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Assume $f \in K[x_1,\dots ,x_m],x_i\in F$

$$p \in K[x_1,\dots, x_m], p = \sum_{i}\phi_ix_1^{{a}_{i1}}x_2^{a_{i2}}\dots x_n^{a_{in}}, L(p) := \phi _lx^l : x^a \prec x^l \forall a\in A(p) \backslash\{l\}$$

$$\textbf{th. 1}: [\forall f_i, f_j \in\{f_1,f_2 \dots,f_n\} = F : \nexists \lambda \in F: L(f_i) = \lambda L(f_j) ] \rightarrow [F - \text{Grobner basis}] (\textbf{It isn't true, following Marcus Ritt's counterexample})$$

$$\textbf{th. 2}: [\forall f_i, f_j \in\{f_1,f_2 \dots,f_n\} = F : \forall i \neq j : L(f_i) \text{ and } L(f_j) \text{ are mutually prime} ] \rightarrow [F - \text{Grobner basis}]$$

Looks like this feature works in general, but i don't know, how to prove it. Is it true, that if left-hand condition is true, then $S$-polynomial $S(f_i,f_j) \overset{F}{\rightsquigarrow} 0$ (reducible to zero), so i could use Buchberger criteria ?

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  • $\begingroup$ Do you have a reference for the theorem? What‘s the definition of $L$? $\endgroup$ May 23, 2019 at 10:25
  • $\begingroup$ idk, i found this statement by myself and want to prove it, should i edit "th." to "st." ? $\endgroup$
    – envy grunt
    May 23, 2019 at 10:27
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    $\begingroup$ No, but then please give the definition of $L$. $\endgroup$ May 23, 2019 at 10:32
  • $\begingroup$ $L(f_i)$ is the "leading" (greatest monomial) $\endgroup$
    – envy grunt
    May 23, 2019 at 15:14
  • $\begingroup$ in lexicographical order i mean $\endgroup$
    – envy grunt
    May 23, 2019 at 15:24

1 Answer 1

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No, for example $F=\{f_1,f_2\}$ with $f_1=x^2y-2y^2+x$ and $f_2=x^3-2xy$ satisfies your criterion, but is not a Gröbner basis, since $S(f_1,f_2)=x^2$.

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  • $\begingroup$ hmm, you're right, but is it true, if we assume that, $\forall i\neq j : L(f_i)$ and $L(f_j)$ are mutually prime? In this case I think, it should work $\endgroup$
    – envy grunt
    May 23, 2019 at 16:17

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