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I've been reading up on Newton's Identities on wikipedia (https://en.wikipedia.org/wiki/Newton's_identities) where the section 'Mathematical Statement' first defines $p_k(x_1,\ldots,x_n)=\sum_{i=1}^n x_i^k = x_1^k+\cdots+x_n^k$ and the elementary symmetric polynomial $$ e_0(x_1, \ldots, x_n) = 1,\\ e_1(x_1, \ldots, x_n) = x_1 + x_2 + \cdots + x_n,\\ e_2(x_1,\ldots,x_n) = \sum_{1 \leq i<j\leq n}x_ix_j,\\ \;\;\vdots \\ e_n(x_1, \ldots, x_n) = x_1 x_2 \cdots x_n,\\ e_k(x_1, \ldots, x_n) = 0, \quad\text{for}\ k>n.\\ $$ Then it states that Newton's identities can be used to obtain the following identities in the ring of symmetric functions $$ e_1 = p_1,\\ 2e_2 = e_1p_1-p_2 = p_1^2-p_2,\\ 3e_3 = e_2p_1 - e_1p_2 + p_3 = \tfrac12 p_1^3-\tfrac32p_1p_2+p_3,\\ 4e_4 = e_3p_1 - e_2p_2 + e_1p_3 - p_4 = \tfrac16p_1^4 - p_1^2p_2 + \tfrac43p_1p_3+\tfrac12p_2^2-p_4,\\ $$ (and so on) where they state that

the left-hand sides never become zero.

My question is: Why is the left-hand side never zero? From the way the section is written, I would have thought that the statement says $e_k$ are non-zero for all $k\leq n$, but I do not understand why this should be the case.

dips_123
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1 Answers1

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Yes, $e_k$ can be zero. In fact, if there are $n$ variables, $x_1,x_2,...,x_n$ meaning that $x_{k}=0$ for $k\geq n+1$, then we have $e_k=0$ for all $k\geq n+1$ by definition of the elementary symmetric polynomials $e_k$.

However, if this happens, Newton identities, in other words Newton-Girard identities still hold. For example, if there are two variables, $x_1=x$ and $x_2=y$, then $$p_3\stackrel{?}{=}p_2e_1-p_1e_1+3e_3$$ $$\small x^3+y^3\stackrel{}{=}(x^2+y^2)(x+y)-(x+y)xy+\color{red}{0}$$ which checks.

Bob Dobbs
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