Let $\mathbb{C}[x,y]$ be the polynomial ring with variables $x,y$ and coefficient in $\mathbb{C}$.

Let $f,g\in \mathbb{C}[x,y]$.

Let $(f,g)$ be the ideal of $\mathbb{C}[x,y]$ generated by $f,g$.

Given $h\in \mathbb{C}[x,y]$, how to determine whether $h\in (f,g)$ or not?

I have tried some examples by the online programming "sagemath".

Are there any methods that can give a proof?

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    $\begingroup$ Have in mind that sage and its sub-components have bugs... $\endgroup$ – joro Apr 27 '15 at 7:05
  • $\begingroup$ Do you really mean complex numbers for the coefficients? If you had suggested rational coefficients, I would not be asking this question, but even to determine the zeroness of a complex number that was arrived at by some analytic process seems to me to pose a serious problem. $\endgroup$ – Lubin Apr 27 '15 at 17:37
  • $\begingroup$ Dear Prof. Lubin, I guess cohomology rings with rational coefficient and complex number coefficient are the same except for tensoring with $\mathbb{C}$. Are they different? $\endgroup$ – QSH Apr 28 '15 at 1:41
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    $\begingroup$ For my question, it depends on how you come upon the $h$ in question. If it arose out of some analytic process, and its coefficients were genuine complex numbers, not known to be in (say) the field generated by the coefficients of $f$ and $g$, I don’t see how you’d be able to know that $h\in(f,g)$. I’m thinking like a computer scientist here, not an algebraic geometer. $\endgroup$ – Lubin Apr 28 '15 at 3:06

You should use "Gröbner basis", (Groebner) . see the book by "Cox D., Little J., O'Shea D.": named "Ideals, Varieties, and Algorithms", for example. In page.82 they have:

Corollary.2. Let $G = \{g_1, \cdots , g_t\}$ be a Groebner basis for an ideal $I \subset k[x_1, \cdots , x_n]$ and let $f \in k[x_1, \cdots , x_n]$. Then $f \in I$ if and only if the remainder on division of $f$ by $G$ is zero.

"Buchberger’s Algorithm", (page.88 of the book), helps you to produce the Groebner basis. Also "CoCoA" can compute Gröbner basis with the command "GBasis(I)" (for special field).

  • $\begingroup$ And how do you prove "CoCoA" computed the correct groebner basis? $\endgroup$ – joro Apr 27 '15 at 16:59

The algorithm is described in these notes by Madhu Sudan. The sage implementation is described in the Sage Manual.

  • $\begingroup$ So you are suggesting the OP to trust the notes and/or the sage implementation? $\endgroup$ – joro Apr 27 '15 at 17:22
  • $\begingroup$ @joro No, I think he should start with the axioms, derive all of commutative algebra, and then implement it in his head. On second thought, a better idea would be to (a) understand the mathematics and (b) see if the behaviour of the software conforms to his understanding. Unless, of course, the OP is an engineer, and just wants the answer. Then, he should just trust the Force, er, I mean, Sage. $\endgroup$ – Igor Rivin Apr 27 '15 at 18:39

Too long for a comment.

All CASes have bugs, so if you are using CAS solution, better run on as many CASes as you can.

Comment suggests to use Groebner basis, but this leads to the question "How do you compute Groebner basis without CAS?"

If your ideal is complicated enough, computing Groebner basis by hand might not be trivial.

  • $\begingroup$ If your ideal is complicated enough, even computing Groebner basis by machine is non-trivial (the complexity of the algorithm is horrible). $\endgroup$ – Igor Rivin Apr 27 '15 at 16:32
  • $\begingroup$ @IgorRivin Indeed, especially proving the groebner basis is correct even if the algorithm claimed to compute it. I doubt a human on paper can prove the (pseudo) primality of sufficiently large number. $\endgroup$ – joro Apr 27 '15 at 16:54
  • $\begingroup$ as for your answer, do I understand that you are suggesting what is suggested in this link: farm4.static.flickr.com/3129/3106491352_9db1b52554_o.gif $\endgroup$ – Igor Rivin Apr 27 '15 at 18:40
  • $\begingroup$ Humans are buggy too. Also, IIRC, verification that a basis is a Groebner basis of some ideal is generally a far easier problem than computing a Groebner basis of that ideal. $\endgroup$ – Hurkyl Dec 24 '16 at 17:07

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