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Can two monic irreducible polynomials over $mathbbZ$, of coprime degrees, have the same splitting field?

Galois group of a degree 5 irreducible polynomial with two complex roots.Splitting field and dimension of irreducible polynomialsThe splitting fields of two irreducible polynomials over $Z / p Z$ both of degree 2 are isomorphicSplitting Field of Irreducible Polynomial over a Finite FieldSplitting field of cyclotomic polynomials over $mathbbF_2$.$5|#textGal(f/mathbbQ)subset S_5 implies textGal(f/mathbbQ)$ contains a $5$-cycle?Let $mathbbK$ the the splitting field of $x^4 -2 x^2 -2$ over $mathbbQ$. Determine all the subgroups of the Galois groupQuestion on splitting field and irreducible polynomials.Same Galois group for two polynomialsWhat quadratic field is contained in the splitting field of $x^5 - 4x - 2$ over $mathbbQ$?

5

$begingroup$

Let $f,g in mathbbZ[X]$ be monic polynomials. It is possible for distinct monic polynomials over $mathbbZ$ to have the same splitting field. For example $f = x^4 - 2$ and $g= x^4+2$ both have splitting field
$mathbbQ(sqrt[leftroot-2uproot24]2, i)$. Here $f,g$ are both irreducible with equal degrees.

Another example is $f=x^n - 1$ and $g=Phi_n(x)$, the $n$'th cyclotomic polynomial. The splitting field of both is the $n$'th cyclotomic field. In this case, $g$ is irreducible and divides $f$. The degree of $g$ is of course $varphi(n)$.

Question: Suppose $f,g$ are monic and irreducible, with coprime degrees. Say $deg f = n$ and $deg g = k$. Let us write $K_f$ and $K_g$ for their splitting fields.

Is it possible for these splitting fields to coincide, $K_f = K_g$? I am particularly interested in the case $k=n-1$.

I have attempted to approach the question using group theory. If we write $G$ for their common Galois group, then via the action of $G$ on the roots of $f$ and $g$ respectively, we would realize $G$ as a transitive subgroup of both $S_n$ and $S_k$. Therefore, as $k$ and $n$ are coprime

1. 
   $nk mid |G|$.


2. $|G| leq k!$


3. 
   $[S_n : G] geq n(n-1)cdots(k+1)$.

Furthermore, if $alpha$ is a root of $f$ and $beta$ is a root of $g$, then their stabilizers $Stab(alpha)$ and $Stab(beta)$ are subgroups of index $n$ and $k$ respectively. As these are coprime by assumption we would have $G = Stab(alpha)Stab(beta)$.

My conjecture is that this is not possible, but I am not able to show it. I seem to remember that, if $l < q$ then $S_l$ is never a transitive subgroup of $S_q$, except the case $S_5 leq S_6$. Here $S_5$ acts transitively on it's $5$-Sylow subgroups by conjugation, and there are $6$ of them. However, I might be wrong as I am unable to recover a reference for this. Of course, if this was true, we would simply strengthen 2. and 3. above to strict inequalities, unless $n=6$ and $k=5$.

I am also unaware whether there exists such a $f$ with degree $6$ and Galois group $S_5$ acting on the roots in this exotic manner.

group-theory polynomials galois-theory algebraic-number-theory irreducible-polynomials

share|cite|improve this question

edited 2 days ago

Sil

5,16421643

asked Mar 7 at 10:02

Emil LaursenEmil Laursen

263

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another approach is to use the Chebotarev density theorem, see this MO-question.
  $endgroup$
  – Dietrich Burde
  Mar 7 at 10:24
  
  

  
  
  
  

add a comment | 

5

$begingroup$

Let $f,g in mathbbZ[X]$ be monic polynomials. It is possible for distinct monic polynomials over $mathbbZ$ to have the same splitting field. For example $f = x^4 - 2$ and $g= x^4+2$ both have splitting field
$mathbbQ(sqrt[leftroot-2uproot24]2, i)$. Here $f,g$ are both irreducible with equal degrees.

Another example is $f=x^n - 1$ and $g=Phi_n(x)$, the $n$'th cyclotomic polynomial. The splitting field of both is the $n$'th cyclotomic field. In this case, $g$ is irreducible and divides $f$. The degree of $g$ is of course $varphi(n)$.

Question: Suppose $f,g$ are monic and irreducible, with coprime degrees. Say $deg f = n$ and $deg g = k$. Let us write $K_f$ and $K_g$ for their splitting fields.

Is it possible for these splitting fields to coincide, $K_f = K_g$? I am particularly interested in the case $k=n-1$.

