Mixing
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
asked 23 hours ago
BBC3
341212
add a comment |
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
asked 23 hours ago
BBC3
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
add a comment |
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
asked 23 hours ago
BBC3
341212
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
asked 23 hours ago
BBC3
341212
asked 23 hours ago
BBC3
341212
edited 23 hours ago
asked 23 hours ago
BBC3
341212
asked 23 hours ago
BBC3
341212
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
add a comment |
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
add a comment |
1 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?
answered 22 hours ago
Richard Martin
1,2588
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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?
answered 22 hours ago
Richard Martin
1,2588
add a comment |
up vote
0
down vote
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?
answered 22 hours ago
Richard Martin
1,2588
add a comment |
up vote
0
down vote
up vote
0
down vote
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?
answered 22 hours ago
Richard Martin
1,2588
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?
answered 22 hours ago
Richard Martin
1,2588
answered 22 hours ago
Richard Martin
1,2588
answered 22 hours ago
Richard Martin
1,2588
answered 22 hours ago
Richard Martin
1,2588
1,2588
add a comment |
add a comment |
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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