
UPDATE: One way to salvage the argument is to take a larger prime $p$, so that a lift $tilde{h}$ splits into six distinct linear factors and one irreducible quadratic factor. As $degtilde{h}=8$ this requires $pgeq6$, hence $p=11$ is the smallest prime that might work. And indeed, surprisingly little fiddling around shows that the lift
$$tilde{h}=x^8+x+3+7(x+8)big((x+3)(x+4)+6(x+1)(x+2)(x+5)(x+10)big)inBbb{Z}[x],$$
satisfies
$$h_p=(x+1)(x+2)(x+3)(x+5)(x+8)(x+10)(x^2+4x+5)inBbb{F}_p[x],$$
which shows that $operatorname{Gal}(tilde{h})$ contains a transposition.
answered yesterday
Servaes
20.6k33789
How did you find $h_7$?
– Ray Bern
yesterday
An educated guess; there is quite often a $cinBbb{F}_p$ such that $x^n+x+c$ is irreducible.
– Servaes
yesterday
A good technique. But I'm afraid I need to complain about $h_2(x)$ not being separable in $Bbb{F}_2[x]$.
– Jyrki Lahtonen
yesterday
@Jyrki You're absolutely right. It seems the argument needs a bigger prime, the computations will likely not be so clean. I'll give it some thought tomorrow.
– Servaes
yesterday
Many ways to use this idea I think. One way is to specify the factors modulo any finite collection of primes, then do a Chinese remainder theorem combination of the products. And only then lift from $Bbb{Z}_{p_1p_2cdots p_k}$ to $Bbb{Z}$. As you pointed out the end result may not look nice :-)
– Jyrki Lahtonen
yesterday
add a comment |
up vote
3
down vote
This is not a complete answer, just a start as you indicate that you don't even know where to begin.
Let $finBbb{Z}[x]$ monic. If its image $f_p$ in $Bbb{F}_p[x]$ is separable and factors as $f_p=prod_{i=1}^kg_k$, then $operatorname{Gal}(f)$ contains an element of cycle type $(deg g_1,ldots,deg g_k)$. So it makes sense to start from an irreducibe polynomial $hinBbb{F}_7[x]$ as then any lift $tilde{h}inBbb{Z}[x]$ already has an element of order $8$ in $operatorname{Gal}(tilde{h})$. An easy first candidate is
$$h_7=x^8+x+3inBbb{F}_7[x].$$
To make sure we also have a transpotion in $operatorname{Gal}(f)$, we choose a lift $tilde{h}inBbb{Z}[x]$ that factors into one quadratic and six linear factors mod $2$. A bit of fiddling around yields, for example
$$tilde{h}=x^8+x+3+7(x^7+x^6+x+1)inBbb{Z}[x],$$
so that
$$h_2=x^8+x^7+x^6=x^6(x^2+x+1)inBbb{F}_2[x].$$
which shows that $operatorname{Gal}(tilde{h})$ contains a transposition. This doesn't quite give you that $operatorname{Gal}(tilde{h})cong S_8$, but gets you on the right track.
EDIT: As pointed out in the comments below $h_2$ is not separable, so this choice of $tilde{h}$ doesn't quite work. I have no doubt the argument can be salvaged, but the result will likely not be as pretty. I'll give it some thought tomorrow.
UPDATE: One way to salvage the argument is to take a larger prime $p$, so that a lift $tilde{h}$ splits into six distinct linear factors and one irreducible quadratic factor. As $degtilde{h}=8$ this requires $pgeq6$, hence $p=11$ is the smallest prime that might work. And indeed, surprisingly little fiddling around shows that the lift
$$tilde{h}=x^8+x+3+7(x+8)big((x+3)(x+4)+6(x+1)(x+2)(x+5)(x+10)big)inBbb{Z}[x],$$
satisfies
$$h_p=(x+1)(x+2)(x+3)(x+5)(x+8)(x+10)(x^2+4x+5)inBbb{F}_p[x],$$
which shows that $operatorname{Gal}(tilde{h})$ contains a transposition.
answered yesterday
Servaes
20.6k33789
How did you find $h_7$?
