Background:
Let $X$ be a set of possibly infinitely many variables, and let $K[X]$ be the polynomial ring with variables in $X$ over some field $K$. Define a monomial order to be a total order $>$ on the monomials of $K[X]$ such that for any monomials $m_1$,$m_2$, and any monomial $n\not=1$, we have $m_1>m_2$ implies $nm_1>nm_2>m_2$.
Question:
I want to show that any monomial order $>$ refines the partial order of divisibility on the monomials of $K[X]$. More specifically, I want to show that if $u=vw$ for some monomials $u,v,w$ with $v\not=1$ and $w\not=1$, then $u>v$ and $u>w$.
My thoughts:
I'd like to make use of the part of the definition that says $nm_2>m_2$, for it would follow that $vw>v$ and $vw>w$. But I need to know there is some monomial strictly greater than $v$ and some monomial strictly greater than $w$. How do I know that $u$ or $v$ are not somehow a maximal element with respect to $>$?
Is this definition of monomial order just bad? Or am I missing something? I'm hoping that I am just missing something.
Think I figured it out. Just work by cases.
First suppose $v=w$. Since $w\not=1$, we cannot have $u=v$. Thus either $u>v$ or $v>u$. If $u>v$ then also $u>w$ so we're done. Suppose now that $v>u$. Then since $w\not=1$ we have $wv>wu$ so $u>wu$, a contradiction. This handles the case of $v=w$.
Suppose now $v\not=w$. WLOG assume $v>w$. Then since $w\not=1$ we have $wv>w$, so $u>w$. Again, since $v,w\not=1$, we cannot have $u=v$, so either $u>v$ or $v>u$. If $u>v$ we are done so suppose $v>u$. Since $w\not=1$ we obtain $wv>wu$, so $u>wu$, a contradiction.
Side note: both times we obtain a contradiction with $u>wu$ because $u$ is not a maximal element ($v>u$) and therefore since $w\not=1$, we have by definition $wu>u$.