I have attempted to approach the question using group theory. If we write $G$ for their common Galois group, then via the action of $G$ on the roots of $f$ and $g$ respectively, we would realize $G$ as a transitive subgroup of both $S_n$ and $S_k$. Therefore, as $k$ and $n$ are coprime

1. 
   $nk mid |G|$.


2. $|G| leq k!$


3. 
   $[S_n : G] geq n(n-1)cdots(k+1)$.

Furthermore, if $alpha$ is a root of $f$ and $beta$ is a root of $g$, then their stabilizers $Stab(alpha)$ and $Stab(beta)$ are subgroups of index $n$ and $k$ respectively. As these are coprime by assumption we would have $G = Stab(alpha)Stab(beta)$.

My conjecture is that this is not possible, but I am not able to show it. I seem to remember that, if $l < q$ then $S_l$ is never a transitive subgroup of $S_q$, except the case $S_5 leq S_6$. Here $S_5$ acts transitively on it's $5$-Sylow subgroups by conjugation, and there are $6$ of them. However, I might be wrong as I am unable to recover a reference for this. Of course, if this was true, we would simply strengthen 2. and 3. above to strict inequalities, unless $n=6$ and $k=5$.

I am also unaware whether there exists such a $f$ with degree $6$ and Galois group $S_5$ acting on the roots in this exotic manner.

group-theory polynomials galois-theory algebraic-number-theory irreducible-polynomials

share|cite|improve this question

edited 2 days ago

Sil

5,16421643

asked Mar 7 at 10:02

Emil LaursenEmil Laursen

263

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another approach is to use the Chebotarev density theorem, see this MO-question.
  $endgroup$
  – Dietrich Burde
  Mar 7 at 10:24
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $f,g in mathbbZ[X]$ be monic polynomials. It is possible for distinct monic polynomials over $mathbbZ$ to have the same splitting field. For example $f = x^4 - 2$ and $g= x^4+2$ both have splitting field
$mathbbQ(sqrt[leftroot-2uproot24]2, i)$. Here $f,g$ are both irreducible with equal degrees.

Another example is $f=x^n - 1$ and $g=Phi_n(x)$, the $n$'th cyclotomic polynomial. The splitting field of both is the $n$'th cyclotomic field. In this case, $g$ is irreducible and divides $f$. The degree of $g$ is of course $varphi(n)$.

Question: Suppose $f,g$ are monic and irreducible, with coprime degrees. Say $deg f = n$ and $deg g = k$. Let us write $K_f$ and $K_g$ for their splitting fields.

Is it possible for these splitting fields to coincide, $K_f = K_g$? I am particularly interested in the case $k=n-1$.

I have attempted to approach the question using group theory. If we write $G$ for their common Galois group, then via the action of $G$ on the roots of $f$ and $g$ respectively, we would realize $G$ as a transitive subgroup of both $S_n$ and $S_k$. Therefore, as $k$ and $n$ are coprime

1. 
   $nk mid |G|$.


2. $|G| leq k!$


3. 
   $[S_n : G] geq n(n-1)cdots(k+1)$.

Furthermore, if $alpha$ is a root of $f$ and $beta$ is a root of $g$, then their stabilizers $Stab(alpha)$ and $Stab(beta)$ are subgroups of index $n$ and $k$ respectively. As these are coprime by assumption we would have $G = Stab(alpha)Stab(beta)$.

My conjecture is that this is not possible, but I am not able to show it. I seem to remember that, if $l < q$ then $S_l$ is never a transitive subgroup of $S_q$, except the case $S_5 leq S_6$. Here $S_5$ acts transitively on it's $5$-Sylow subgroups by conjugation, and there are $6$ of them. However, I might be wrong as I am unable to recover a reference for this. Of course, if this was true, we would simply strengthen 2. and 3. above to strict inequalities, unless $n=6$ and $k=5$.

I am also unaware whether there exists such a $f$ with degree $6$ and Galois group $S_5$ acting on the roots in this exotic manner.

group-theory polynomials galois-theory algebraic-number-theory irreducible-polynomials

share|cite|improve this question

edited 2 days ago

Sil

5,16421643

asked Mar 7 at 10:02

Emil LaursenEmil Laursen

263

$endgroup$

share|cite|improve this question

edited 2 days ago

Sil

5,16421643

asked Mar 7 at 10:02

Emil LaursenEmil Laursen

263

share|cite|improve this question

edited 2 days ago

Sil

5,16421643

asked Mar 7 at 10:02

Emil LaursenEmil Laursen

263

edited 2 days ago

Sil

5,16421643

edited 2 days ago

Sil

5,16421643

asked Mar 7 at 10:02

Emil LaursenEmil Laursen

263

asked Mar 7 at 10:02

Emil LaursenEmil Laursen

263