– Ray Bern
yesterday
An educated guess; there is quite often a $cinBbb{F}_p$ such that $x^n+x+c$ is irreducible.
– Servaes
yesterday
A good technique. But I'm afraid I need to complain about $h_2(x)$ not being separable in $Bbb{F}_2[x]$.
– Jyrki Lahtonen
yesterday
@Jyrki You're absolutely right. It seems the argument needs a bigger prime, the computations will likely not be so clean. I'll give it some thought tomorrow.
– Servaes
yesterday
Many ways to use this idea I think. One way is to specify the factors modulo any finite collection of primes, then do a Chinese remainder theorem combination of the products. And only then lift from $Bbb{Z}_{p_1p_2cdots p_k}$ to $Bbb{Z}$. As you pointed out the end result may not look nice :-)
– Jyrki Lahtonen
yesterday
add a comment |
up vote
3
down vote
up vote
3
down vote
This is not a complete answer, just a start as you indicate that you don't even know where to begin.
Let $finBbb{Z}[x]$ monic. If its image $f_p$ in $Bbb{F}_p[x]$ is separable and factors as $f_p=prod_{i=1}^kg_k$, then $operatorname{Gal}(f)$ contains an element of cycle type $(deg g_1,ldots,deg g_k)$. So it makes sense to start from an irreducibe polynomial $hinBbb{F}_7[x]$ as then any lift $tilde{h}inBbb{Z}[x]$ already has an element of order $8$ in $operatorname{Gal}(tilde{h})$. An easy first candidate is
$$h_7=x^8+x+3inBbb{F}_7[x].$$
To make sure we also have a transpotion in $operatorname{Gal}(f)$, we choose a lift $tilde{h}inBbb{Z}[x]$ that factors into one quadratic and six linear factors mod $2$. A bit of fiddling around yields, for example
$$tilde{h}=x^8+x+3+7(x^7+x^6+x+1)inBbb{Z}[x],$$
so that
$$h_2=x^8+x^7+x^6=x^6(x^2+x+1)inBbb{F}_2[x].$$
which shows that $operatorname{Gal}(tilde{h})$ contains a transposition. This doesn't quite give you that $operatorname{Gal}(tilde{h})cong S_8$, but gets you on the right track.
EDIT: As pointed out in the comments below $h_2$ is not separable, so this choice of $tilde{h}$ doesn't quite work. I have no doubt the argument can be salvaged, but the result will likely not be as pretty. I'll give it some thought tomorrow.
UPDATE: One way to salvage the argument is to take a larger prime $p$, so that a lift $tilde{h}$ splits into six distinct linear factors and one irreducible quadratic factor. As $degtilde{h}=8$ this requires $pgeq6$, hence $p=11$ is the smallest prime that might work. And indeed, surprisingly little fiddling around shows that the lift
$$tilde{h}=x^8+x+3+7(x+8)big((x+3)(x+4)+6(x+1)(x+2)(x+5)(x+10)big)inBbb{Z}[x],$$
satisfies
$$h_p=(x+1)(x+2)(x+3)(x+5)(x+8)(x+10)(x^2+4x+5)inBbb{F}_p[x],$$
which shows that $operatorname{Gal}(tilde{h})$ contains a transposition.
answered yesterday
Servaes
20.6k33789
This is not a complete answer, just a start as you indicate that you don't even know where to begin.
Let $finBbb{Z}[x]$ monic. If its image $f_p$ in $Bbb{F}_p[x]$ is separable and factors as $f_p=prod_{i=1}^kg_k$, then $operatorname{Gal}(f)$ contains an element of cycle type $(deg g_1,ldots,deg g_k)$. So it makes sense to start from an irreducibe polynomial $hinBbb{F}_7[x]$ as then any lift $tilde{h}inBbb{Z}[x]$ already has an element of order $8$ in $operatorname{Gal}(tilde{h})$. An easy first candidate is
$$h_7=x^8+x+3inBbb{F}_7[x].$$
To make sure we also have a transpotion in $operatorname{Gal}(f)$, we choose a lift $tilde{h}inBbb{Z}[x]$ that factors into one quadratic and six linear factors mod $2$. A bit of fiddling around yields, for example
$$tilde{h}=x^8+x+3+7(x^7+x^6+x+1)inBbb{Z}[x],$$
so that
$$h_2=x^8+x^7+x^6=x^6(x^2+x+1)inBbb{F}_2[x].$$
which shows that $operatorname{Gal}(tilde{h})$ contains a transposition. This doesn't quite give you that $operatorname{Gal}(tilde{h})cong S_8$, but gets you on the right track.
