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Galois group is isomorphic to $S_5$?
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Let f be an irreducible polynomial of degree $5$ in $mathbbQ[x]$. Suppose that in $mathbbC$, $f$ has exactly two nonreal roots. Then the Galois group of the splitting field of $f$ is isomorphic to $S_5$.
My effort: Let $G$ be the Galois group. The complex conjugation map $mathbbsigma (x+iy)=x-iy$ is non-trivial element of $G$ of order $2$. Let $alpha_1,alpha_2, alpha_3$ be the real roots of $f(x)$ and $beta_1$ and $beta_2$ be the nonreal roots. Then, $sigma$ sends $beta_1$ to $beta_2$ while keeping other roots fixed. Now if we can construct an element of order $5$ in $G$, then we can proceed in the direction of showing that $G cong S_5$.
I know that any such element $tau$ of $G$ (if exists) will permute all the roots of $f$. But I am not sure which permutation of roots of $f$ will give me the correct candidate for the element $tau$?
Thanks!
galois-theory splitting-field
asked Jul 14 at 16:40
Shubham Namdeo
331212
add a comment |Â
up vote
0
down vote
favorite
Let f be an irreducible polynomial of degree $5$ in $mathbbQ[x]$. Suppose that in $mathbbC$, $f$ has exactly two nonreal roots. Then the Galois group of the splitting field of $f$ is isomorphic to $S_5$.
My effort: Let $G$ be the Galois group. The complex conjugation map $mathbbsigma (x+iy)=x-iy$ is non-trivial element of $G$ of order $2$. Let $alpha_1,alpha_2, alpha_3$ be the real roots of $f(x)$ and $beta_1$ and $beta_2$ be the nonreal roots. Then, $sigma$ sends $beta_1$ to $beta_2$ while keeping other roots fixed. Now if we can construct an element of order $5$ in $G$, then we can proceed in the direction of showing that $G cong S_5$.
I know that any such element $tau$ of $G$ (if exists) will permute all the roots of $f$. But I am not sure which permutation of roots of $f$ will give me the correct candidate for the element $tau$?
Thanks!
galois-theory splitting-field
asked Jul 14 at 16:40
Shubham Namdeo
331212
1
What's the question?
– Lord Shark the Unknown
Jul 14 at 16:41
1
Demanding others solve a problem rather than answer an actual question makes it look like this is a homework problem. In fact it is a standard type of problem for students learning Galois theory.
– KCd
Jul 14 at 16:48
@KCd please have a look. I have put my efforts along with question.
– Shubham Namdeo
Jul 14 at 16:57
The Galois group of an irreducible polynomial over a field of characteristic zero will act transitively on its zeros.
– Lord Shark the Unknown
Jul 14 at 17:01
@LordSharktheUnknown after that how to proceed?
– Shubham Namdeo
Jul 14 at 17:04
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let f be an irreducible polynomial of degree $5$ in $mathbbQ[x]$. Suppose that in $mathbbC$, $f$ has exactly two nonreal roots. Then the Galois group of the splitting field of $f$ is isomorphic to $S_5$.
My effort: Let $G$ be the Galois group. The complex conjugation map $mathbbsigma (x+iy)=x-iy$ is non-trivial element of $G$ of order $2$. Let $alpha_1,alpha_2, alpha_3$ be the real roots of $f(x)$ and $beta_1$ and $beta_2$ be the nonreal roots. Then, $sigma$ sends $beta_1$ to $beta_2$ while keeping other roots fixed. Now if we can construct an element of order $5$ in $G$, then we can proceed in the direction of showing that $G cong S_5$.
I know that any such element $tau$ of $G$ (if exists) will permute all the roots of $f$. But I am not sure which permutation of roots of $f$ will give me the correct candidate for the element $tau$?
Thanks!
galois-theory splitting-field
asked Jul 14 at 16:40
Shubham Namdeo
331212
Let f be an irreducible polynomial of degree $5$ in $mathbbQ[x]$. Suppose that in $mathbbC$, $f$ has exactly two nonreal roots. Then the Galois group of the splitting field of $f$ is isomorphic to $S_5$.
