*Algebra*

## Section 16.1. Symmetric Functions

Exercise 16.1.1

- The polynomial is not symmetric because it is not invariant under a tranposition; the other polynomial in the orbit is \(u_2^2 u_1 + u_3^2 u_2 + u_1^2 u_3\). But it is invariant under a 3-cycle, so the orbit has size 2.
- The polynomial is symmetric. According to the systematic procedure, we first set \(u_3 = 0\), leaving \((u_1 + u_2)u_2 u_1\), which is just the product \(s_1 s_2\) with \(n = 2\). For \(n = 3\), we have \(s_1 s_2 = (u_1 + u_2 + u_3)(u_1 u_2 + u_2 u_3 + u_3 u_1)\). We know this differs from our target polynomial by a multiple of \(s_3 = u_1 u_2 u_3\). Since the target polynomial and \(s_1 s_2\) both have degree 3, the multiple of \(s_3\) must be a constant multiple. By inspection, the target polynomial contains the term \(u_1 u_2 u_3\) twice and \(s_1 s_2\) contains it three times. Therefore \((u_1 + u_2)(u_2 + u_3)(u_3 + u_1) = s_1 s_2 - s_3\).
- Similarly to (a), this polynomial is invariant under 3-cycles but not transpositions. (This is more obvious once we replace \(u_1 - u_3\) by \(-(u_3 - u_1)\).) The other polynomial in the orbit is \((u_2 - u_1)(u_1 - u_3)(u_2 - u_3)\), which is the additive inverse of the original polynomial.
- Similarly to (c), this polynomial is invariant under 3-cycles but not transpositions, and a transposition converts it to its additive inverse.
- It is obvious that this polynomial is symmetric. Instead of the systematic procedure described earlier in the text, which requires a number of steps equal to the number of variables, we will take a different approach. Observe that \[ s_1^3 = (u_1 + \ldots + u_n)^3 = \sum_i u_i^3 + 3\sum_{i < j} (u_i^2 u_j + u_i u_j^2) + 6\sum_{i < j < k} u_i u_j u_k \] To eliminate the second term we look at: \[ s_1 s_2 = (\sum_i u_i)(\sum_{i < j} u_i u_j) = \sum_{i < j} (u_i^2 u_j + u_i u_j^2) + 3\sum_{i < j < k} u_i u_j u_k \] therefore, \[ s_1^3 - 3s_1 s_2 = \sum_i u_i^3 - 3\sum_{i < j < k} u_i u_j u_k \] so \[ \sum_i u_i^3 = s_1^3 - 3s_1 s_2 + 3s_3 \]

Exercise 16.1.2 As mentioned in the text, the orbit sums form one basis. Another basis is given by monomials evaluated at elementary symmetric polynomials; this follows from Theorem 16.1.6 and the fact that monomials form a basis for polynomials.

Exercise 16.1.3

We will explicitly compute the coefficient of each monomial in \(P = ks_k + \sum_{i=1}^k (-1)^i w_i s_{k-i}\) when expanded in terms of the \(u_i\)'s, and show that all such coefficients vanish, therefore \(P = 0\). (Note that \(s_i = 0\) for \(i > n\), and \(s_0 = 1\), and \(P\) is equal to \((-1)^k\) times the polynomial given in the problem text.)

Let \(d_1, \ldots, d_n\) be nonnegative integers. We now compute the coefficient \(c\) of \(\prod_{j=1}^n u_j^{d_j}\). First, observe that \(P\) is homogeneous of degree \(k\), so if \(\sum_j d_j \ne k\), then \(c = 0\).

Now, consider a term in \(w_i\), namely some \(u_j^i\). When this is multiplied by a term in \(s_{k-i}\), either the degree of \(u_j\) in the latter is 0, or it is 1. If it is 0, then in the product term, \(u_j\) has exponent \(i\), \(k-i\) other variables have exponent 1, and the remaining \(n-i-1\) variables have exponent 0, for a total of \(k-i+1\) variables with nonzero exponents. Otherwise, in the product term, \(u_j\) has exponent \(i+1\), \(k-i-1\) other variables have exponent 1, and the remaining \(n-i\) variables have exponent 0, for a total of \(k-i\) variables with nonzero exponents. Therefore, (1) if more than one of the \(d_j\)'s exceeds 1, then \(c\) vanishes; and (2) a monomial in which exactly \(i\) variables have nonzero exponents only occurs in \(w_{k-i}s_i\) and \(w_{k-i+1} s_{i-1}\).

Suppose all \(d_j\)'s are equal to 0 or 1. Then exactly \(k\) of them are equal to 1 and the monomial occurs \(k\) times in \(ks_k\) and \(k\) times in \(s_{k-1}w_1\). These two terms have opposite signs in \(P\), so \(c = 0\).

Suppose on the other hand that there is some \(j\) such that \(d_j\) exceeds 1. We have already argued that there can only be one such \(j\) or else \(c = 0\). Thus, \(k - d_j\) other variables have exponent 1, and a total of \(k - d_j + 1\) variables have nonzero exponent. This monomial therefore only occurs in \(w_{d_j - 1} s_{k- d_j + 1}\) and \(w_{d_j} s_{k - d_j}\). In the former, it occurs once, and in the latter it also occurs once. These two terms have opposite sign in \(P\), so \(c = 0\).

Yes, if the ground ring for the symmetric polynomials is a field. Newton's identity with \(k = 1\) shows that \(s_1 = w_1\). Substituting this into the identity with \(k = 2\) shows that \(2s_2\) lies in the ideal generated by \(w_1, w_2\), so if 2 has an inverse, then \(s_2\) lies in this ideal. Substituting the expressions for \(s_1\) and \(s_2\) into the identity with \(k = 3\) shows that \(3s_3\) lies in the ideal generated by \(w_1, w_2, w_3\), and so on. Thus, the subring generated by \(w_1, \ldots, w_n\) contains \(s_1, \ldots, s_n\), so it is the entire ring of symmetric functions.

If the ground ring is not a field, then the power sums might not generate the ring of symmetric functions. The simplest example is that in \(\Z/2\Z\) with \(n = 2\), there is no way to write \(s_2 = u_1 u_2\) in terms of the power sums \(w_1 = u_1 + u_2, w_2 = u_1^2 + u_2^2\), since \(w_1(1, 1) = w_2(1, 1) = 0\) but \(s_2(1, 1) \ne 0\). If 2 had an inverse, we would write \(s_2 = \frac{1}{2}(w_1^2 - w_2)\).