EDIT: As pointed out in the comments below $h_2$ is not separable, so this choice of $tilde{h}$ doesn't quite work. I have no doubt the argument can be salvaged, but the result will likely not be as pretty. I'll give it some thought tomorrow.
UPDATE: One way to salvage the argument is to take a larger prime $p$, so that a lift $tilde{h}$ splits into six distinct linear factors and one irreducible quadratic factor. As $degtilde{h}=8$ this requires $pgeq6$, hence $p=11$ is the smallest prime that might work. And indeed, surprisingly little fiddling around shows that the lift
$$tilde{h}=x^8+x+3+7(x+8)big((x+3)(x+4)+6(x+1)(x+2)(x+5)(x+10)big)inBbb{Z}[x],$$
satisfies
$$h_p=(x+1)(x+2)(x+3)(x+5)(x+8)(x+10)(x^2+4x+5)inBbb{F}_p[x],$$
which shows that $operatorname{Gal}(tilde{h})$ contains a transposition.
answered yesterday
Servaes
20.6k33789
edited 17 hours ago
answered yesterday
Servaes
20.6k33789
answered yesterday
Servaes
20.6k33789
answered yesterday
Servaes
20.6k33789
20.6k33789
How did you find $h_7$?
– Ray Bern
yesterday
An educated guess; there is quite often a $cinBbb{F}_p$ such that $x^n+x+c$ is irreducible.
– Servaes
yesterday
A good technique. But I'm afraid I need to complain about $h_2(x)$ not being separable in $Bbb{F}_2[x]$.
– Jyrki Lahtonen
yesterday
@Jyrki You're absolutely right. It seems the argument needs a bigger prime, the computations will likely not be so clean. I'll give it some thought tomorrow.
– Servaes
yesterday
Many ways to use this idea I think. One way is to specify the factors modulo any finite collection of primes, then do a Chinese remainder theorem combination of the products. And only then lift from $Bbb{Z}_{p_1p_2cdots p_k}$ to $Bbb{Z}$. As you pointed out the end result may not look nice :-)
– Jyrki Lahtonen
yesterday
add a comment |
How did you find $h_7$?
– Ray Bern
yesterday
An educated guess; there is quite often a $cinBbb{F}_p$ such that $x^n+x+c$ is irreducible.
– Servaes
yesterday
A good technique. But I'm afraid I need to complain about $h_2(x)$ not being separable in $Bbb{F}_2[x]$.
– Jyrki Lahtonen
yesterday
@Jyrki You're absolutely right. It seems the argument needs a bigger prime, the computations will likely not be so clean. I'll give it some thought tomorrow.
– Servaes
yesterday
Many ways to use this idea I think. One way is to specify the factors modulo any finite collection of primes, then do a Chinese remainder theorem combination of the products. And only then lift from $Bbb{Z}_{p_1p_2cdots p_k}$ to $Bbb{Z}$. As you pointed out the end result may not look nice :-)
– Jyrki Lahtonen
yesterday
How did you find $h_7$?
– Ray Bern
yesterday
How did you find $h_7$?
– Ray Bern
yesterday
An educated guess; there is quite often a $cinBbb{F}_p$ such that $x^n+x+c$ is irreducible.
– Servaes
yesterday
An educated guess; there is quite often a $cinBbb{F}_p$ such that $x^n+x+c$ is irreducible.
– Servaes
yesterday
A good technique. But I'm afraid I need to complain about $h_2(x)$ not being separable in $Bbb{F}_2[x]$.
– Jyrki Lahtonen
yesterday
A good technique. But I'm afraid I need to complain about $h_2(x)$ not being separable in $Bbb{F}_2[x]$.
– Jyrki Lahtonen
yesterday
@Jyrki You're absolutely right. It seems the argument needs a bigger prime, the computations will likely not be so clean. I'll give it some thought tomorrow.
