Exercise 2.2.5

Proof.

(a)

Let $h\in F[u_1,u_2,\cdots ,u_n],h\ne 0$, and $c{u}_1^{b_1}{u}_2^{b_2}\cdots u_n^{b_n}$ a term of $h$.

The leading term of a product is the product of the leading term of the factors (Ex 2.2.2), and the leading term of ${\sigma }_r$ is $x_1{x}_2\cdots x_r$ (Ex 2.2.1), so the leading term of $c{\sigma }_1^{b_1}{\sigma }_2^{b_2}\cdots {\sigma }_n^{b_n}$ is

$$\begin{array}{llll}\hfill \mathrm{LT}(c{\sigma }_1^{b_1}{\sigma }_2^{b_2}\cdots {\sigma }_n^{b_n})& =c{(x_1)}^{b_1}{(x_1{x}_2)}^{b_2}\cdots {(x_1{x}_2\cdots x_n)}^{b_n}& \hfill & \\ \hfill & =c{x}_1^{b_1+b_2+\cdots +b_n}{x}_2^{b_2+\cdots +b_n}\cdots x_n^{b_n}.& \hfill & \\ \hfill & & \hfill \end{array}$$

(b)

If $a_i,b_i\in \mathbb{Z}$, the system of equations $$\begin{array}{lll}\hfill b_1+b_2+\cdots +b_n=a_1,& & \hfill \\ \hfill b_2+\cdots +b_n=a_2,& & \hfill \\ \hfill \text{⋯}& & \hfill \\ \hfill b_n=a_n,& & \hfill \end{array}$$

is equivalent to

$$\begin{array}{llll}\hfill b_1& =a_1-a_2,& \hfill & \\ \hfill b_2& =a_2-a_3,& \hfill & \\ \hfill & \cdots & \hfill & \\ \hfill b_{n-1}& =a_n-a_{n-1},& \hfill & \\ \hfill b_n& =a_n.& \hfill & \end{array}$$

So the application $f:{\mathbb{Z}}^n\to {\mathbb{Z}}^n$ defined by

$$(b_1,b_2,\cdots ,b_n)\mapsto (b_1+b_2+\cdots +b_n,b_2+\cdots +b_n,\cdots ,b_n)$$

is bijective (one-to-one and onto).

(c)

As $h\ne 0$, there exists a term $c{u}_1^{b_1}{u}_2^{b_2}\cdots u_n^{b_n}$ of $h$ such that the leading term $c{x}_1^{a_1}\cdots x_n^{a_n}$ of $c{\sigma }_1^{b_1}{\sigma }_2^{b_2}\cdots {\sigma }_n^{b_n}$ is maximal. Then every other term $c^{\prime }{u}_1^{b_1^{\prime }}{u}_2^{b_2^{\prime }}\cdots u_n^{b_n^{\prime }}$ of $h$ verifies $(b_1^{\prime },b_2^{\prime },\cdots ,b_n^{\prime })\ne (b_1,b_2,\cdots ,b_n)$ and the leading term $c^{\prime }{x}_1^{a_1^{\prime }}\cdots x_n^{a_n^{\prime }}$ of $c^{\prime }{\sigma }_1^{b_1^{\prime }}{\sigma }_2^{b_2^{\prime }}\cdots {\sigma }_n^{b_n^{\prime }}$ is less than $c{x}_1^{a_1}\cdots x_n^{a_n}$ : it can not be greater because this term is maximal, and $(a_1,a_2,\cdots ,a_n)\ne (a_1^{\prime },a_2^{\prime },\cdots ,a_n^{\prime })$, since the application $f$ in (b) is bijective. The graded lexicographic order defined on the monomials $x_1^{a_1}\cdots x_n^{a_n}$ being a total order, $x_1^{a_1}\cdots x_n^{a_n}>x_1^{a_1^{\prime }}\cdots x_n^{a_n^{\prime }}$.

So $c{x}_1^{a_1}\cdots x_n^{a_n}$ is greater than the leading terms of every other term $c^{\prime }{\sigma }_1^{a_1^{\prime }}{\sigma }_2^{a_2^{\prime }}\cdots {\sigma }_n^{a_n^{\prime }}$ of $h({\sigma }_1,\cdots ,{\sigma }_n)\ne 0$, so is a fortiori greater than every other term of $h({\sigma }_1,\cdots ,{\sigma }_n)$.

It can’t be cancelled in the sum of these terms, and consequently $h({\sigma }_1,\cdots ,{\sigma }_n)\ne 0$.