Mixing
On coefficients of $(t-x_1)(t-x_2)..(t-x_i)$
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I was reading Emil Artin's Galois Theory (2nd edition). On pp.39-40, Artin defines a number of things as follows.
Let $k$ be a field and $E=k(x_1,ldots,x_n)$ be the field of all rational functions of $n$ indeterminates $x_1,ldots,x_n$. Define a polynomial
$$
f(t)=(t-x_1)cdots(t-x_n)=t^n+a_1t^{n-1}+cdots+a_n,tag{3}
$$
where $a_1=-sum_ix_i,,a_2=sum_{i<j}x_ix_j$,... are elementary symmetric functions in $x_1,ldots,x_n$. Then he puts $F_n(t)=f(t)$ and also for $i=n-1,,n-2,ldots,1$,
$$
F_i(t)=frac{f(t)}{(t-x_{i+1})(t-x_{i+2})cdots(t-x_n)}=frac{F_{i+1}(t)}{t-x_{i+1}}.tag{5}
$$
(If I am not mistaken, this simply means that $F_i(t)=(t-x_1)(t-x_2)..(t-x_i)$ for each $i$, but what matters below is the division on the RHS of $(5)$.) He then writes that (emphasis mine):
Performing the division [in $(5)$] we see that $F_i(t)$ is a polynomial in $t$ of degree $i$ whose highest coefficient is $1$ and whose coefficients are polynomials in the variables $a_1,a_2,ldots,a_i$ and $x_{i+1},x_{i+2},ldots,x_n$. Only integer enter as coefficients in these expressions.
Can someone explain the bold part? I have thought about using Euclidean algorithm, but the $a_i$s are polynomials in $x_1,ldots,x_n$ rather than constants. I am not sure how the division is done.
abstract-algebra symmetric-polynomials
asked Jan 23 at 5:01
William McGonagallWilliam McGonagall
1547
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add a comment |
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I was reading Emil Artin's Galois Theory (2nd edition). On pp.39-40, Artin defines a number of things as follows.
Let $k$ be a field and $E=k(x_1,ldots,x_n)$ be the field of all rational functions of $n$ indeterminates $x_1,ldots,x_n$. Define a polynomial
$$
f(t)=(t-x_1)cdots(t-x_n)=t^n+a_1t^{n-1}+cdots+a_n,tag{3}
$$
where $a_1=-sum_ix_i,,a_2=sum_{i<j}x_ix_j$,... are elementary symmetric functions in $x_1,ldots,x_n$. Then he puts $F_n(t)=f(t)$ and also for $i=n-1,,n-2,ldots,1$,
$$
F_i(t)=frac{f(t)}{(t-x_{i+1})(t-x_{i+2})cdots(t-x_n)}=frac{F_{i+1}(t)}{t-x_{i+1}}.tag{5}
$$
(If I am not mistaken, this simply means that $F_i(t)=(t-x_1)(t-x_2)..(t-x_i)$ for each $i$, but what matters below is the division on the RHS of $(5)$.) He then writes that (emphasis mine):
Performing the division [in $(5)$] we see that $F_i(t)$ is a polynomial in $t$ of degree $i$ whose highest coefficient is $1$ and whose coefficients are polynomials in the variables $a_1,a_2,ldots,a_i$ and $x_{i+1},x_{i+2},ldots,x_n$. Only integer enter as coefficients in these expressions.
Can someone explain the bold part? I have thought about using Euclidean algorithm, but the $a_i$s are polynomials in $x_1,ldots,x_n$ rather than constants. I am not sure how the division is done.
abstract-algebra symmetric-polynomials
asked Jan 23 at 5:01
William McGonagallWilliam McGonagall
1547
$endgroup$
add a comment |
$begingroup$
I was reading Emil Artin's Galois Theory (2nd edition). On pp.39-40, Artin defines a number of things as follows.
Let $k$ be a field and $E=k(x_1,ldots,x_n)$ be the field of all rational functions of $n$ indeterminates $x_1,ldots,x_n$. Define a polynomial
$$
f(t)=(t-x_1)cdots(t-x_n)=t^n+a_1t^{n-1}+cdots+a_n,tag{3}
$$
where $a_1=-sum_ix_i,,a_2=sum_{i<j}x_ix_j$,... are elementary symmetric functions in $x_1,ldots,x_n$. Then he puts $F_n(t)=f(t)$ and also for $i=n-1,,n-2,ldots,1$,
$$
F_i(t)=frac{f(t)}{(t-x_{i+1})(t-x_{i+2})cdots(t-x_n)}=frac{F_{i+1}(t)}{t-x_{i+1}}.tag{5}
$$
(If I am not mistaken, this simply means that $F_i(t)=(t-x_1)(t-x_2)..(t-x_i)$ for each $i$, but what matters below is the division on the RHS of $(5)$.) He then writes that (emphasis mine):
Performing the division [in $(5)$] we see that $F_i(t)$ is a polynomial in $t$ of degree $i$ whose highest coefficient is $1$ and whose coefficients are polynomials in the variables $a_1,a_2,ldots,a_i$ and $x_{i+1},x_{i+2},ldots,x_n$. Only integer enter as coefficients in these expressions.
Can someone explain the bold part? I have thought about using Euclidean algorithm, but the $a_i$s are polynomials in $x_1,ldots,x_n$ rather than constants. I am not sure how the division is done.
abstract-algebra symmetric-polynomials
asked Jan 23 at 5:01
William McGonagallWilliam McGonagall
1547
$endgroup$
I was reading Emil Artin's Galois Theory (2nd edition). On pp.39-40, Artin defines a number of things as follows.