– Servaes
yesterday
@Jyrki You're absolutely right. It seems the argument needs a bigger prime, the computations will likely not be so clean. I'll give it some thought tomorrow.
– Servaes
yesterday
Many ways to use this idea I think. One way is to specify the factors modulo any finite collection of primes, then do a Chinese remainder theorem combination of the products. And only then lift from $Bbb{Z}_{p_1p_2cdots p_k}$ to $Bbb{Z}$. As you pointed out the end result may not look nice :-)
– Jyrki Lahtonen
yesterday
Many ways to use this idea I think. One way is to specify the factors modulo any finite collection of primes, then do a Chinese remainder theorem combination of the products. And only then lift from $Bbb{Z}_{p_1p_2cdots p_k}$ to $Bbb{Z}$. As you pointed out the end result may not look nice :-)
– Jyrki Lahtonen
yesterday
add a comment |
up vote
0
down vote
This example was produced in the first place by randomly generating polynomials and checking the factorizations with computer help (it only took a modest number of polynomials before finding a satisfactory one), so one might (reasonably) object that this answer isn't actually a "construction". But any method is going to have to show somehow that $operatorname{red}_7 f$ is irreducible, and doing this naively is labor-intensive, as here are $588 + 112 + 24 + 7 = 631$ irreducible polynomials over $Bbb F_7$ of degree $1 leq d leq 4$. If one has a faster way of generating irreducible polynomials over $Bbb F_7$ of degree $8$ that we can then factor over $Bbb F_2, Bbb F_3$ (which is much faster to do manually, see below), one might be able to optimize considerably here.
Like Servaes' approach, the method here uses Dedekind's Theorem to show that $operatorname{Gal}(F / Bbb Q)$ contains certain cycle types. In particular, we'll use that a transitive subgroup of $S_n$ that contains an $(n - 1)$-cycle and a transposition is $S_n$ itself.
Take
$$f(x) := x^8 + 4 x^7 + 3 x^6 + 3 x^5 + 3 x^4 + 5 x^3 + x^2 + 4 x + 5 .$$
Factoring $operatorname{red}_p f$ over $Bbb F_p$ for the below $p$ gives:
$operatorname{red}_7 f$ is irreducible over $f$, so $f$ satisfies the given hypothesis and by Dedekind's Theorem $operatorname{Gal}(F / Bbb Q)$ acts transitively on the roots of $f$.
$operatorname{red}_{2} f = p_3 hat p_3 p_2$ for irreducible (and distinct) polynomials of respective degrees $3, 3, 2$. Again by Dedekind's Theorem $operatorname{Gal}(F / Bbb Q)$ contains a product $sigma$ of cycle type $(3, 3, 2)$, so $sigma^3 in operatorname{Gal}(F / Bbb Q)$ is a transposition.
$operatorname{red}_{3} f = q_7 q_1$ for irreducible polynomials $q_d$, so Dedekind's Theorem this time gives us that $operatorname{Gal}(F / Bbb Q)$ contains a $7$-cycle.
After checking the irreducibility of $operatorname{red}_7 f$ as discussed above, the most intensive verification is checking that $q_7$ is irreducible over $Bbb F_3$, but there are only $8 + 3 + 3 = 14$ irreducible polynomials of degree $1 leq d leq 3$ irreducible over $Bbb F_3$.
answered 14 hours ago
Travis
58.7k765142
add a comment |
up vote
0
down vote
This example was produced in the first place by randomly generating polynomials and checking the factorizations with computer help (it only took a modest number of polynomials before finding a satisfactory one), so one might (reasonably) object that this answer isn't actually a "construction". But any method is going to have to show somehow that $operatorname{red}_7 f$ is irreducible, and doing this naively is labor-intensive, as here are $588 + 112 + 24 + 7 = 631$ irreducible polynomials over $Bbb F_7$ of degree $1 leq d leq 4$. If one has a faster way of generating irreducible polynomials over $Bbb F_7$ of degree $8$ that we can then factor over $Bbb F_2, Bbb F_3$ (which is much faster to do manually, see below), one might be able to optimize considerably here.
