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I was working a problem in which I needed to reduce two polynomials in variables c,p,t into a single polynomial in c,t. Note that t appears only once, at the end of the $q_2$ definition.

$$q_1\equiv 1 + 2 c^2 + 3 c^3 + 3 c^4 + 3 c^5 + c^6 + (2 c + c^2 + 2 c^3 + c^4) p + (c + 5 c^2 + 6 c^3 + 12 c^4 + 6 c^5) p^2 + (1 + 4 c^2 + 4 c^3) p^3 + (4 c + 3 c^2 + 18 c^3 + 15 c^4) p^4 + (2 c + 6 c^2) p^5 + (1 + 12 c^2 + 20 c^3) p^6 + 4 c p^7 + (3 c + 15 c^2) p^8 + p^9 + 6 c p^{10} + p^{12}$$

$$q_2\equiv 16 + 32 c^2 + 48 c^3 + 48 c^4 + 48 c^5 + 16 c^6 + (16 c + 16 c^2 + 16 c^3) p + (16 c + 48 c^2 + 80 c^3 + 160 c^4 + 80 c^5) p^2 + 32 c^2 p^3 + (16 c + 32 c^2 + 192 c^3 + 160 c^4) p^4 + 16 c p^5 + (96 c^2 + 160 c^3) p^6 + (16 c + 80 c^2) p^8 + 16 c p^{10} - t$$

A user on the Mathematica SE gave the following magical line of code that returns exactly what I needed:

GroebnerBasis[{q1, q2}, {p, c}][[1]]

returns:

$$4096 + 8192 c^2 + 12288 c^3 + 12288 c^4 + 12288 c^5 + 4096 c^6 - 768 t - 256 c^2 t + 256 c^3 t + 256 c^4 t + 48 t^2 - 16 c^2 t^2 - t^3$$

Perfect! And I've verified that it's correct. But I have no idea how this result was calculated, and the online explanations I read about the Groebner Basis were a bit over my head because I've never studied Ring Theory.

Can anyone explain to me how this result might have been calculated? I don't need the specifics of how Mathematica does it, just what it is an how one can calculate it. How does one eliminate p from a pair of polynomials like this? What is the Groebner Basis as the concept relates to a pair of polynomials? Can this be explained without Ring Theory?

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In Linear Algebra the common way to directly solve a system of linear equations is by row reduction of the augmented matrix representing the system of equations, often called Gaussian Elimination. It transforms the augmented matrix into a form that represents and equivalent system of equations but with maximal elimination of variables. The creation of a Groebner basis for a set of polynomials is an analogous process. It creates the 'smallest' set of polynomials that have the same set of zeros as the original set of polynomials. That is, the set of common roots of the elements of the Groebner basis is the same as the set of common roots (zeroes) of the original set of polynomials. The polynomials in the Groebner basis are usually much simpler than the original set. This is the best I can do without going into a little commutative ring theory. I hope it helps.

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  • $\begingroup$ If your interested in exploring further, the book "Topics in Commutative Ring Theory" by John Watkins has an introductory treatment of this subject. $\endgroup$ – John McGee Dec 25 '15 at 12:47

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