置换群$G$不变多项式的基

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三元轮换对称多项式的基本定理及其应用 林才雄,吴康    豆丁
定义 1 (三元轮换对称多项式) 若关于变元 $x, y, z$ 的多项式 $f(x, y, z)$ 是数环 $R$ 上的一个三元多项式, 且满足关系 $f(x, y, z)=f(y, z, x)=f(z, x, y)$, 则称多项式 $f(x, y, z)$ 是 关于变元 $x, y, z$ 的三元轮换对称多项式.
例如, $f(x, y, z)=x^2 y+y^2 z+z^2 x$ 是整数环上的一个三元轮换对称多项式.
定理 1（对称多项式基本定理） 数环 $R$ 上的每一个 $n$ 元对称多项式 $f(x_1, x_2, \cdots,x_n)$ 都可以表示成初等对称多项式 $\sigma_1, \sigma_2, \cdots, \sigma_n$ 的系数在 $R$ 中的多项式, 并且这种表示法是唯一的.
定理 2（三元轮换对称多项式基本定理）关于变元 $x, y, z$ 在数环 $R$ 上的任一三元轮换对称多项式 $f(x, y, z)$, 都可以表示成三元初等对称多项式 $\sigma_1, \sigma_2, \sigma_3$ 外加一个轮换对称式的系数在 $R$ 中的多项式, 即可以用
$$
\sigma_1=\sum x, \sigma_2=\sum x y, \sigma_3=\sum x y z, \sigma_{31}=\sum x^2 y
$$
表示出来. 我们称 $(*)$ 式为三元初等轮换对称式.
证明：我们对 $f(x, y, z)$ 的次数 $n$ 进行数学归纳法.
(1) 当 $n=1$ 时, $f(x, y, z)=a \sum x$, 则令 $g\left(\sigma_1, \sigma_2, \sigma_3, \sigma_{31}\right)=a \sigma_1$ 即可;
当 $n=2$ 时, $f(x, y, z)=a \sum x^2+b \sum x y+c \sum x$, 则令
$g\left(\sigma_1, \sigma_2, \sigma_3, \sigma_{31}\right)=a \sigma_1^2+(b-2 a) \sigma_2+c \sigma_3$ 即可;
当 $n=3$ 时, $f(x, y, z)=a \sum x^3+b \sum x^2 y+c \sum x y^2+d x y z+e \sum x^2+p \sum x y+\mathrm{q} \sum x$,
则令
$$g\left(\sigma_1, \sigma_2, \sigma_3, \sigma_{31}\right)=a \sigma_1^3+e \sigma_1^2+(c-3 a) \sigma_1 \sigma_2+q \sigma_1+(p-2 e) \sigma_2+(3 a-3 c+d) \sigma_3+(b-c) \sigma_{31}$$即可;
(2) 假设命题对 $f(x, y, z)$ 的次数 $n \leqslant k-1, k \geqslant 4, k \in N$ 时成立, 则当次数为 $n=k$ 时, 设多项式 $f(x, y, z)$ 按字典排序法的最前项是 $\lambda x^{\alpha_1} y^{\alpha_2} z^{\alpha_3}$, 则 $\alpha_1+\alpha_2+\alpha_3=k$, 且 $\alpha_1 \geqslant \alpha_2 \geqslant \alpha_3$. 不妨假定 $f(x, y, z)$ 只含有 $\lambda x^{\alpha_1} y^{\alpha_2} z^{\alpha_3}$ 且 $\alpha_1 \geqslant \alpha_2 \geqslant \alpha_3$, 因为
$$
\begin{gathered}
\sum x^{\alpha_1} y^{\alpha_2} z^{\alpha_3}+\sum x^{\alpha_1} y^{\alpha_1 z^{\alpha_2}}=\left(x^{\alpha_1} y^{\alpha_2} z^{\alpha_3}+y^{\alpha_1} z^{\alpha_2} x^{\alpha_3}+z^{\alpha_1} x^{\alpha_2} y^{\alpha_3}\right)+\left(x^{\alpha_1} y^{\alpha_3 z^{\alpha_2}}+y^{\alpha_1} z^{\alpha_3} x^{\alpha_2}+z^{\alpha_1} x^{\alpha_3} y^{\alpha_2}\right) \\
=\sum x^{\alpha_1} y^{\alpha_1} z^{\alpha_3}\left(y^{\alpha_2-\alpha_3}+z^{\alpha_2-\alpha_3}\right)=\sum x^{\alpha_2-\alpha_3}\cdot \sum x^{\alpha_1 y^{\alpha_1} z^{\alpha_3}}-\sum x^{\alpha_1+\alpha_2-\alpha_3} y^{\alpha_1} z^{\alpha_3},
\end{gathered}
$$
其中 $\sum x^{\alpha_2-\alpha_3}, \sum x^{\alpha_1} y^{\alpha_3} z^{\alpha_3}, \sum x^{\alpha_1+\alpha_2-\alpha_3} y^{\alpha_3} z^{\alpha_3}$ 都是对称多项式, 所以由对称多项式的基本定理可得, 存在一个数环 $R$ 上的多项式 $g(x, y, z)$, 使得
$$