Like Servaes' approach, the method here uses Dedekind's Theorem to show that $operatorname{Gal}(F / Bbb Q)$ contains certain cycle types. In particular, we'll use that a transitive subgroup of $S_n$ that contains an $(n - 1)$-cycle and a transposition is $S_n$ itself.
Take
$$f(x) := x^8 + 4 x^7 + 3 x^6 + 3 x^5 + 3 x^4 + 5 x^3 + x^2 + 4 x + 5 .$$
Factoring $operatorname{red}_p f$ over $Bbb F_p$ for the below $p$ gives:
$operatorname{red}_7 f$ is irreducible over $f$, so $f$ satisfies the given hypothesis and by Dedekind's Theorem $operatorname{Gal}(F / Bbb Q)$ acts transitively on the roots of $f$.
$operatorname{red}_{2} f = p_3 hat p_3 p_2$ for irreducible (and distinct) polynomials of respective degrees $3, 3, 2$. Again by Dedekind's Theorem $operatorname{Gal}(F / Bbb Q)$ contains a product $sigma$ of cycle type $(3, 3, 2)$, so $sigma^3 in operatorname{Gal}(F / Bbb Q)$ is a transposition.
$operatorname{red}_{3} f = q_7 q_1$ for irreducible polynomials $q_d$, so Dedekind's Theorem this time gives us that $operatorname{Gal}(F / Bbb Q)$ contains a $7$-cycle.
After checking the irreducibility of $operatorname{red}_7 f$ as discussed above, the most intensive verification is checking that $q_7$ is irreducible over $Bbb F_3$, but there are only $8 + 3 + 3 = 14$ irreducible polynomials of degree $1 leq d leq 3$ irreducible over $Bbb F_3$.
answered 14 hours ago
Travis
58.7k765142
add a comment |
up vote
0
down vote
up vote
0
down vote
This example was produced in the first place by randomly generating polynomials and checking the factorizations with computer help (it only took a modest number of polynomials before finding a satisfactory one), so one might (reasonably) object that this answer isn't actually a "construction". But any method is going to have to show somehow that $operatorname{red}_7 f$ is irreducible, and doing this naively is labor-intensive, as here are $588 + 112 + 24 + 7 = 631$ irreducible polynomials over $Bbb F_7$ of degree $1 leq d leq 4$. If one has a faster way of generating irreducible polynomials over $Bbb F_7$ of degree $8$ that we can then factor over $Bbb F_2, Bbb F_3$ (which is much faster to do manually, see below), one might be able to optimize considerably here.
Like Servaes' approach, the method here uses Dedekind's Theorem to show that $operatorname{Gal}(F / Bbb Q)$ contains certain cycle types. In particular, we'll use that a transitive subgroup of $S_n$ that contains an $(n - 1)$-cycle and a transposition is $S_n$ itself.
Take
$$f(x) := x^8 + 4 x^7 + 3 x^6 + 3 x^5 + 3 x^4 + 5 x^3 + x^2 + 4 x + 5 .$$
Factoring $operatorname{red}_p f$ over $Bbb F_p$ for the below $p$ gives:
$operatorname{red}_7 f$ is irreducible over $f$, so $f$ satisfies the given hypothesis and by Dedekind's Theorem $operatorname{Gal}(F / Bbb Q)$ acts transitively on the roots of $f$.
$operatorname{red}_{2} f = p_3 hat p_3 p_2$ for irreducible (and distinct) polynomials of respective degrees $3, 3, 2$. Again by Dedekind's Theorem $operatorname{Gal}(F / Bbb Q)$ contains a product $sigma$ of cycle type $(3, 3, 2)$, so $sigma^3 in operatorname{Gal}(F / Bbb Q)$ is a transposition.
$operatorname{red}_{3} f = q_7 q_1$ for irreducible polynomials $q_d$, so Dedekind's Theorem this time gives us that $operatorname{Gal}(F / Bbb Q)$ contains a $7$-cycle.
