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To show: Galois group of a polynomial is not an alternating group .
Clash Royale CLAN TAG#URR8PPP
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I have a polynomial of this type: $p(t)+l(t)s(a,t)$ in $F_q(a)[t]$, where a=(a_0,...,a_m)$ are specialization.
I could show that this polynomial is doubly transitive, by making $a_0=0$ (specialization technique for irreducibility and assumptions on f and g).
Now I need to show that the discriminant of this polynomial is not a square. Here I do not know how to proceed. Could any one please help me
polynomials galois-theory discriminant
asked Jul 24 at 3:49
user152394
957
 |Â
show 1 more comment
up vote
0
down vote
favorite
I have a polynomial of this type: $p(t)+l(t)s(a,t)$ in $F_q(a)[t]$, where a=(a_0,...,a_m)$ are specialization.
I could show that this polynomial is doubly transitive, by making $a_0=0$ (specialization technique for irreducibility and assumptions on f and g).
Now I need to show that the discriminant of this polynomial is not a square. Here I do not know how to proceed. Could any one please help me
polynomials galois-theory discriminant
asked Jul 24 at 3:49
user152394
957
1
It may be my limitation and lack of exposure to specializations, but why couldn't you get an alternating group here? Say, if $p(t)$ is a cubic, irreducible over $BbbF_q$, and $ell(t)=0$. Then the splitting field is $BbbF_q^3(a)$ and the Galois group is cyclic of order three, i.e. $A_3$. It is also easy make $A_2$ appear - make your polynomial a reducible quadratic. I know I'm nitpicking, but...
– Jyrki Lahtonen
Jul 24 at 4:21
so your polynomial can be any polynomial in $Bbb F_q(a)[t]$ ?
– mercio
Jul 24 at 7:32
@JyrkiLahtonen, I thought the polynomial $p(t)+l(t)s(a,t)$, is linear in $F(a,t)$. (By taking this polynomial as a polynomial in $a_0$,), hence irreducible,
– user152394
Jul 24 at 10:43
@JyrkiLahtonen, the author further assumes p(t) and l(t) are relatively prime polynomials. and the doubly irreducibility is proved similarly
– user152394
Jul 24 at 10:51
@JyrkiLahtonen, is there a general method to show the discriminant is not a square
– user152394
Jul 24 at 10:53
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a polynomial of this type: $p(t)+l(t)s(a,t)$ in $F_q(a)[t]$, where a=(a_0,...,a_m)$ are specialization.
I could show that this polynomial is doubly transitive, by making $a_0=0$ (specialization technique for irreducibility and assumptions on f and g).
Now I need to show that the discriminant of this polynomial is not a square. Here I do not know how to proceed. Could any one please help me
polynomials galois-theory discriminant
asked Jul 24 at 3:49
user152394
957
I have a polynomial of this type: $p(t)+l(t)s(a,t)$ in $F_q(a)[t]$, where a=(a_0,...,a_m)$ are specialization.
I could show that this polynomial is doubly transitive, by making $a_0=0$ (specialization technique for irreducibility and assumptions on f and g).
Now I need to show that the discriminant of this polynomial is not a square. Here I do not know how to proceed. Could any one please help me
polynomials galois-theory discriminant
asked Jul 24 at 3:49
user152394
957
asked Jul 24 at 3:49
user152394
957
asked Jul 24 at 3:49
user152394
957
asked Jul 24 at 3:49
user152394
957
957
1
It may be my limitation and lack of exposure to specializations, but why couldn't you get an alternating group here? Say, if $p(t)$ is a cubic, irreducible over $BbbF_q$, and $ell(t)=0$. Then the splitting field is $BbbF_q^3(a)$ and the Galois group is cyclic of order three, i.e. $A_3$. It is also easy make $A_2$ appear - make your polynomial a reducible quadratic. I know I'm nitpicking, but...
– Jyrki Lahtonen
Jul 24 at 4:21
so your polynomial can be any polynomial in $Bbb F_q(a)[t]$ ?
– mercio
Jul 24 at 7:32
@JyrkiLahtonen, I thought the polynomial $p(t)+l(t)s(a,t)$, is linear in $F(a,t)$. (By taking this polynomial as a polynomial in $a_0$,), hence irreducible,
– user152394
Jul 24 at 10:43
@JyrkiLahtonen, the author further assumes p(t) and l(t) are relatively prime polynomials. and the doubly irreducibility is proved similarly
– user152394
Jul 24 at 10:51
@JyrkiLahtonen, is there a general method to show the discriminant is not a square
– user152394
Jul 24 at 10:53
 |Â
show 1 more comment
1
It may be my limitation and lack of exposure to specializations, but why couldn't you get an alternating group here? Say, if $p(t)$ is a cubic, irreducible over $BbbF_q$, and $ell(t)=0$. Then the splitting field is $BbbF_q^3(a)$ and the Galois group is cyclic of order three, i.e. $A_3$. It is also easy make $A_2$ appear - make your polynomial a reducible quadratic. I know I'm nitpicking, but...
