Existence of polynomials in $mathbb{Q}[x]$ and $mathbb{Z}[x]$ with same splitting fields.











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I've been asked to prove the following statement in a Galois Theory Seminar after being introduced to Dedekind's Theorem. (I assume this could potentially help getting the answer.)




Let $f(x)$ be a a monic polynomial of degree $N$ in $mathbb{Q}[x]$ and let $E_f$ be its splitting field over $mathbb{Q}$. Then, there exists a monic polynomial $p(x) in mathbb{Z}[x]$ of degree $N$ that has the same splitting field $E_f$.




I can't find any way to prove it.



Any help is welcome!



EDIT:
There used to be what I though was a counter example that has been answered already.






polynomials galois-theory splitting-field







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  • This is not a counter-example. The problem is that you're assuming that $f$ and $p$ have the same roots. This is not necessary in this case since any polynomial with integral coefficients and rational roots has the splitting field of $mathbb{Q}$. So, you could take $p=1$ and get the same splitting field as $f$.
    – Michael Burr
    23 hours ago












  • BBC3, all integers are also elements of $Bbb{Q}$.
    – Jyrki Lahtonen
    23 hours ago










  • You're both right! I was taking $mathbb{Z}$ as its splitting field instead which makes no sense at all. I'll edit the question to the proof only.
    – BBC3
    23 hours ago






  • 2




    Are you possibly taking the same course as the asker of this question. My comments there settle your question also.
    – Jyrki Lahtonen
    22 hours ago















up vote
1
down vote

favorite












I've been asked to prove the following statement in a Galois Theory Seminar after being introduced to Dedekind's Theorem. (I assume this could potentially help getting the answer.)




Let $f(x)$ be a a monic polynomial of degree $N$ in $mathbb{Q}[x]$ and let $E_f$ be its splitting field over $mathbb{Q}$. Then, there exists a monic polynomial $p(x) in mathbb{Z}[x]$ of degree $N$ that has the same splitting field $E_f$.




I can't find any way to prove it.



Any help is welcome!



EDIT:
There used to be what I though was a counter example that has been answered already.






polynomials galois-theory splitting-field







share|cite|improve this question
























  • This is not a counter-example. The problem is that you're assuming that $f$ and $p$ have the same roots. This is not necessary in this case since any polynomial with integral coefficients and rational roots has the splitting field of $mathbb{Q}$. So, you could take $p=1$ and get the same splitting field as $f$.
    – Michael Burr
    23 hours ago












  • BBC3, all integers are also elements of $Bbb{Q}$.
    – Jyrki Lahtonen
    23 hours ago










  • You're both right! I was taking $mathbb{Z}$ as its splitting field instead which makes no sense at all. I'll edit the question to the proof only.
    – BBC3
    23 hours ago






  • 2




    Are you possibly taking the same course as the asker of this question. My comments there settle your question also.
    – Jyrki Lahtonen
    22 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I've been asked to prove the following statement in a Galois Theory Seminar after being introduced to Dedekind's Theorem. (I assume this could potentially help getting the answer.)




Let $f(x)$ be a a monic polynomial of degree $N$ in $mathbb{Q}[x]$ and let $E_f$ be its splitting field over $mathbb{Q}$. Then, there exists a monic polynomial $p(x) in mathbb{Z}[x]$ of degree $N$ that has the same splitting field $E_f$.




I can't find any way to prove it.



Any help is welcome!



EDIT:
There used to be what I though was a counter example that has been answered already.






polynomials galois-theory splitting-field







share|cite|improve this question















I've been asked to prove the following statement in a Galois Theory Seminar after being introduced to Dedekind's Theorem. (I assume this could potentially help getting the answer.)




Let $f(x)$ be a a monic polynomial of degree $N$ in $mathbb{Q}[x]$ and let $E_f$ be its splitting field over $mathbb{Q}$. Then, there exists a monic polynomial $p(x) in mathbb{Z}[x]$ of degree $N$ that has the same splitting field $E_f$.




I can't find any way to prove it.



Any help is welcome!



