Brian Bi

## Section 16.1. Symmetric Functions

Exercise 16.1.1

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.
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 $$(u_1 + u_2)(u_2 + u_3)(u_3 + u_1) = s_1 s_2 - s_3$$.
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.
4. Similarly to (c), this polynomial is invariant under 3-cycles but not transpositions, and a transposition converts it to its additive inverse.
5. 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

1. 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 $$c = 0$$.

2. 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 $$s_2 = \frac{1}{2}(w_1^2 - w_2)$$.