Let $k$ be a field and $E=k(x_1,ldots,x_n)$ be the field of all rational functions of $n$ indeterminates $x_1,ldots,x_n$. Define a polynomial
$$
f(t)=(t-x_1)cdots(t-x_n)=t^n+a_1t^{n-1}+cdots+a_n,tag{3}
$$
where $a_1=-sum_ix_i,,a_2=sum_{i<j}x_ix_j$,... are elementary symmetric functions in $x_1,ldots,x_n$. Then he puts $F_n(t)=f(t)$ and also for $i=n-1,,n-2,ldots,1$,
$$
F_i(t)=frac{f(t)}{(t-x_{i+1})(t-x_{i+2})cdots(t-x_n)}=frac{F_{i+1}(t)}{t-x_{i+1}}.tag{5}
$$
(If I am not mistaken, this simply means that $F_i(t)=(t-x_1)(t-x_2)..(t-x_i)$ for each $i$, but what matters below is the division on the RHS of $(5)$.) He then writes that (emphasis mine):
Performing the division [in $(5)$] we see that $F_i(t)$ is a polynomial in $t$ of degree $i$ whose highest coefficient is $1$ and whose coefficients are polynomials in the variables $a_1,a_2,ldots,a_i$ and $x_{i+1},x_{i+2},ldots,x_n$. Only integer enter as coefficients in these expressions.
Can someone explain the bold part? I have thought about using Euclidean algorithm, but the $a_i$s are polynomials in $x_1,ldots,x_n$ rather than constants. I am not sure how the division is done.
abstract-algebra symmetric-polynomials
abstract-algebra symmetric-polynomials
asked Jan 23 at 5:01
William McGonagallWilliam McGonagall
1547
asked Jan 23 at 5:01
William McGonagallWilliam McGonagall
1547
edited Jan 26 at 1:17
William McGonagall
asked Jan 23 at 5:01
William McGonagallWilliam McGonagall
1547
asked Jan 23 at 5:01
William McGonagallWilliam McGonagall
1547
asked Jan 23 at 5:01
William McGonagallWilliam McGonagall
1547
1547
add a comment |
add a comment |
1 Answer
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You just perform polynomial division as if the symbols were numbers. For example, you can check that
$$
(t^3+a_1t^2+a_2t+a_3)div(t-x_3)=t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]
$$
and
$$
left(t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]right)div(t-x_2)=t+(a_1+x_2+x_3).
$$
answered Jan 23 at 5:19
David HillDavid Hill
9,2261619
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$begingroup$
That makes a lot of sense. Thank you very much!
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– William McGonagall
Jan 23 at 5:35
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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$begingroup$
You just perform polynomial division as if the symbols were numbers. For example, you can check that
$$
(t^3+a_1t^2+a_2t+a_3)div(t-x_3)=t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]
$$
and
$$
left(t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]right)div(t-x_2)=t+(a_1+x_2+x_3).
$$
answered Jan 23 at 5:19
David HillDavid Hill
9,2261619
$endgroup$
$begingroup$
That makes a lot of sense. Thank you very much!
$endgroup$
– William McGonagall
Jan 23 at 5:35
add a comment |
$begingroup$
You just perform polynomial division as if the symbols were numbers. For example, you can check that
$$
(t^3+a_1t^2+a_2t+a_3)div(t-x_3)=t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]
$$
and
$$
left(t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]right)div(t-x_2)=t+(a_1+x_2+x_3).
$$
answered Jan 23 at 5:19
David HillDavid Hill
9,2261619
$endgroup$
$begingroup$
That makes a lot of sense. Thank you very much!
$endgroup$
– William McGonagall
Jan 23 at 5:35
add a comment |
$begingroup$
You just perform polynomial division as if the symbols were numbers. For example, you can check that
$$
(t^3+a_1t^2+a_2t+a_3)div(t-x_3)=t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]
$$
and
$$
left(t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]right)div(t-x_2)=t+(a_1+x_2+x_3).
$$
answered Jan 23 at 5:19
David HillDavid Hill
9,2261619
$endgroup$
You just perform polynomial division as if the symbols were numbers. For example, you can check that
$$
(t^3+a_1t^2+a_2t+a_3)div(t-x_3)=t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]
$$
and
$$
left(t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]right)div(t-x_2)=t+(a_1+x_2+x_3).
$$
answered Jan 23 at 5:19
David HillDavid Hill
9,2261619
answered Jan 23 at 5:19
David HillDavid Hill
9,2261619
answered Jan 23 at 5:19
David HillDavid Hill
9,2261619
answered Jan 23 at 5:19
David HillDavid Hill
9,2261619
9,2261619
$begingroup$
That makes a lot of sense. Thank you very much!
$endgroup$
– William McGonagall
Jan 23 at 5:35
add a comment |
$begingroup$
That makes a lot of sense. Thank you very much!
$endgroup$
– William McGonagall
Jan 23 at 5:35
$begingroup$
That makes a lot of sense. Thank you very much!
$endgroup$
– William McGonagall
Jan 23 at 5:35
$begingroup$
That makes a lot of sense. Thank you very much!
$endgroup$
– William McGonagall
Jan 23 at 5:35
add a comment |
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