Exercise 2.2.1

Proof. We show that the leading term of ${\sigma }_r$ for the graded lexicographic order is $x_1{x}_2\cdots x_r$.

Let $x_{i_1}{x}_{i_2}\cdots x_{i_r}(i_1<i_2<\cdots <i_r)$ any term of ${\sigma }_r$, distinct of $x_1{x}_2\cdots x_r$. We must show that $x_1{x}_2\cdots x_r>x_{i_1}{x}_{i_2}\cdots x_{i_r}$.

If $i_1>1$, then $x_1$ as no occurence in $x_{i_1}{x}_{i_2}\cdots x_{i_r}$. Its exponent is 0 in the right monomial, and $1$ in the left monomial, so

and the proof is done in this case.

If $i_1=1$, let $j(1<j<n)$ the first subscript such that $i_j\ne j$. Then

Such a subscript exists, otherwise $x_1{x}_2\cdots x_r=x_{i_1}{x}_{i_2}\cdots x_{i_r}$. As $i_j>i_{j-1}=j-1$, $i_j\ge j$, and as $i_j\ne j,i_j>j$, so the exponent of $x_j$ is $0$ in the right monomial.

Therefore

So the leading term of ${\sigma }_r$ is $x_1{x}_2\cdots x_r$. □