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I need to calculate a Grobner basis of $$I=(u-x^2,v-y^2,w-xy)$$ with lexicographic order $x>y>u>v>w$.

I am trying to proceed with Buchberger's algorithm, but i probably didn't quite understand how to apply the multivariate division algorithm.

  • Let $f_1:=u-x^2$, $f_2:=v-y^2$, $f_3:=w-xy$.

  • Denote by $g_i$ the leading term of $f_i$ with respect to the given ordering $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; g_1:=x^2$, $g_2:=y^2$, $g_3:=xy$.

  • Denote by $a_{ij}$ the least common multiple of $g_i$ and $g_j$.$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;a_{1,2}=x^2y^2$, $a_{1,3}=x^2y$, $a_{2,3}=xy^2$

  • Choose two polynomials in G and let $S_{ij} = (a_{ij} / g_i) f_i − (a_{ij} / g_j) f_j. \;\;\;$I chose $S_{1,2}=y^2(u-x^2)-x^2(v-y^2)$

But now, from multivariate division algorithm of $S_{ij}$ by $f_1,f_2,f_3$ always gives $r=0$. What am I doing wrong?

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  • $\begingroup$ (u-x^2)*y^2+u*(v-y^2)-(w+x*y)*(w-x*y) $\endgroup$ Commented Dec 4, 2020 at 15:53

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We calculate $S_{1,3}=\frac{x^2y}{-x^2}f_1-\frac{x^2y}{-xy}f_2=xw-yu$ . Now by the lexicographic order the leading term of $S_{1,3}$ is given by $xw$ but this is not divisible by the leading terms of any of the three given polynomials. Now we initiate the multivariate division algorithm with the following data:
1)$r^{(0)}=0$, $p^{(0)}=S_{1,3}$
Now the leading term of $p$ is given by $xw$ and this is not divisible by any of the leading terms of the $f_i$ so following the algorithm we have
2)$p^{(1)}=p^{(0)}-lt(p^{(0)})=-yu$ and $r^{(1)}=r^{(0)}+lt(p^{(0)})=xw$
Now we notice again that the leading term of $p^{(1)}$ is not divisible by any of the leading terms of the $f_i$ so doing another step in the algorithm we obtain:
3)$p^{(2)}=p^{(1)}-lt(p^{(1)})=-yu+yu=0$ and $r^{(2)}=r^{(1)}+lt(p^{(1)})=xw-yu=S_{1,3}$
Now our $p^{(2)}=0$ hence we stop the algorithm and add the rest $r=S_{1,3}$ to the divisors and start from scratch.
Note that in the multivariate division divisibility only depends on the leading terms. I hope this clarifies some of your problems! The full Gröbner basis is given by $[x^2 - u, xy - w, xv - yw, xw - yu, y^2 - v, uv - w^2]$. Just adding this to check the results of your further calculations :)
Lg Mo

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  • $\begingroup$ I proceeded as suggested: I arrived at $$(u-x^2,v-y^2,w-xy,xw-yu,xv-yw)$$ So, $$g_1=-x^2\;g_2=-y^2\;g_3=-xy\;g_4=xw\;g_5=xv$$ Then I calculate the different $a_{ij}$, in particular $$a_{4,5}=xvw$$. Then $$S_{4,5}=w^2y-uvy$$ and for multivariate division algorithm I should add $w^2y-uvy$ in Groebner basis. But I want only $w^2-uv$.....What is wrong? $\endgroup$ Commented Dec 4, 2020 at 21:44
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    $\begingroup$ There is nothing wrong with your answer! In general a groebner basis is NOT unique. We only obtain uniqueness is we demand the groebner basis to be reduced(see for example homepage.univie.ac.at/dietrich.burde/papers/… Prop 2.5.8 for a proof of the uniqueness and Def 2.5.7 for the term "reduced"). After looking a bit into the script you will see, that the basis i gave is indeed reduced and yours is not. That does not mean your answer is wrong. The basis you give is still a gröbner basis, just not reduced. $\endgroup$
    – Mo145
    Commented Dec 4, 2020 at 22:14
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    $\begingroup$ It's just that most computer algebra programms compute the reduced version and since i calculated mine using sagemath(if you are interested in using computer algebra programms see for example sagemath.org) it looked different from yours. $\endgroup$
    – Mo145
    Commented Dec 4, 2020 at 22:14
  • $\begingroup$ I hope that clarifies your concerns! Lg Mo $\endgroup$
    – Mo145
    Commented Dec 4, 2020 at 22:15
  • $\begingroup$ Ah! Ok!!!! Thank you! $\endgroup$ Commented Dec 4, 2020 at 22:39

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