g\left(\sigma_1, \sigma_2, \sigma_3\right)=\sum x^{\alpha_2-\alpha_3}\cdot\sum x^{\alpha_1} y^{\alpha_1} z^{\alpha_3}-\sum x^{\alpha_1+\alpha_2-\alpha_3} y^{\alpha_3} z^{\alpha_3},
$$
下面, 我们分两种情况来讨论:
(1)若 $\alpha_1-\alpha_2 \geqslant \alpha_2-\alpha_3$, 则令
$$
\rho_1\left(\sigma_1, \sigma_2, \sigma_3, \sigma_{31}\right)=\lambda \sigma_1^{\left(\alpha_1-\alpha_2\right)-\left(\alpha_2-\alpha_1\right)} \sigma_3^{\alpha^\alpha} \sigma_{31}^{\alpha_2-\alpha_1} \text {, }
$$
此时有, $\rho_1\left(\sigma_1, \sigma_2, \sigma_3, \sigma_{31}\right)$ 的最前项为
$$
\lambda x^{\left(\alpha_1-\alpha_2\right)-\left(\alpha_2-\alpha_1\right)}(x y z)^{\alpha_3}\left(x^2 y\right)^{\alpha_2-\alpha_3}=\lambda x^{\alpha_1 \alpha_1 \alpha_2 z^{\alpha_3}} .
$$
(2)若 $\alpha_1-\alpha_2<\alpha_2-\alpha_3$, 则令
$$
\rho_1\left(\sigma_1, \sigma_2, \sigma_3, \sigma_{31}\right)=\lambda \sigma_2^{\left(\alpha_2-\alpha_{)}\right)-\left(\alpha_1-\alpha_2\right)} \sigma_3^{\alpha_1} \sigma_{31}^{\alpha_1-\alpha_2},
$$
此时有, $\rho_1\left(\sigma_1, \sigma_2, \sigma_3, \sigma_{31}\right)$ 的最前项为
$$
\lambda(x y)^{\left(\alpha_2-\alpha_3\right)-\left(\alpha_1-\alpha_2\right)}(x y z)^{\alpha_3}\left(x^2 y\right)^{\alpha_1-\alpha_2}=\lambda x^{\alpha_1 y^{\alpha_2} z^{\alpha_3}} .
$$
我们对多项式 $f_1(x, y, z)$ 继续上述步骤, 则经过有限步之后, 存在某个多项式 $f_m(x, y, z)=f_{m-1}(x, y, z)-\rho_m\left(\sigma_1, \sigma_2, \sigma_3, \sigma_{31}\right), m \in N^{+}$是 $k-1$ 次的, 于是, 由归纳假设可得, 存在多项式 $h(x, y, z, u)$, 使得
$$
f_m(x, y, z)=f_{m-1}(x, y, z)-\rho_m\left(\sigma_1, \sigma_2, \sigma_3, \sigma_{31}\right)=h\left(\sigma_1, \sigma_2, \sigma_3, \sigma_{31}\right)
$$于是有\begin{aligned} f(x, y, z) & =f_{1}(x, y, z)+\rho_{1}\left(\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{31}\right) \\ & =f_{2}(x, y, z)+\rho_{2}\left(\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{31}\right)+\rho_{1}\left(\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{31}\right) \\ & =\cdots=f_{m}(x, y, z)+\rho_{m}\left(\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{31}\right)+\cdots+\rho_{1}\left(\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{31}\right) \\ & =h\left(\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{31}\right)+\rho_{m}\left(\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{31}\right)+\cdots+\rho_{1}\left(\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{31}\right)\end{aligned}所以, $f(x, y, z)$ 表示成关于 $\sigma_1, \sigma_2, \sigma_3, \sigma_{31}$ 的多项式. 即命题对 $\mathrm{n}=k$ 时也成立.
故根据(1)、(2), 由数学归纳法, 可知:
关于变元 $x, y, z$ 在数环 $R$ 上的任一三元轮换对称多项式 $f(x, y, z)$, 都可以表示成三元初等对称多项式 $\sigma_1, \sigma_2, \sigma_3$ 外加一个轮换对称式的系数在 $R$ 中的多项式, 即可以用 $\sigma_1=\sum x, \sigma_2=\sum x y, \sigma_3=\sum x y z, \sigma_{31}=\sum x^2 y$ 表示出来.

