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I have a polynomial ring $S=k[x_1,\dots,x_n]$ positively graded, that is, each indeterminate has degree $>0$, and I suppose moreover that $\deg(x_1)>1$. In $S$ I have an ideal $P$ which is prime, homogenous (with respect to the given grading), and such that $x_1\notin P$.

My question is the following: if $\mathcal K$ is a minimal system of homogenous generators of $P$, is it true that $\mathcal K$ is also a Gröbner basis of $P$, with respect to a suitable term order? I have reason to believe that the answer is yes, because this should be the natural step to complete a certain proof. The only thing I know is that to carry out this proof I should use Buchberger's Algorithm.

So, here is a first attempt: let $d_i$ be the degrees of $x_i$, and denote $\le$ the term order which corresponds to the Matrix having as top rows $$\left(\begin{array}{cccc} d_1& d_2 &\dots& d_n\\ -1 &0&\dots &0 \end{array}\right),$$ and that goes on in a suitable way (i.e., so that the determinant is $\ne0$). An information I know is that, if $g_1,\dots,g_t$ is a homogenous, minimal Gröbner basis of $P$, then $x_1$ cannot divide any of the leading terms $LT_\le(g_i)$ of the $g_i$'s: this is true because $x_1$ dividing some $LT(g_i)$ would imply $x_1|g_i$, and since $x_1\notin P$ this would contradict the minimality of the Gröbner basis. So, in order to prove that the answer to my question is yes, I would begin with an easy case where $\mathcal K=\{f_1,f_2\}$ is a minimal system of homogeneous generators with just two elements. (If there's only one element the answer is trivially yes.) Now I would like to use Buchberger's Algorithm as follows: denoting $x^{\alpha_i}:=LT(f_i)$ for $i=1,2$, there is only one S-polynomial to consider: $$S_{12}:=\frac{x^{\alpha_1}}{\gcd(x^{\alpha_1},x^{\alpha_2})}f_2-\frac{x^{\alpha_2}}{\gcd(x^{\alpha_1},x^{\alpha_2})}f_1.$$ If is suppose that $\{f_1,f_2\}$ is not a Gröbner basis, it means that for each division of $S_{12}$ by $f_1$ and $f_2$ $$S_{12}=g_1f_1+g_2f_2+r$$ the remainder $r$ is not $0$. And by division recall that we mean that

  • for each $i\in\{1,2\}$, we have $LT(g_if_i)\le LT(S_{12})$,

  • $LT(r)\notin\big(LT(f_1),LT(f_2)\big)$.

I am trying to find a contradiction, but so far I haven't succeeded. Could anyone give me some hints? Thank you very much in advance for your help!

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    $\begingroup$ "... with respect to a suitable term order..." Does that mean you want to have the set first and then want to construct term order such that this specific set becomes a Groeber basis? $\endgroup$ – Dirk May 9 '17 at 10:56
  • $\begingroup$ Thank you. Yes, I have the set of generators and I hope to find such a term order. I am hoping that an order given by a matrix like the one described above can do the trick. Or maybe some other order which somehow "isolates" the indeterminate $x_1$. $\endgroup$ – Miles Eagle May 9 '17 at 11:33

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