As the title anticipates, I have a problem with Exercise 2.4.9 by Cox et al.
If $I=\langle x^{\alpha(1)},\dots,x^{\alpha(s)} \rangle$ is a monomial ideal, prove that a polynomial $f$ is in $I$ if and only if the remainder of $f$ on division by $x^{\alpha(1)},\dots,x^{\alpha(s)}$ is zero
Here, $I$ is an ideal of the polynomial ring $\mathbb{K}[x_1,\dots,x_n]$. Well, multivariable Euclidean division algorithm strongly depends on the chosen monomial ordering and on the order of the $s$-tuple $(x^{\alpha(1)},\dots,x^{\alpha(s)})$. Changing one of them (the ordering or the order of the $s$-tuple), the remainder on division can be $0$ or not. Hence, I'm stuck on the real content of the exercise. It seems to be very imprecise in my opinion.
My presentment is that the correct statement is the following:
If $I=\langle x^{\alpha(1)},\dots,x^{\alpha(s)} \rangle$ is a monomial ideal, prove that a polynomial $f$ is in $I$ if and only if for each monomial ordering on $\mathbb{K}[x_1,\dots,x_n]$ and any permutation of the $s$-tuple $(x^{\alpha(1)},\dots,x^{\alpha(s)})$ it results that the remainder of $f$ on division by $x^{\alpha(1)},\dots,x^{\alpha(s)}$ is zero.
For the proof: assume that $f \in I$ and let a monomial ordering on $\mathbb{K}[x_1,\dots,x_n]$ and any permutation of the $s$-tuple $(x^{\alpha(1)},\dots,x^{\alpha(s)})$ be fixed. By Lemma 3 we have that $f$ is a linear combination of $x^{\alpha(1)},\dots,x^{\alpha(s)}$ and, thus, the remainder of the division is zero.
Conversely, if for each monomial ordering on $\mathbb{K}[x_1,\dots,x_n]$ and any permutation of the $s$-tuple $(x^{\alpha(1)},\dots,x^{\alpha(s)})$ it results that the remainder of $f$ on division by $x^{\alpha(1)},\dots,x^{\alpha(s)}$ is zero, then we get an expression of the form $f=h_1\alpha^{(1)}+\dots+h_s\alpha^{(s)}$, so that $f \in I$ by Lemma 3.
However, such a claim seems to be very strong, and I'm not so convincted that the proof I gave for the first implication is correct, because nobody ensures me that changing the order of the monomials gives the same remainder. Actually, I'm very undecided if the authors wanted to say the following claim:
If $I=\langle x^{\alpha(1)},\dots,x^{\alpha(s)} \rangle$ is a monomial ideal, prove that a polynomial $f$ is in $I$ if and only if for each monomial ordering on $\mathbb{K}[x_1,\dots,x_n]$ there exists a permutation of the $s$-tuple $(x^{\alpha(1)},\dots,x^{\alpha(s)})$ such that the remainder of $f$ on division by such a permuted $s$-tuple is zero.
The previous proof seems to be solved in such a case by choosing the increasing ordering of the $s$-tuple with respect to chosen monomial ordering.
Finally, there is a possible weaker statement, i.e.
If $I=\langle x^{\alpha(1)},\dots,x^{\alpha(s)} \rangle$ is a monomial ideal, prove that a polynomial $f$ is in $I$ if and only if there exist a monomial ordering on $\mathbb{K}[x_1,\dots,x_n]$ and a permutation of the $s$-tuple $(x^{\alpha(1)},\dots,x^{\alpha(s)})$ such that the remainder of $f$ on division by such a permuted $s$-tuple is zero.
Here, the problem is again in the left implication. In fact, how to get such a monomial ordering.
Which is the correct interpretation of the statement?
The correct interpretation is the first one that you wrote, namely that $f \in I$ if and only if for every monomial ordering and every ordering of the tuple $(x^{\alpha(1)},\dots,x^{\alpha(s)})$, the remainder of $f$ on division by $x^{\alpha(1)},\dots,x^{\alpha(s)}$ is zero. In fact, this explains the imprecision in the statement of the exercise: whether $f \in I$ or not can be detected by the division algorithm, no matter what monomial ordering you choose, no matter how you list the monomials $x^{\alpha(1)},\dots,x^{\alpha(s)}$.
As for the proof (of the nontrivial direction), you can simply proceed by contrapositive. If the division algorithm leaves a nonzero remainder, then $f$ can be written as \begin{equation*} f = q_1x^{\alpha(1)}+\dots+q_sx^{\alpha(s)}+r, \end{equation*} where $r$ is a linear combination of monomials, none of which is divisible by any of $x^{\alpha(1)},\dots,x^{\alpha(s)}$. This means that some terms of $f$ are not in $I$, which implies that $f \notin I$ (by Lemma $3$).
If you prefer a direct proof instead, the one that you gave is correct, although you need to make sure that you understand why the division algorithm yields a zero remainder in this case (if this isn't already obvious to you). Remember: there are examples of polynomials $f$ that are linear combinations of polynomials $g_1,g_2$, but whose remainder on division by $g_1,g_2$ is $f$ itself. In the case at hand however, all the terms of $f$ are divisible by one of $x^{\alpha(1)},\dots,x^{\alpha(s)}$ (by Lemma $3$), so at every step of the division algorithm, the leading term of the dividend is subtracted, with no new terms introduced. Therefore, the terms of $f$ are subtracted one at a time until there is none left, leaving a zero remainder.