Relation of complete homogeneous symmetric polynomials and the elementary symmetric polynomials












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I was reading about the symmetric polynomials and saw the following relation:




$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$




The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
SECOND EDITION I. G. MACDONALD, page: 21.



(Question) I was wondering if this relation still holds for any length $k$, i.e.




$$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$




P.S. I tried cases and it holds for any case.






abstract-algebra commutative-algebra symmetric-polynomials







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$endgroup$

















    4












    $begingroup$


    I was reading about the symmetric polynomials and saw the following relation:




    $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$




    The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
    SECOND EDITION I. G. MACDONALD, page: 21.



    (Question) I was wondering if this relation still holds for any length $k$, i.e.




    $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$




    P.S. I tried cases and it holds for any case.






    abstract-algebra commutative-algebra symmetric-polynomials







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      I was reading about the symmetric polynomials and saw the following relation:




      $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$




      The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
      SECOND EDITION I. G. MACDONALD, page: 21.



      (Question) I was wondering if this relation still holds for any length $k$, i.e.




      $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$




      P.S. I tried cases and it holds for any case.






      abstract-algebra commutative-algebra symmetric-polynomials







      share|cite|improve this question











      $endgroup$




      I was reading about the symmetric polynomials and saw the following relation:




      $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{n})=0text{ for } m>0$$




      The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
      SECOND EDITION I. G. MACDONALD, page: 21.



      (Question) I was wondering if this relation still holds for any length $k$, i.e.




      $$sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},ldots ,X_{n})h_{{m-i}}(X_{1},ldots ,X_{k})=0text{ for }k in {1,...,n}$$




      P.S. I tried cases and it holds for any case.






      abstract-algebra commutative-algebra symmetric-polynomials




      abstract-algebra commutative-algebra symmetric-polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 17 at 11:06







      Max

















      asked Feb 3 at 10:11









      MaxMax

      1,0041319




      1,0041319






















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          $begingroup$

          We have
          $$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
          $$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
          $$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
          If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.






          share|cite|improve this answer











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            $begingroup$

            We have
            $$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
            $$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
            $$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
            If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.






            share|cite|improve this answer











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              2












              $begingroup$

              We have
              $$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
              $$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
              $$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
              If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                We have
                $$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
                $$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
                $$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
                If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.






                share|cite|improve this answer











                $endgroup$



                We have
                $$h_p(X_1,ldots,X_n)=sum h_r(X_1,ldots,X_k)h_{p-r}(X_{k+1},ldots,X_n)$$
                $$e_p(X_1,ldots,X_n)=sum e_r(X_1,ldots,X_k)e_{p-r}(X_{k+1},ldots,X_n)$$
                $$sum_{i=0}^msum_{p=0}^isum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,ldots,X_k)e_p(X_{k+1},ldots,X_n)h_q(X_{k+1},ldots,X_n)$$
                If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 3 at 19:11

























                answered Feb 3 at 18:48









                Matt SamuelMatt Samuel

                39.3k63870




                39.3k63870






























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