Question on symmetrical integral polynomials
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I am reading a proof of the transcendence of $pi$ but I am stucked with a problem on polynomials. The proof begins like that:
"Suppose ${beta _1},{beta _2}, ldots ,{beta _m}$ are the roots of an equation
$d{x^m} + {d_1}{x^{m - 1}} + ldots + {d_m} = 0$
with integral coefficients. Any symmetrical integral polynomial in
${dbeta _1},{dbeta _2}, ldots ,{dbeta _m}$
is an integral polynomial in
${d_1},{d_2}, ldots ,{d_m}$
and is therefore an integer."
Do someone have a proof please?
Of course it has to do with the fact that the roots have denominators that divide $d$, and they are either rational, or of the form $frac{a pm sqrt{b}}{c}$ (real or complex conjugate). And I can easily convince myself that in a symmetric polynomial of the roots the terms $left(sqrt{b}right)^n$ cancel each other; still I would like to find a nice proof.
symmetric-polynomials
asked Jan 5 at 7:09
Marco PerinMarco Perin
12
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add a comment |
$begingroup$
I am reading a proof of the transcendence of $pi$ but I am stucked with a problem on polynomials. The proof begins like that:
"Suppose ${beta _1},{beta _2}, ldots ,{beta _m}$ are the roots of an equation
$d{x^m} + {d_1}{x^{m - 1}} + ldots + {d_m} = 0$
with integral coefficients. Any symmetrical integral polynomial in
${dbeta _1},{dbeta _2}, ldots ,{dbeta _m}$
is an integral polynomial in
${d_1},{d_2}, ldots ,{d_m}$
and is therefore an integer."
Do someone have a proof please?
Of course it has to do with the fact that the roots have denominators that divide $d$, and they are either rational, or of the form $frac{a pm sqrt{b}}{c}$ (real or complex conjugate). And I can easily convince myself that in a symmetric polynomial of the roots the terms $left(sqrt{b}right)^n$ cancel each other; still I would like to find a nice proof.
symmetric-polynomials
asked Jan 5 at 7:09
Marco PerinMarco Perin
12
New contributor
Marco Perin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
$endgroup$
– Robert Lewis
Jan 5 at 7:27
1
$begingroup$
Yes sorry, it is $m$ everywhere.
$endgroup$
– Marco Perin
Jan 5 at 7:57
add a comment |
$begingroup$
I am reading a proof of the transcendence of $pi$ but I am stucked with a problem on polynomials. The proof begins like that:
"Suppose ${beta _1},{beta _2}, ldots ,{beta _m}$ are the roots of an equation
$d{x^m} + {d_1}{x^{m - 1}} + ldots + {d_m} = 0$
with integral coefficients. Any symmetrical integral polynomial in
${dbeta _1},{dbeta _2}, ldots ,{dbeta _m}$
is an integral polynomial in
${d_1},{d_2}, ldots ,{d_m}$
and is therefore an integer."
Do someone have a proof please?
Of course it has to do with the fact that the roots have denominators that divide $d$, and they are either rational, or of the form $frac{a pm sqrt{b}}{c}$ (real or complex conjugate). And I can easily convince myself that in a symmetric polynomial of the roots the terms $left(sqrt{b}right)^n$ cancel each other; still I would like to find a nice proof.
symmetric-polynomials
asked Jan 5 at 7:09
Marco PerinMarco Perin
12
New contributor
Marco Perin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am reading a proof of the transcendence of $pi$ but I am stucked with a problem on polynomials. The proof begins like that:
"Suppose ${beta _1},{beta _2}, ldots ,{beta _m}$ are the roots of an equation
$d{x^m} + {d_1}{x^{m - 1}} + ldots + {d_m} = 0$
with integral coefficients. Any symmetrical integral polynomial in
${dbeta _1},{dbeta _2}, ldots ,{dbeta _m}$
is an integral polynomial in
${d_1},{d_2}, ldots ,{d_m}$
and is therefore an integer."
Do someone have a proof please?
Of course it has to do with the fact that the roots have denominators that divide $d$, and they are either rational, or of the form $frac{a pm sqrt{b}}{c}$ (real or complex conjugate). And I can easily convince myself that in a symmetric polynomial of the roots the terms $left(sqrt{b}right)^n$ cancel each other; still I would like to find a nice proof.
symmetric-polynomials
symmetric-polynomials
asked Jan 5 at 7:09
Marco PerinMarco Perin
12
New contributor
Marco Perin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 5 at 7:09
Marco PerinMarco Perin
12
New contributor
Marco Perin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 5 at 12:30
Marco Perin
asked Jan 5 at 7:09
Marco PerinMarco Perin
12
New contributor
Marco Perin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 5 at 7:09
Marco PerinMarco Perin
12
asked Jan 5 at 7:09
Marco PerinMarco Perin
12
12
New contributor
Marco Perin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Marco Perin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Marco Perin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
$endgroup$
– Robert Lewis
Jan 5 at 7:27
1
$begingroup$
Yes sorry, it is $m$ everywhere.
$endgroup$
– Marco Perin
Jan 5 at 7:57
add a comment |
$begingroup$
I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
$endgroup$
– Robert Lewis
Jan 5 at 7:27
1
$begingroup$
Yes sorry, it is $m$ everywhere.
$endgroup$
– Marco Perin
Jan 5 at 7:57
$begingroup$
I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
$endgroup$
– Robert Lewis
Jan 5 at 7:27
$begingroup$
I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
$endgroup$
– Robert Lewis
Jan 5 at 7:27
1
1
$begingroup$
Yes sorry, it is $m$ everywhere.
$endgroup$
– Marco Perin
Jan 5 at 7:57
$begingroup$
Yes sorry, it is $m$ everywhere.
$endgroup$
– Marco Perin
Jan 5 at 7:57
add a comment |
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$begingroup$
I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
$endgroup$
– Robert Lewis
Jan 5 at 7:27
1
$begingroup$
Yes sorry, it is $m$ everywhere.
$endgroup$
– Marco Perin
Jan 5 at 7:57