# Constructing the sum of two pairs of symmetric polynomials

I have a question about generating a certain set of symmetric polynomials. I believe that what I'm looking for is known result, but I'm not 100% sure.

Suppose that I have two sets of indeterminates $X_1,\ldots,X_n$ and $Y_1, \ldots, Y_n$, of the same (finite) cardinality. Now let $e_i(A,B,C,\ldots)$ denote the $i$th elementary symmetric polynomial on $A,B,C,\ldots$ and define, $$A_i = e_i(X_1,\ldots,X_n) + e_i(Y_1,\ldots,Y_n)$$

Now define the following:

1. $$B_{i} = X_i + Y_i$$
2. $$C_{i} = \left(\prod_{j\neq i} X_j\right) + \left(\prod_{j\neq i} Y_j \right)$$

Note that $\sum_{i=1}^n B_i = A_1, \sum_{i=1}^n C_i = A_{n-1}$.

I was wondering the following:

If I am given $B_i, C_i, \forall i$ and $A_n$, can I construct $A_i$ for all $i \in\{2,\ldots,n-2\}$?

I'm not sure if it is possible, but I think we should be able to construct a basis for the symmetric polynomials with the $B_i, C_i$ and $A_n$, and as such we should be able to get $A_i, \forall i$. I must admit, I don't have much experience with these type of problems, so I might be barking up the wrong tree.

Thanks!

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What you want is not possible. Suppose $n>3$, and consider making $A_2=\sum_{i < j}(X_iX_j+Y_iY_j)$. All polynomials available are homogeneous in the total degree, and the degree of the $C_i$ is too high to be of any use, so you'll have to make do with the $B_i$. But the only way to get a monomial $X_iX_j$ is to multiply $B_iB_j$; however that gives you mixed terms $X_iY_j$ and $X_jY_i$ as well, which you don't want. There is no way to get only the terms in $A_2$.

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Thanks a lot! I think I sort of guessed this result (I stared at the $n=4$ case for a while), but I wasn't totally confident. –  Tarun Chitra Dec 1 '11 at 5:14

Explicitly: Let the $X_i$ be $0,0,a,b$ and the $Y_i$ be $0,0,-a,-b$. Then all the $B_i$ and $C_i$ are $0$ but $A_2$ is $2ab$, which varies.

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