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例 2、设 $x^3-2 x^2+x-3=0$ 的三个根分别为 $x_1, x_2, x_3$, 求值:
(1) $f\left(x_1, x_2, x_3\right)=x_1^2 x_2+x_2^2 x_3+x_3^2 x_1$;
(2) $f\left(x_1, x_2, x_3\right)=x_1^3 x_2+x_2^3 x_3+x_3^3 x_1$;
(3) $f\left(x_1, x_2, x_3\right)=\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_1}$.
解: 由题设可得, $\sigma_1=\sum x_1=2, \sigma_2=\sum x_1 x_2=1, \sigma_3=\sum x_1 x_2 x_3=3$,
(1) 显然, $f\left(x_1, x_2, x_3\right)=x_1^2 x_2+x_2^2 x_3+x_3^2 x_1=\sum x_1^2 x_2$,
构造 $f\left(x_1, x_2, x_3\right)=\sum x_1^2 x_3$, 则有
$$
\begin{gathered}
\sum x_1^2 x_3+\sum x_1^2 x_2=\sigma_1 \sigma_2-3 \sigma_3=-7 \\
\sum x_1^2 x_3 \cdot \sum x_1^2 x_2=\sigma_1^3 \sigma_3+\sigma_2^3-6 \sigma_1 \sigma_2 \sigma_3+9 \sigma_3^2=70
\end{gathered}
$$
于是由韦达定理, 可解得 $\sum x_1^2 x_2=\frac{-7 \pm i \sqrt{231}}{2}, \sum x_1^2 x_3=\frac{-7 \mp i \sqrt{231}}{2}$, 即 $f\left(x_1, x_2, x_3\right)=\sum x_1^2 x_2=\frac{-7 \pm i \sqrt{231}}{2}$.
(2) 由(1) 可知 $\sigma_{31}=\sum x_1^2 x_2=\frac{-7 \pm i \sqrt{231}}{2}$, 由例 1 知,
$$
\begin{aligned}
& f\left(x_1, x_2, x_3\right)=\sum x_1{ }^3 x_2=\sigma_1 \sigma_{31}+\sigma_1 \sigma_3-\sigma_2^2 \\
&=2 \times \frac{-7 \mp i \sqrt{231}}{2}+2 \times 3-1^2=-2 \pm i \sqrt{231} .
\end{aligned}
$$
(3) 因为 $f\left(x_1, x_2, x_3\right)=\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_1}=\frac{x_1^2 x_3}{x_1 x_2 x_3}+\frac{x_1 x_2^2}{x_1 x_2 x_3}+\frac{x_2 x_3^2}{x_1 x_2 x_3}$
$$
=\frac{1}{\sigma_3} \sum x_1^2 x_3
$$
由(1)可知, 有 $\sum x_1^2 x_3=\frac{-7 \mp i \sqrt{231}}{2}$,
因此可得 $f\left(x_1, x_2, x_3\right)=\sum \frac{x_1}{x_2}=\frac{1}{3} \times \frac{-7 \mp i \sqrt{231}}{2}=\frac{-7 \mp i \sqrt{231}}{6}$.

