Find polynomial given splitting field
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Let $finmathbb{Q}[x]$ a monic polynomial such that $f$ has degree $n$. Let $E_f$ be the splitting field of $f$ over $mathbb{Q}$. I would like to show that there exists a monic polynomial in $mathbb{Z}[x]$ of degree $n$ such that it has the same splitting field. I don't even know how to tackle this problem. Any help would be appreciated.
Edit: as has been pointed out, it would suffice to prove that $f(x)$ and $q^n=f(x/q)$ have the same splitting field for any integer $q$. This is clear since the roots of $f$ in $E_f$ are the same than those pf $f(x/q)$ except for multiplying by a rational constant. Am I right?
field-theory galois-theory splitting-field
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Let $finmathbb{Q}[x]$ a monic polynomial such that $f$ has degree $n$. Let $E_f$ be the splitting field of $f$ over $mathbb{Q}$. I would like to show that there exists a monic polynomial in $mathbb{Z}[x]$ of degree $n$ such that it has the same splitting field. I don't even know how to tackle this problem. Any help would be appreciated.
Edit: as has been pointed out, it would suffice to prove that $f(x)$ and $q^n=f(x/q)$ have the same splitting field for any integer $q$. This is clear since the roots of $f$ in $E_f$ are the same than those pf $f(x/q)$ except for multiplying by a rational constant. Am I right?
field-theory galois-theory splitting-field
1
Show that for any integer $q$ the splitting fields of $f(x)$ and $q^nf(x/q)$ are the same. With a suitable choice of $q$ the latter is in $Bbb{Z}[x]$.
– Jyrki Lahtonen
Nov 18 at 20:09
But it would not be monic
– Ray Bern
Nov 18 at 20:11
Thank you. You're right
– Ray Bern
Nov 18 at 20:16
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $finmathbb{Q}[x]$ a monic polynomial such that $f$ has degree $n$. Let $E_f$ be the splitting field of $f$ over $mathbb{Q}$. I would like to show that there exists a monic polynomial in $mathbb{Z}[x]$ of degree $n$ such that it has the same splitting field. I don't even know how to tackle this problem. Any help would be appreciated.
Edit: as has been pointed out, it would suffice to prove that $f(x)$ and $q^n=f(x/q)$ have the same splitting field for any integer $q$. This is clear since the roots of $f$ in $E_f$ are the same than those pf $f(x/q)$ except for multiplying by a rational constant. Am I right?
field-theory galois-theory splitting-field
Let $finmathbb{Q}[x]$ a monic polynomial such that $f$ has degree $n$. Let $E_f$ be the splitting field of $f$ over $mathbb{Q}$. I would like to show that there exists a monic polynomial in $mathbb{Z}[x]$ of degree $n$ such that it has the same splitting field. I don't even know how to tackle this problem. Any help would be appreciated.
Edit: as has been pointed out, it would suffice to prove that $f(x)$ and $q^n=f(x/q)$ have the same splitting field for any integer $q$. This is clear since the roots of $f$ in $E_f$ are the same than those pf $f(x/q)$ except for multiplying by a rational constant. Am I right?
field-theory galois-theory splitting-field
field-theory galois-theory splitting-field
edited Nov 18 at 20:22
asked Nov 18 at 20:00
Ray Bern
1109
1109
1
Show that for any integer $q$ the splitting fields of $f(x)$ and $q^nf(x/q)$ are the same. With a suitable choice of $q$ the latter is in $Bbb{Z}[x]$.
– Jyrki Lahtonen
Nov 18 at 20:09
But it would not be monic
– Ray Bern
Nov 18 at 20:11
Thank you. You're right
– Ray Bern
Nov 18 at 20:16
add a comment |
1
Show that for any integer $q$ the splitting fields of $f(x)$ and $q^nf(x/q)$ are the same. With a suitable choice of $q$ the latter is in $Bbb{Z}[x]$.
– Jyrki Lahtonen
Nov 18 at 20:09
But it would not be monic
– Ray Bern
Nov 18 at 20:11
Thank you. You're right
– Ray Bern
Nov 18 at 20:16
1
1
Show that for any integer $q$ the splitting fields of $f(x)$ and $q^nf(x/q)$ are the same. With a suitable choice of $q$ the latter is in $Bbb{Z}[x]$.
– Jyrki Lahtonen
Nov 18 at 20:09
Show that for any integer $q$ the splitting fields of $f(x)$ and $q^nf(x/q)$ are the same. With a suitable choice of $q$ the latter is in $Bbb{Z}[x]$.
– Jyrki Lahtonen
Nov 18 at 20:09
But it would not be monic
– Ray Bern
Nov 18 at 20:11
But it would not be monic
– Ray Bern
Nov 18 at 20:11
Thank you. You're right
– Ray Bern
Nov 18 at 20:16
Thank you. You're right
– Ray Bern
Nov 18 at 20:16
add a comment |
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Show that for any integer $q$ the splitting fields of $f(x)$ and $q^nf(x/q)$ are the same. With a suitable choice of $q$ the latter is in $Bbb{Z}[x]$.
– Jyrki Lahtonen
Nov 18 at 20:09
But it would not be monic
– Ray Bern
Nov 18 at 20:11
Thank you. You're right
– Ray Bern
Nov 18 at 20:16