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By imposing certain weights $\mathbf{w}$ on the variables, say, of a polynomial ring $k[x_1,\ldots, x_n]$, I read that we may obtain the initial ideal $in_{\mathbf{w}}(I)$ of an ideal $I$ with respect to this weight.

Is it true that for any fixed weight $\mathbf{w}$ imposed on the variables, we have $in_{\mathbf{w}}(I)=in_{>}(I)$ for some monomial ordering $>$, where $>$ is thought to be lex or graded reverse lex, with the variables shuffled around?

I thought I read this somewhere but I cannot find the reference. Thanks for your time.

Edit: The initial ideal $in(I)$ of an ideal $I$ is thought to be an ideal generated by the leading terms of $f$, where $f\in I$.

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I think this link may answer the original question from above. It relates weighted monomial orders to universal Grobner basis. – math-visitor May 26 '12 at 0:10
up vote 1 down vote accepted

I know this is an old post, but I stumbled upon it trying to answer my own question, and I found it interesting, so I thought I should chime in.

The short answer would be yes. The problem is with ties on your weighting. For example, in $k[x,y,z]$ and $\omega = (3,4,5)$, the monomials $x^3$ and $yz$ have the same weight, so you have to pass to a second term order to break the tie. So what you need is not a single weighting, but actually 3 linearly independent weight vectors to guarantee all ties can be broken.

So once you have all your tiebreaking vectors, you simply have to finagle the equations until you have $x_{i_1}<x_{i_2}<\cdots<x_{i_n}$, and since you have a term order, you know that $x_{i_j}M<x_{i_k}M$ for all $j<k$ and all monomials $M$. The only 'danger' in thinking this way might be that one could be interested in the weighting itself; if you just want to know the order of the monomials, then you are all set.

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