My effort: Let $G$ be the Galois group. The complex conjugation map $mathbbsigma (x+iy)=x-iy$ is non-trivial element of $G$ of order $2$. Let $alpha_1,alpha_2, alpha_3$ be the real roots of $f(x)$ and $beta_1$ and $beta_2$ be the nonreal roots. Then, $sigma$ sends $beta_1$ to $beta_2$ while keeping other roots fixed. Now if we can construct an element of order $5$ in $G$, then we can proceed in the direction of showing that $G cong S_5$.
I know that any such element $tau$ of $G$ (if exists) will permute all the roots of $f$. But I am not sure which permutation of roots of $f$ will give me the correct candidate for the element $tau$?
Thanks!
galois-theory splitting-field
asked Jul 14 at 16:40
Shubham Namdeo
331212
edited Jul 14 at 16:52
asked Jul 14 at 16:40
Shubham Namdeo
331212
asked Jul 14 at 16:40
Shubham Namdeo
331212
asked Jul 14 at 16:40
Shubham Namdeo
331212
331212
1
What's the question?
– Lord Shark the Unknown
Jul 14 at 16:41
1
Demanding others solve a problem rather than answer an actual question makes it look like this is a homework problem. In fact it is a standard type of problem for students learning Galois theory.
– KCd
Jul 14 at 16:48
@KCd please have a look. I have put my efforts along with question.
– Shubham Namdeo
Jul 14 at 16:57
The Galois group of an irreducible polynomial over a field of characteristic zero will act transitively on its zeros.
– Lord Shark the Unknown
Jul 14 at 17:01
@LordSharktheUnknown after that how to proceed?
– Shubham Namdeo
Jul 14 at 17:04
add a comment |Â
1
What's the question?
– Lord Shark the Unknown
Jul 14 at 16:41
1
Demanding others solve a problem rather than answer an actual question makes it look like this is a homework problem. In fact it is a standard type of problem for students learning Galois theory.
– KCd
Jul 14 at 16:48
@KCd please have a look. I have put my efforts along with question.
– Shubham Namdeo
Jul 14 at 16:57
The Galois group of an irreducible polynomial over a field of characteristic zero will act transitively on its zeros.
– Lord Shark the Unknown
Jul 14 at 17:01
@LordSharktheUnknown after that how to proceed?
– Shubham Namdeo
Jul 14 at 17:04
1
1
What's the question?
– Lord Shark the Unknown
Jul 14 at 16:41
What's the question?
– Lord Shark the Unknown
Jul 14 at 16:41
1
1
Demanding others solve a problem rather than answer an actual question makes it look like this is a homework problem. In fact it is a standard type of problem for students learning Galois theory.
– KCd
Jul 14 at 16:48
Demanding others solve a problem rather than answer an actual question makes it look like this is a homework problem. In fact it is a standard type of problem for students learning Galois theory.
– KCd
Jul 14 at 16:48
@KCd please have a look. I have put my efforts along with question.
– Shubham Namdeo
Jul 14 at 16:57
@KCd please have a look. I have put my efforts along with question.
– Shubham Namdeo
Jul 14 at 16:57
The Galois group of an irreducible polynomial over a field of characteristic zero will act transitively on its zeros.
– Lord Shark the Unknown
Jul 14 at 17:01
The Galois group of an irreducible polynomial over a field of characteristic zero will act transitively on its zeros.
– Lord Shark the Unknown
Jul 14 at 17:01
@LordSharktheUnknown after that how to proceed?
– Shubham Namdeo
Jul 14 at 17:04
@LordSharktheUnknown after that how to proceed?
– Shubham Namdeo
Jul 14 at 17:04
add a comment |Â
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1
What's the question?
– Lord Shark the Unknown
Jul 14 at 16:41
1
Demanding others solve a problem rather than answer an actual question makes it look like this is a homework problem. In fact it is a standard type of problem for students learning Galois theory.
– KCd
Jul 14 at 16:48
@KCd please have a look. I have put my efforts along with question.
– Shubham Namdeo
Jul 14 at 16:57
The Galois group of an irreducible polynomial over a field of characteristic zero will act transitively on its zeros.
– Lord Shark the Unknown
Jul 14 at 17:01
@LordSharktheUnknown after that how to proceed?
– Shubham Namdeo
Jul 14 at 17:04