On coefficients of $(t-x_1)(t-x_2)..(t-x_i)$












0












$begingroup$


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







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    0












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






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      edited Jan 26 at 1:17







      William McGonagall

















      asked Jan 23 at 5:01









      William McGonagallWilliam McGonagall

      1547




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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
          $$
          left(t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]right)div(t-x_2)=t+(a_1+x_2+x_3).
          $$






          share|cite|improve this answer









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          • $begingroup$
            That makes a lot of sense. Thank you very much!
            $endgroup$
            – William McGonagall
            Jan 23 at 5:35











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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
          $$
          left(t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]right)div(t-x_2)=t+(a_1+x_2+x_3).
          $$






          share|cite|improve this answer









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          • $begingroup$
            That makes a lot of sense. Thank you very much!
            $endgroup$
            – William McGonagall
            Jan 23 at 5:35
















          1












          $begingroup$

          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
          $$
          left(t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]right)div(t-x_2)=t+(a_1+x_2+x_3).
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That makes a lot of sense. Thank you very much!
            $endgroup$
            – William McGonagall
            Jan 23 at 5:35














          1












          1








          1





          $begingroup$

          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
          $$
          left(t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]right)div(t-x_2)=t+(a_1+x_2+x_3).
          $$






          share|cite|improve this answer









          $endgroup$



          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
          $$
          left(t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]right)div(t-x_2)=t+(a_1+x_2+x_3).
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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