Question on symmetrical integral polynomials
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 yesterday
Marco Perin
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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 yesterday
Marco 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.
I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
– Robert Lewis
yesterday
1
Yes sorry, it is $m$ everywhere.
– Marco Perin
yesterday
add a comment |
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 yesterday
Marco 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.
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 yesterday
Marco 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 yesterday
Marco 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 yesterday
asked yesterday
Marco 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 yesterday
Marco Perin
12
asked yesterday
Marco 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.
I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
– Robert Lewis
yesterday
1
Yes sorry, it is $m$ everywhere.
– Marco Perin
yesterday
add a comment |
I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
– Robert Lewis
yesterday
1
Yes sorry, it is $m$ everywhere.
– Marco Perin
yesterday
I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
– Robert Lewis
yesterday
I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
– Robert Lewis
yesterday
1
1
Yes sorry, it is $m$ everywhere.
– Marco Perin
yesterday
Yes sorry, it is $m$ everywhere.
– Marco Perin
yesterday
add a comment |
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Marco Perin is a new contributor. Be nice, and check out our Code of Conduct.
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I assume $m = n$; am I right; otherwise, the degree of the polynomial and the number of roots don't match up! Cheers!
– Robert Lewis
yesterday
1
Yes sorry, it is $m$ everywhere.
– Marco Perin
yesterday