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I am trying to compute manually a Gröbner basis for $I=\langle f=x_3-x_1^5,g=x_2-x_1^3\rangle$ with the lexicographic order. After the third iteration I get, $$h_1=x_1^2x_2-x_3$$ $$h_2=x_1x_3-x_2^2$$ $$h_3=x_1x_2^3-x_3^2$$ $$h_4=x_3(x_1^4-x_2^3)$$ $$h_5=x_3^3-x_2^5$$ macaulay2 says that the basis should be $\{g,h_1,h_2,h_3,h_5\}$, so my question is, how can I prove that $h_4\in\langle g,h_1,h_2,h_3,h_5\rangle$ if the division algorithm doesn't work? This would be also usefull to prove that all the $S$-polynomials on the next iteration are 0.

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    $\begingroup$ I obtain $x_2h_2-x_1^3x_3g=x_3x_1^4-x_2^3\in I$. Is there typo in $h_4$ ? $\endgroup$ – Dietrich Burde Sep 16 '15 at 19:58
  • $\begingroup$ You can change the order with macaulay2, I changed it to lex $\endgroup$ – Smurf Sep 16 '15 at 20:45
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    $\begingroup$ I think you've made an error. Sage says that $h_4$ is not in your ideal $I$. (This agrees with your observation that dividing $h_4$ by $\{g,h_1,h_2,h_3,h_5\}$ leaves a nonzero remainder.) $\endgroup$ – André 3000 Sep 16 '15 at 23:25
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I found my error computing $h_4$, $$h_4=S(f,h_3)=x_1^4x_3^2-x_2^3x_3$$ after that I used the division algorithm with $h_2$ and I get on the first iteration that, $$x_3(x_1^3x_3h_2-x_2^2g)=h_4$$ so it is, $$h_4\equiv 0\mod I$$

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