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MathWorld–Fundamental Theorem of Symmetric Functions
There is a generalization of this theorem to polynomial invariants of permutation groups $G$, which states that any polynomial invariant $f$ in $R[X_1,...,X_n]$ can be represented as a finite linear combination of special $G$-invariant orbit polynomials with symmetric functions as coefficients, i.e.,
$$f=\sum_{t\text{ special}}p_t(\sigma_1,...,\sigma_n)\text{orbit}_G(t),$$
where $p_t$ in $R[X_1,...,X_n]$,
$$\text{orbit}_G(t)=\sum_{s\in\set{\pi(t)|\pi\in G}}s,$$
and $\sigma_1, ..., \sigma_n$ are elementary symmetric functions, and $t=X_1^{e_1}\cdots X_n^{e_n}$ are special terms. Furthermore, any special term $t$ has a total degree $\le{n(n-1)\over2}$, and a maximal variable degree $\le n-1$.

References
Göbel, M. "Computing Bases for Permutation-Invariant Polynomials." J. Symb. Comput. 19, 285-291, 1995.
Computing Bases for Permutation-Invariant Polynomials.pdf (224.74 KB, Downloads: 37)

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Göbel, M. "On the Number of Special Permutation-Invariant Orbits and Terms." Appl. Algebra Eng. Comm. Comput. 8, 505-509, 1997.

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1#所得的结论与Analogues of the elementary symmetric polynomials for the alternating group完全相同
@Ewan Delanoy 的回答:

You're almost there. It suffices to add a fourth polynomial :
\begin{array}{lcl}
P_1 &=& X_1+X_2+X_3 \\
P_2 &=& X_1X_2+X_1X_3+X_2X_3 \\
P_3 &=& X_1X_2X_3 \\
P_4 &=& X_1^2X_2+X_2^2X_3+X_3^2X_1
\end{array}Suppose that $X=(X_1,X_2,X_3)$ and $Y=(Y_1,Y_2,Y_3)$ satisfy $P_n(X)=P_n(Y)$ for $1\leq n \leq 4$. Using the first three equalities only, we already know that there is a $\sigma\in {\mathfrak S}_n$ such that $X=Y_\sigma$ (by which I mean that $X_i=Y_{\sigma(i)}$ for $1\leq i \leq 3$).

If $\sigma$ is already even, we are done. Otherwise, $\sigma$ is a transposition, say
$\sigma=(1,2)$. Then the last identity becomes $P_4(Y_2,Y_1,Y_3)=P_4(Y_1,Y_2,Y_3)$. Now the polynomial $D=P_4(Y_2,Y_1,Y_3)-P_4(Y_1,Y_2,Y_3)$ factorizes as
$$
D=(Y_2-Y_1)(Y_3-Y_1)(Y_3-Y_2)
$$
So at least two $Y_i$'s are equal. It is easy to deduce from here that there are other permutations $\gamma$ satisfying $X=Y_\gamma$, some of which are even.