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So e₀ = 1 when the number of variables is at least 1. What if the number of variables is 0? Is e₀ equal to 0 in that case? Michael Hardy ( talk) 23:26, 7 September 2009 (UTC) reply

Well, if you define the elementary symmetric polynomials by

${\displaystyle e_{k}(X_{1},\ldots ,X_{n})=\sum _{P\subseteq \{1,\ldots ,n\}:|P|=k}\left(\prod _{i\in P}X_{i}\right),}$

then when n and k are 0, we have a sum over the subsets of the empty set, and the only term in the sum is a product over the empty set, which is by convention 1.

Forgetfulfunctor ( talk) 00:35, 8 September 2009 (UTC) reply

Why is ${\displaystyle e_{2}(X_{1},X_{2},\dots ,X_{n})=\textstyle \sum _{1\leq j<k\leq n}X_{j}X_{k}}$ instead of ${\displaystyle e_{2}(X_{1},X_{2},\dots ,X_{n})=\textstyle \sum _{1\leq j\leq k\leq n}X_{j}X_{k}}$? KlappCK ( talk) 14:35, 13 May 2011 (UTC) reply

Because per definition elementary symmetric polynomials are sums of products of distinct variables; or equivalently sums of square-free monomials. If you do want all monomials ofa given degree as terms, you get the complete homogeneous symmetric polynomials instead, which are quite interesting as well. Marc van Leeuwen ( talk) 04:55, 14 May 2011 (UTC) reply

Marc van Leeuwen, thanks for the insight. I think it would be beneficial to (in some way) emphasize this fact in the header, as I apparently overlooked that distinction the first time through. KlappCK ( talk) 17:03, 16 May 2011 (UTC) reply

Hi,

the following sentence is very difficult to understand:

"Clearly Q is a polynomial in the symmetric polynomials. Moreover, ${\displaystyle x_{n}\,\!}$ occurs with exponent ${\displaystyle i_{n}}$, since it occurs with exponent ${\displaystyle i_{n}-i_{n-1}\,\!}$ in the first term, ${\displaystyle i_{n-1}-i_{n-2}\,\!}$ in the second term, and so on, down to ${\displaystyle i_{1}\,\!}$ times in the final term. Moreover, ${\displaystyle x_{n}^{i_{n}}\,\!}$ only occurs in a single monomial in Q, and in that monomial, ${\displaystyle x_{n-1}\,\!}$ occurs with exponent ${\displaystyle i_{n-1}\,\!}$, since it does not include a term from ${\displaystyle s_{1}\,\!}$, and it occurs in the rest of Q with exponent ${\displaystyle i_{n-1}-i_{n-2}\,\!}$ in the second term, ${\displaystyle i_{n-2}-i_{n-3}\,\!}$ in the third term, and so on, down to ${\displaystyle i_{1}\,\!}$ times in the final term. And so on for the remaining variables. This shows that ${\displaystyle P-Q\,\!}$ has monomials that are smaller than the one just eliminated. We can now continue the process until nothing remains in P."

I believe that "it" here refers to ${\displaystyle x_{n-1}\,\!}$, but even so it is difficult to understand what's meant. What are the terms here? Wisapi ( talk) 16:22, 26 May 2011 (UTC) reply

I think I have made this proof more readable now. -- Darij ( talk) 01:27, 29 October 2011 (UTC) reply

I'm feeling stupid for asking this since I have supposedly proofread all of this page a year or so ago. But why do we know in the First Proof that "${\displaystyle R(X_{1},\ldots ,X_{n})}$ is a symmetric polynomial in X₁, ..., X_n, of the same degree as ${\displaystyle P_{\mbox{lacunary}}}$"? I mean the "same degree" part. Theoretically, couldn't it happen that ${\displaystyle {\tilde {Q}}}$ is a polynomial of huge degree, but all of its high-degree terms cancel out when it is applied to ${\displaystyle (\sigma _{1,n-1},\ldots ,\sigma _{n-1,n-1})}$, whereas applying it to ${\displaystyle (\sigma _{1,n},\ldots ,\sigma _{n-1,n})}$ does not cancel them out? This is of course made impossible by the algebraic independency of the elementary symmetric polynomials, but that's not supposed to be known in this proof. -- Darij ( talk) 01:30, 28 March 2013 (UTC) reply