Exercise 2.2.12

Proof.

(a)

Here we suppose that the exponents $a_i$ are distinct

If $\sigma ,\tau \in S_n$ and $\sigma \ne \tau$, then $x_{\sigma (1)}^{a_1}\cdots x_{\sigma (n)}^{a_n}\ne x_{\tau (1)}^{a_1}\cdots x_{\tau (n)}^{a_n}$.

Then ${\mathrm{\Sigma }}_n{x}_1^{a_1}\cdots x_n^{a_n}=\underset{\sigma \in S_n}{\sum <!--\; FUNCTION\; APPLICATION\; -->}{x}_{\sigma (1)}^{a_1}\cdots x_{\sigma (n)}^{a_n}$ has $n!=|S_n|$ terms.

(b)

Now we suppose that the exponents have same value on $I_1=\text{[[}1,l_1\text{]]}$ and on each interval $I_k=\text{[[}{l}_1+\cdots +l_{k-1}+1,l_1+\cdots +l_k\text{]]},(k=2,\cdots ,r)$, with distinct constants on each interval.

The terms of ${\mathrm{\Sigma }}_n{x}_1^{a_1}\cdots x_n^{a_n}$ are the terms of the image of the application

$$\phi :\begin{array}{ccc}\hfill S_n\hfill & \hfill \to \hfill & \hfill F[x_1,\cdots ,x_n]\hfill \\ \hfill \sigma \hfill & \hfill \mapsto \hfill & \hfill x_{\sigma (1)}^{a_1}\cdots x_{\sigma (n)}^{a_n}=\sigma \cdot (x_1^{a_1}\cdots x_n^{a_n}).\hfill \end{array}$$

This image is the orbit ${\mathcal{}O}_t$ of $t=x_1^{a_1}\cdots x_n^{a_n}$ for the group operation defined by $(\sigma ,f)\mapsto \sigma \cdot f$.

As $|{\mathcal{}O}_t|=|S_n|∕|{\mathrm{Stab}}_{S_n}(t)|$, it is sufficient to compute the cardinality of this stabilizer $S={\mathrm{Stab}}_{S_n}(t)$, stabilizer in $S_n$ of $x_1^{a_1}\cdots x_n^{a_n}$:

$$S=\{\sigma \in S_n|x_{\sigma (1)}^{a_1}\cdots x_{\sigma (n)}^{a_n}=x_1^{a_1}\cdots x_n^{a_n}\}.$$

Note that $\sigma \in S$ iff $\sigma$ applies $I_k$ on itself :

$$\sigma (I_k)=I_k,k=1,\dots ,r.$$

Let $\psi$ be the application

$$\psi :\begin{array}{ccc}\hfill S\hfill & \hfill \to \hfill & \hfill S(I_1)\times S(I_2)\times \cdots S(I_r)\hfill \\ \hfill \sigma \hfill & \hfill \mapsto \hfill & \hfill ({\sigma }_1,{\sigma }_2,\cdots ,{\sigma }_r)\hfill \end{array}$$

where ${\sigma }_k=\sigma {\text{|}}_{I_k}$ is the restriction of $\sigma$ to $I_k$.

$\psi$ is bijective, so

$$|S|=l_1!l_2!\cdots l_r!.$$

So the number of terms in ${\mathrm{\Sigma }}_n{x}_1^{a_1}\cdots x_n^{a_n}$, equal to the cardinality of the orbit of the monomial $t$, is equal to

$$|{\mathcal{}O}_t|=|S_n|∕|{\mathrm{Stab}}_{S_n}(x_1^{a_1}\cdots x_n^{a_n})|=\frac{n!}{l_1!l_2!\cdots l_r!}$$