After checking the irreducibility of $operatorname{red}_7 f$ as discussed above, the most intensive verification is checking that $q_7$ is irreducible over $Bbb F_3$, but there are only $8 + 3 + 3 = 14$ irreducible polynomials of degree $1 leq d leq 3$ irreducible over $Bbb F_3$.
answered 14 hours ago
Travis
58.7k765142
This example was produced in the first place by randomly generating polynomials and checking the factorizations with computer help (it only took a modest number of polynomials before finding a satisfactory one), so one might (reasonably) object that this answer isn't actually a "construction". But any method is going to have to show somehow that $operatorname{red}_7 f$ is irreducible, and doing this naively is labor-intensive, as here are $588 + 112 + 24 + 7 = 631$ irreducible polynomials over $Bbb F_7$ of degree $1 leq d leq 4$. If one has a faster way of generating irreducible polynomials over $Bbb F_7$ of degree $8$ that we can then factor over $Bbb F_2, Bbb F_3$ (which is much faster to do manually, see below), one might be able to optimize considerably here.
Like Servaes' approach, the method here uses Dedekind's Theorem to show that $operatorname{Gal}(F / Bbb Q)$ contains certain cycle types. In particular, we'll use that a transitive subgroup of $S_n$ that contains an $(n - 1)$-cycle and a transposition is $S_n$ itself.
Take
$$f(x) := x^8 + 4 x^7 + 3 x^6 + 3 x^5 + 3 x^4 + 5 x^3 + x^2 + 4 x + 5 .$$
Factoring $operatorname{red}_p f$ over $Bbb F_p$ for the below $p$ gives:
$operatorname{red}_7 f$ is irreducible over $f$, so $f$ satisfies the given hypothesis and by Dedekind's Theorem $operatorname{Gal}(F / Bbb Q)$ acts transitively on the roots of $f$.
$operatorname{red}_{2} f = p_3 hat p_3 p_2$ for irreducible (and distinct) polynomials of respective degrees $3, 3, 2$. Again by Dedekind's Theorem $operatorname{Gal}(F / Bbb Q)$ contains a product $sigma$ of cycle type $(3, 3, 2)$, so $sigma^3 in operatorname{Gal}(F / Bbb Q)$ is a transposition.
$operatorname{red}_{3} f = q_7 q_1$ for irreducible polynomials $q_d$, so Dedekind's Theorem this time gives us that $operatorname{Gal}(F / Bbb Q)$ contains a $7$-cycle.
After checking the irreducibility of $operatorname{red}_7 f$ as discussed above, the most intensive verification is checking that $q_7$ is irreducible over $Bbb F_3$, but there are only $8 + 3 + 3 = 14$ irreducible polynomials of degree $1 leq d leq 3$ irreducible over $Bbb F_3$.
answered 14 hours ago
Travis
58.7k765142
answered 14 hours ago
Travis
58.7k765142
answered 14 hours ago
Travis
58.7k765142
answered 14 hours ago
Travis
58.7k765142
58.7k765142
add a comment |
add a comment |
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Replace $S_8$ by $S_3$. Take $f in mathbb{Z}[x], deg(f) = 3$ with $f in mathbb{Z}/(p)[x] $ irreducible. Thus $f in mathbb{Z}[x]$ is irreducible and $R = mathbb{Z}[x]/(f(x))$ is an integral domain. Then let $g(y) = frac{f(y)}{y-x} in R[y]$ and assume $g(y) in R/(q)[y]$ is irreducible (note we can change $gbmod q$ without changing $f bmod p$), thus $g(y) in R[y]$ is irreducible and $R[y]/(g(y))$ is an integral domain. Whence $F = Frac(R[y]/(g(y)))$ is the splitting field of $f(x) in mathbb{Q}[x]$ and $[F:mathbb{Q}] = 6$ implies $Gal(F/mathbb{Q}) = S_3$.
– reuns
yesterday
2
Are you familiar with Dedekind's theorem on the relation between irreducible factors of $f$ mod primes and cycle types of the elements of $operatorname{Gal}(f)$? This would suggest to start with an irreducible polynomial in $Bbb{Z}/7Bbb{Z}[X]$ and lift it to $Bbb{Z}[X]$; this will give you a Galois group with a $7$-cycle. Then it does not take much to get all of $S_8$, e.g. a transposition will do.
– Servaes
yesterday
*In the above it should say $8$-cycle in stead of $7$-cycle.
– Servaes
yesterday