Edit: question has been changed from 'lexicographic' (cf. "D K"'s answer below) to 'degree' minimality.

Let $x_1,x_2,x_3$ be indeterminates. Fix an integer $k> 3$. Consider the set $M$ of all monomials of the form $x_1^{i_1}.x_2^{i_2}.x_3^{i_3}$ where each $i_j\in \mathbb{N}$ with $i_j\geq 1$ and $i_1+i_2+i_3 = k$.

Fix an arbitrary $F\in M$ and write $F = x_1^{a}.x_2^{b}.x_3^{c}$.

Question When does there exist a triple $(w_1, w_2, w_3)$, with each $w_i\in \mathbb{N}$ and $w_i\geq 1$, such that $$a.w_1 + b.w_2 + c.w_3 < i_1.w_1 + i_2.w_2 + i_3.w_3, \forall G = x_1^{i_1}.x_2^{i_2}.x_3^{i_3}\in M \;\text{with} \; G\neq F ?$$ In the exponential case, same question replaced by $a^{w_1}+ b^{w_2}+ c^{w_3} < i_1^{w_1}+i_2^{w_2}+ i_3^{w_3}$.

Positive example in additive case: $k = 4$ on three indeterminates. Then if $F$ corresponds to the element of $B_4$ with exponents $(2,1,1)$, then for any $w_2, w_3>1$, $w = (1,w_2,w_3)$ has the desired property.

Sometimes this question has a negative answer: e.g. k=4 on only two indeterminates. if we take F to correspond to the element in $B_4$ with exponents $(2,2)$ then we cannot achieve both $2w_1 + 2w_2 < w_1 + 3 w_2$ and $2w_1+2w_2 < 3w_1+w_2$.

Same question for $n$-variables $x_1,\ldots, x_n$ and $k > n.$

Apologies if this question has been answered; (general and specific) references also appreciated.


2 Answers 2


I haven't completely checked, but examples suggest that $F$ must be, up to reordering the $x_i$'s, the minimal lexicographic element in $B_k$ (in general, you can also replace $B_k$ with a subset of $B_k$, and then take the minimal element of this new set)

Example: take $k=5$. Then $(a,b,c)$ must have one of $a,b,$ or $c$ equal to $3$, i.e. no exponent of 2 is allowed in $F$ for a positive answer.

Explicitly, for the affirmative: we can assume without loss of generality that $F = xyz^3$. Then we get that $w_3 < w_1$, $w_3 < w_2$ and $2w_3 < w_1+w_2$, which is infinitely possible. Take, e.g. $w = (2,2,1)$. On the other hand if $F = xy^2z^2$ there is no solution since we must satisfy $w_3 < w_2$ and $w_2 < w_3$.

  • 2
    $\begingroup$ ok, I think that's correct in the general case: for $n\geq 2, k>n$, then if $F$ is, up to reordering, the minimal element of $B_k$ in the lexicographic ordering, then there are infinitely many $w's$ for which you have a positive answer to your question. It's probably true that this condition is also sufficient, i.e. if $F$ is not minimal, then there are no solutions. $\endgroup$
    – user34821
    Jun 11, 2013 at 9:28

It is clearly not the case, as the optimal choice for $a,b,c$ once the weights are fixed is always to put $k$ on the minimal weight and $0$ on the others.

You can see that if you start from a different configuration, the total weight will decrease when you move to this one, as you transfer "mass" from "heavy" to "light".

So if $a,b,c$ are not $k,0,0$ (in any order), then it cannot be a minimal choice, no matter what the weights are.

This is still true in the exponentiation setting.

  • $\begingroup$ thank you, though i have now changed the question. and yes, you are completely correct that in the original (and wrong) version of the question, we won't get a desired lexicographic ordering unless the chosen $f$ was already the minimal wrt lex. ordering. $\endgroup$
    – Freddy
    Jun 9, 2013 at 15:42
  • $\begingroup$ as a very easy example of positivity for the new question: take $k=4$ for three indeterminates. Then the exponents of the three elements of $B_k$ have coordinates $(2,1,1), (1,2,1),$ and $(1,1,2)$. Say we take $F\in B_k$ to be the element with $(2,1,1)$. Then in fact there are infinitely-many good $w's$, namely $w = (1, w_2, w_3)$ where $w_2,w_3>1$ $\endgroup$
    – Freddy
    Jun 9, 2013 at 16:05

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