I've come across an interesting connection between irreducible polynomials and polynomial constraints. For example, consider the basic quadratic:

$$af^2 + bf + c = 0$$

If we're working in a ring, and $f$ is assumed to be irreducible, then $c$ must be some multiple of $f$:

$$c = mf$$

Furthermore, substituting this back into the original equation, an additional constraint is discovered for $b$:

$$b + m = qf$$

Substituting this back into the quadratic equation, we find that $a=-q$.

So we now have a system of equations:

$$af^2 + bf +c = 0$$ $$c - mf = 0$$ $$b + m + af = 0$$

and I'm starting to wonder if I can translate my assumption of irreducibility into polynomial constraints that can be handled by a Groebner basis calculation.

For example, if the ring is $\mathbb{Z}$, and $f^2-6f-7=0$ is solved by 7 (irreducible), then $c$ is a multiple of 7, its multiple is -1, and $b$ (-6) plus that multiple is also a multiple of 7 (again -1), and $a$ is the negative of that multiple.

I'm working in a more complex system where (for example) $dx(nd)$ has to be a multiple of some irreducible $f$ with a zero $x$-derivative, so I'm led to conclude that $nd$ must have the form $fq+c$ where $c$ has a zero $x$-derivative. Now I'm thinking of adding $nd=fq+c$ and $c_x=0$ to my system and doing a differential Groebner basis calculation (maybe).

Can I find an exhaustive list of polynomial constraints triggered by the assumption of irreducibility? Exhaustive in the sense of a finite basis for an ideal.

Has anyone seen anything like this before? Any references?


closed as unclear what you're asking by Felipe Voloch, Boris Bukh, John Pardon, Jeremy Rouse, Per Alexandersson Sep 4 '15 at 21:58

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  • $\begingroup$ What do you mean by a constraint? Can you give a definition, and then formulate a short self-contained question. As it is phrased, it is not clear what you are asking. In addition: 1) You appear to divide by f in the step starting with "Furthemore". The element f might be not invertible. 2) Then you start talk about Groebner bases --- so, what kind of ring are you in? $\endgroup$ – Boris Bukh Sep 4 '15 at 19:02
  • $\begingroup$ @Boris Bukh: A constraint is an equation like $c=mf$, an additional polynomial that wouldn't otherwise be implied (though $c=mf$ isn't a good example, since it's implied even without irreducibility). My question is: Can anyone provide a reference that expounds on this theory? 1) I don't divide by $f$, since it factors out, but I do assume that it isn't a zero-divisor, and 2) Let's say I'm in Q[x], for the first example, or Q[x,t] for the second one $\endgroup$ – Brent Baccala Sep 4 '15 at 20:06
  • $\begingroup$ I've posted a follow-up question, maybe a better one, here: mathoverflow.net/questions/217402/… $\endgroup$ – Brent Baccala Sep 4 '15 at 20:26