– Jyrki Lahtonen
Jul 24 at 4:21
so your polynomial can be any polynomial in $Bbb F_q(a)[t]$ ?
– mercio
Jul 24 at 7:32
@JyrkiLahtonen, I thought the polynomial $p(t)+l(t)s(a,t)$, is linear in $F(a,t)$. (By taking this polynomial as a polynomial in $a_0$,), hence irreducible,
– user152394
Jul 24 at 10:43
@JyrkiLahtonen, the author further assumes p(t) and l(t) are relatively prime polynomials. and the doubly irreducibility is proved similarly
– user152394
Jul 24 at 10:51
@JyrkiLahtonen, is there a general method to show the discriminant is not a square
– user152394
Jul 24 at 10:53
1
1
It may be my limitation and lack of exposure to specializations, but why couldn't you get an alternating group here? Say, if $p(t)$ is a cubic, irreducible over $BbbF_q$, and $ell(t)=0$. Then the splitting field is $BbbF_q^3(a)$ and the Galois group is cyclic of order three, i.e. $A_3$. It is also easy make $A_2$ appear - make your polynomial a reducible quadratic. I know I'm nitpicking, but...
– Jyrki Lahtonen
Jul 24 at 4:21
It may be my limitation and lack of exposure to specializations, but why couldn't you get an alternating group here? Say, if $p(t)$ is a cubic, irreducible over $BbbF_q$, and $ell(t)=0$. Then the splitting field is $BbbF_q^3(a)$ and the Galois group is cyclic of order three, i.e. $A_3$. It is also easy make $A_2$ appear - make your polynomial a reducible quadratic. I know I'm nitpicking, but...
– Jyrki Lahtonen
Jul 24 at 4:21
so your polynomial can be any polynomial in $Bbb F_q(a)[t]$ ?
– mercio
Jul 24 at 7:32
so your polynomial can be any polynomial in $Bbb F_q(a)[t]$ ?
– mercio
Jul 24 at 7:32
@JyrkiLahtonen, I thought the polynomial $p(t)+l(t)s(a,t)$, is linear in $F(a,t)$. (By taking this polynomial as a polynomial in $a_0$,), hence irreducible,
– user152394
Jul 24 at 10:43
@JyrkiLahtonen, I thought the polynomial $p(t)+l(t)s(a,t)$, is linear in $F(a,t)$. (By taking this polynomial as a polynomial in $a_0$,), hence irreducible,
– user152394
Jul 24 at 10:43
@JyrkiLahtonen, the author further assumes p(t) and l(t) are relatively prime polynomials. and the doubly irreducibility is proved similarly
– user152394
Jul 24 at 10:51
@JyrkiLahtonen, the author further assumes p(t) and l(t) are relatively prime polynomials. and the doubly irreducibility is proved similarly
– user152394
Jul 24 at 10:51
@JyrkiLahtonen, is there a general method to show the discriminant is not a square
– user152394
Jul 24 at 10:53
@JyrkiLahtonen, is there a general method to show the discriminant is not a square
– user152394
Jul 24 at 10:53
 |Â
show 1 more comment
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1
It may be my limitation and lack of exposure to specializations, but why couldn't you get an alternating group here? Say, if $p(t)$ is a cubic, irreducible over $BbbF_q$, and $ell(t)=0$. Then the splitting field is $BbbF_q^3(a)$ and the Galois group is cyclic of order three, i.e. $A_3$. It is also easy make $A_2$ appear - make your polynomial a reducible quadratic. I know I'm nitpicking, but...
– Jyrki Lahtonen
Jul 24 at 4:21
so your polynomial can be any polynomial in $Bbb F_q(a)[t]$ ?
– mercio
Jul 24 at 7:32
@JyrkiLahtonen, I thought the polynomial $p(t)+l(t)s(a,t)$, is linear in $F(a,t)$. (By taking this polynomial as a polynomial in $a_0$,), hence irreducible,
– user152394
Jul 24 at 10:43
@JyrkiLahtonen, the author further assumes p(t) and l(t) are relatively prime polynomials. and the doubly irreducibility is proved similarly
– user152394
Jul 24 at 10:51
@JyrkiLahtonen, is there a general method to show the discriminant is not a square
– user152394
Jul 24 at 10:53