EDIT:
There used to be what I though was a counter example that has been answered already.






polynomials galois-theory splitting-field




polynomials galois-theory splitting-field






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 23 hours ago

























asked 23 hours ago









BBC3

341212




341212












  • This is not a counter-example. The problem is that you're assuming that $f$ and $p$ have the same roots. This is not necessary in this case since any polynomial with integral coefficients and rational roots has the splitting field of $mathbb{Q}$. So, you could take $p=1$ and get the same splitting field as $f$.
    – Michael Burr
    23 hours ago












  • BBC3, all integers are also elements of $Bbb{Q}$.
    – Jyrki Lahtonen
    23 hours ago










  • You're both right! I was taking $mathbb{Z}$ as its splitting field instead which makes no sense at all. I'll edit the question to the proof only.
    – BBC3
    23 hours ago






  • 2




    Are you possibly taking the same course as the asker of this question. My comments there settle your question also.
    – Jyrki Lahtonen
    22 hours ago


















  • This is not a counter-example. The problem is that you're assuming that $f$ and $p$ have the same roots. This is not necessary in this case since any polynomial with integral coefficients and rational roots has the splitting field of $mathbb{Q}$. So, you could take $p=1$ and get the same splitting field as $f$.
    – Michael Burr
    23 hours ago












  • BBC3, all integers are also elements of $Bbb{Q}$.
    – Jyrki Lahtonen
    23 hours ago










  • You're both right! I was taking $mathbb{Z}$ as its splitting field instead which makes no sense at all. I'll edit the question to the proof only.
    – BBC3
    23 hours ago






  • 2




    Are you possibly taking the same course as the asker of this question. My comments there settle your question also.
    – Jyrki Lahtonen
    22 hours ago
















This is not a counter-example. The problem is that you're assuming that $f$ and $p$ have the same roots. This is not necessary in this case since any polynomial with integral coefficients and rational roots has the splitting field of $mathbb{Q}$. So, you could take $p=1$ and get the same splitting field as $f$.
– Michael Burr
23 hours ago






This is not a counter-example. The problem is that you're assuming that $f$ and $p$ have the same roots. This is not necessary in this case since any polynomial with integral coefficients and rational roots has the splitting field of $mathbb{Q}$. So, you could take $p=1$ and get the same splitting field as $f$.
– Michael Burr
23 hours ago














BBC3, all integers are also elements of $Bbb{Q}$.
– Jyrki Lahtonen
23 hours ago




BBC3, all integers are also elements of $Bbb{Q}$.
– Jyrki Lahtonen
23 hours ago












You're both right! I was taking $mathbb{Z}$ as its splitting field instead which makes no sense at all. I'll edit the question to the proof only.
– BBC3
23 hours ago




You're both right! I was taking $mathbb{Z}$ as its splitting field instead which makes no sense at all. I'll edit the question to the proof only.
– BBC3
23 hours ago




2




2




Are you possibly taking the same course as the asker of this question. My comments there settle your question also.
– Jyrki Lahtonen
22 hours ago




Are you possibly taking the same course as the asker of this question. My comments there settle your question also.
– Jyrki Lahtonen
22 hours ago










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Let $x^n + a_1 x^{n_1}+ cdots + a_n=0$, and write $y=Nx$; then $y^n + a_1 N y^{n_1}+ cdots + a_n N^n =0$. Write $a_j=p_j/q_j$ and $N=prod_j q_j$ and you are done. This is basically Gauss's lemma isn't it?






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    Let $x^n + a_1 x^{n_1}+ cdots + a_n=0$, and write $y=Nx$; then $y^n + a_1 N y^{n_1}+ cdots + a_n N^n =0$. Write $a_j=p_j/q_j$ and $N=prod_j q_j$ and you are done. This is basically Gauss's lemma isn't it?






    share|cite|improve this answer

























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      Let $x^n + a_1 x^{n_1}+ cdots + a_n=0$, and write $y=Nx$; then $y^n + a_1 N y^{n_1}+ cdots + a_n N^n =0$. Write $a_j=p_j/q_j$ and $N=prod_j q_j$ and you are done. This is basically Gauss's lemma isn't it?






      share|cite|improve this answer























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        up vote
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        Let $x^n + a_1 x^{n_1}+ cdots + a_n=0$, and write $y=Nx$; then $y^n + a_1 N y^{n_1}+ cdots + a_n N^n =0$. Write $a_j=p_j/q_j$ and $N=prod_j q_j$ and you are done. This is basically Gauss's lemma isn't it?






        share|cite|improve this answer












        Let $x^n + a_1 x^{n_1}+ cdots + a_n=0$, and write $y=Nx$; then $y^n + a_1 N y^{n_1}+ cdots + a_n N^n =0$. Write $a_j=p_j/q_j$ and $N=prod_j q_j$ and you are done. This is basically Gauss's lemma isn't it?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 22 hours ago









        Richard Martin

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