Exercise 2.2.2

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

This exercise will study the order relation defined in (2.5). Given an exponent vector $\alpha = (a_1,\ldots,a_n)$, where each $a_i \geq 0$ is an integer, let $x^\alpha$ denote the monomial
$$x^{\alpha} = x_1^{a_1}\cdots x_n^{a_n}.$$
If $\alpha$ and $\beta$ are exponent vectors, note that $x^{\alpha} x^{\beta} = x^{\alpha+\beta}$. Also, the leading term of a nonzero polynomial $f \in F[x_1,\ldots,x_n]$ will be denoted {
ormalfont  \scriptsize LT}$(f)$.
\begin{enumerate}
\item[(a)] Suppose that $x^{\alpha}> x^{\beta}$, and let $x^{\gamma}$ be any monomial. Prove that $x^{\alpha+\gamma} > x^{\beta+\gamma}$.
\item[(b)] Suppose that $x^{\alpha} > x^{\beta}$ and $x^{\gamma} > x^{\delta}$. Prove that $x^{\alpha + \gamma} > x^{\beta + \delta}$.
\item[(c)] Let $f,g \in F(x_1,\ldots,x_n]$ be nonzero. Prove that {
ormalfont  \scriptsize LT}$(fg) =$ {
ormalfont  \scriptsize LT}$(f)${
ormalfont  \scriptsize LT}$(g)$.
\end{enumerate}

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers \def\dcro{\mbox{]\hspace{-.15em}]}} ewcommand{\be} {\begin{enumerate}} ewcommand{\ee} {\end{enumerate}} ewcommand{\deb}{\begin{eqnarray*}} ewcommand{\fin}{\end{eqnarray*}} ewcommand{\ssi} {si et seulement si } ewcommand{\D}{\mathrm{d}} ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{\U}{\mathbb{U}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\im}{\,\mathrm{Im}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{\Gal}{\mathrm{Gal}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \begin{proof} \begin{enumerate} \item[(a)] Let $\alpha = (a_1,a_2,\cdots, a_n),\beta = (b_1,b_2,\cdots,b_n),\gamma = (c_1,c_2,\cdots, c_n)$ and suppose that $x^{\alpha}>x^{\beta}$. Then $a_1+a_2+\cdots+a_n \geq b_1+b_2+\cdots+b_n$, otherwise $x^{\alpha} b_1+b_2+\cdots+b_n$, then $(a_1+c_1)+ \cdots+(a_n+c_n) > (b_1 + c_1)+\cdots + (a_n+c_n)$, thus $x^{\alpha+\gamma}>x^{\beta+\gamma}$. We suppose now that $a_1+a_2+\cdots+a_n = b_1+b_2+\cdots+b_n$. By definition of the graded lexicographical order, $a_1 \geq b_1$, otherwise $x^{\alpha}b_1$, then $a_1+c_1>b_1+c_1$, which implies $x^{\alpha+\gamma}>x^{\beta+\gamma}$. It remains the case where $a_1=b_1$. Let $j\ (jx^{\beta}$, such a subscript exists, otherwise $x^{\alpha}=x^{\beta}$. If $a_j < b_j$, we would have $x^{\alpha} b_j$. Then $a_1 +c_1 = b_1+c_1, \cdots, a_{j-1}+c_{j-1} = b_{j-1} + c_{j-1}$ and $a_j + c_j > b_j + c_j$, so $$x^{\alpha+\gamma}>x^{\beta+\gamma}.$$ Conclusion : $$x^{\alpha} > x^{\beta} \Rightarrow x^{\alpha+\gamma}>x^{\beta+\gamma}.$$ \item[(b)] If $x^{\alpha}> x^{\beta}$ and $x^{\gamma} > x^{\delta}$, then by (a), $$x^{\alpha+\gamma}> x^{\beta+\gamma},$$ $$x^{\beta+\gamma}>x^{\beta+\delta}.$$ So, by transitivity $$x^{\alpha+\gamma}>x^{\beta+\delta}.$$ \item[(c)] Let $c x^{\alpha} = $ { \scriptsize LT}$(f), d x^{\beta} = $ { \scriptsize LT}$(g)$. By definition of the leading term, for every term $ux^\gamma$ in $f$, distinct of {\scriptsize LT}$(f)$, \begin{align*} x^{\alpha}> x^{\gamma}, \end{align*} and for every term $vx^\delta$ in $g$, distinct of {\scriptsize LT}$(g)$, \begin{align*} x^{\beta} > x^{\delta}. \end{align*} Every monomial in $fg$ distinct of $cd x^{\alpha+\beta}$ is of the form $gx^{\gamma+ \delta}$, where $\beta,\gamma$ verify $\alpha \geq \gamma, \beta > \delta$, or $\alpha>\gamma, \beta \geq \delta$. In both cases, by (a) and (b), $$x^{\alpha+\beta}>x^{\gamma+\delta}.$$ Therefore $cd x^{\alpha+\beta}$ is the leading term of $fg$, so \begin{center} { \scriptsize LT}$(fg)$ = { \scriptsize LT}$(f)${ \scriptsize LT}$(g)$. \end{center} \end{enumerate} \end{proof}

Exercise 2.2.2

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Exercise 2.2.2

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