Mixing
if $a$ is an algebraic integer and $min mathbb{Z}$ then $a+m$ is an algebraic integer.
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I tried to use newton binomial, if $p_a(x) = sum_0^nb_kx^k$ , then $b_k(a+m)^k =b_ka^k +b_ksum_0^{k-1}a^im^{k-i}$, denote $T_k=b_ksum_0^{k-1}a^im^{k-i}$, one gets that $p_{a}(x)-(sum_1^nT_k ) - mb_0$ has $(m+a)$ as a root, but the problem is that $T_k$ are not guaranteed to be integers. I also worked out the $acdot m$ must be an algebraic integer (by taking $p_a(x)=sum_0^nb_kx^k$ and define: $sum_0^n m^{n-k}b_kx^k$ ), yet I didn't succeed to use it in order to manipulate $p_{a}(x)-(sum_1^nT_k ) - mb_0$ back to $mathbb{Z}[x]$
number-theory algebraic-number-theory
asked Jan 14 at 4:17
dandan
559613
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add a comment |
$begingroup$
I tried to use newton binomial, if $p_a(x) = sum_0^nb_kx^k$ , then $b_k(a+m)^k =b_ka^k +b_ksum_0^{k-1}a^im^{k-i}$, denote $T_k=b_ksum_0^{k-1}a^im^{k-i}$, one gets that $p_{a}(x)-(sum_1^nT_k ) - mb_0$ has $(m+a)$ as a root, but the problem is that $T_k$ are not guaranteed to be integers. I also worked out the $acdot m$ must be an algebraic integer (by taking $p_a(x)=sum_0^nb_kx^k$ and define: $sum_0^n m^{n-k}b_kx^k$ ), yet I didn't succeed to use it in order to manipulate $p_{a}(x)-(sum_1^nT_k ) - mb_0$ back to $mathbb{Z}[x]$
number-theory algebraic-number-theory
asked Jan 14 at 4:17
dandan
559613
$endgroup$
add a comment |
$begingroup$
I tried to use newton binomial, if $p_a(x) = sum_0^nb_kx^k$ , then $b_k(a+m)^k =b_ka^k +b_ksum_0^{k-1}a^im^{k-i}$, denote $T_k=b_ksum_0^{k-1}a^im^{k-i}$, one gets that $p_{a}(x)-(sum_1^nT_k ) - mb_0$ has $(m+a)$ as a root, but the problem is that $T_k$ are not guaranteed to be integers. I also worked out the $acdot m$ must be an algebraic integer (by taking $p_a(x)=sum_0^nb_kx^k$ and define: $sum_0^n m^{n-k}b_kx^k$ ), yet I didn't succeed to use it in order to manipulate $p_{a}(x)-(sum_1^nT_k ) - mb_0$ back to $mathbb{Z}[x]$
number-theory algebraic-number-theory
asked Jan 14 at 4:17
dandan
559613
$endgroup$
I tried to use newton binomial, if $p_a(x) = sum_0^nb_kx^k$ , then $b_k(a+m)^k =b_ka^k +b_ksum_0^{k-1}a^im^{k-i}$, denote $T_k=b_ksum_0^{k-1}a^im^{k-i}$, one gets that $p_{a}(x)-(sum_1^nT_k ) - mb_0$ has $(m+a)$ as a root, but the problem is that $T_k$ are not guaranteed to be integers. I also worked out the $acdot m$ must be an algebraic integer (by taking $p_a(x)=sum_0^nb_kx^k$ and define: $sum_0^n m^{n-k}b_kx^k$ ), yet I didn't succeed to use it in order to manipulate $p_{a}(x)-(sum_1^nT_k ) - mb_0$ back to $mathbb{Z}[x]$
number-theory algebraic-number-theory
number-theory algebraic-number-theory
asked Jan 14 at 4:17
dandan
559613
asked Jan 14 at 4:17
dandan
559613
asked Jan 14 at 4:17
dandan
559613
asked Jan 14 at 4:17
dandan
559613
asked Jan 14 at 4:17
dandan
559613
559613
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint: If $P(x)=sum_{k=0}^n b_k X^k$ has $a$ as a root, then
$$P(X-m)=sum_{k=0}^n b_k (X-m)^k$$
has $a+m$ as a root. As a polynomial this does have integer coefficients.
Your mistake was thinking about this as an expression in $a$ not as a polynomial. You need the coefficient of $a^j$ to be an integer, not $T_k$.
answered Jan 14 at 4:22
N. S.N. S.
104k7112208
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add a comment |
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1 Answer
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active
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1 Answer
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active
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$begingroup$
Hint: If $P(x)=sum_{k=0}^n b_k X^k$ has $a$ as a root, then
$$P(X-m)=sum_{k=0}^n b_k (X-m)^k$$
has $a+m$ as a root. As a polynomial this does have integer coefficients.
Your mistake was thinking about this as an expression in $a$ not as a polynomial. You need the coefficient of $a^j$ to be an integer, not $T_k$.
answered Jan 14 at 4:22
N. S.N. S.
104k7112208
$endgroup$
add a comment |
$begingroup$
Hint: If $P(x)=sum_{k=0}^n b_k X^k$ has $a$ as a root, then
$$P(X-m)=sum_{k=0}^n b_k (X-m)^k$$
has $a+m$ as a root. As a polynomial this does have integer coefficients.
Your mistake was thinking about this as an expression in $a$ not as a polynomial. You need the coefficient of $a^j$ to be an integer, not $T_k$.
answered Jan 14 at 4:22
N. S.N. S.
104k7112208
$endgroup$
add a comment |
$begingroup$
Hint: If $P(x)=sum_{k=0}^n b_k X^k$ has $a$ as a root, then
$$P(X-m)=sum_{k=0}^n b_k (X-m)^k$$
has $a+m$ as a root. As a polynomial this does have integer coefficients.
Your mistake was thinking about this as an expression in $a$ not as a polynomial. You need the coefficient of $a^j$ to be an integer, not $T_k$.
answered Jan 14 at 4:22
N. S.N. S.
104k7112208
$endgroup$
Hint: If $P(x)=sum_{k=0}^n b_k X^k$ has $a$ as a root, then
$$P(X-m)=sum_{k=0}^n b_k (X-m)^k$$
has $a+m$ as a root. As a polynomial this does have integer coefficients.
Your mistake was thinking about this as an expression in $a$ not as a polynomial. You need the coefficient of $a^j$ to be an integer, not $T_k$.
answered Jan 14 at 4:22
N. S.N. S.
104k7112208
answered Jan 14 at 4:22
N. S.N. S.
104k7112208
answered Jan 14 at 4:22
N. S.N. S.
104k7112208
answered Jan 14 at 4:22
N. S.N. S.
104k7112208
104k7112208
add a comment |
add a comment |
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