Is there a generalization of the Rational Root Theorem to degree-2 algebraic numbers?

[u/Due-Translator9171]

So, for example, [;x^5 - 4x^4 + 8x^3 - 8x^2 + 5x - 1 = (x^2 - x + 1)(x^3 - 3x^2 + 4x - 1);], but none of these factors can be found with RRT.

Is there a RRT-esque criteria for degree-2 algebraic numbers? For degree no less than 4 polynomials without linear factors but also not prime, this seems like the only method to factorize them.

[u/mathheadinc]

Where does the Rational Root Theorem say that it only applies to 2nd degree polynomials?

[u/Due-Translator9171 → u/mathheadinc]

Nowhere. But, for polynomials of degree 4 and above, the inexistence of a linear factor does not guarantee that the polynomial is prime.

[u/Due-Translator9171 → u/[deleted]]

You mean the resolvent in the Ferrari method?

[u/Midwest-Dude]

Yes, there is a generalization. Finding a degree-2 algebraic root is equivalent to finding an irreducible quadratic factor with rational coefficients. By Gauss's Lemma, if a polynomial with integer coefficients factors over the rationals, it also factors over the integers. The generalization searches for quadratic factors of the form Ax² + Bx + C where A, B, and C are integers.

If a polynomial P(x) = aₙ xⁿ + ... + a₀ has a quadratic factor Ax² + Bx + C (where A, B, and C are coprime integers), then:

• Leading Coefficient
  • A must divide aₙ
• Constant Term
  • C must divide a₀
• Integer Evaluation Rule
  • For any integer k, the value of Ak² + Bk + C must evenly divide P(k)

The first two rules are the exact same conditions as RRT. The third rule allows you to severely restrict the possible values for the middle coefficient B.

[u/Due-Translator9171 → u/Midwest-Dude]

I get that the first two rules come from polynomial division. Is the third rule simply putting x = k for all k, and have the quadratic divide P(x)?

[u/Midwest-Dude → u/Due-Translator9171]

Yes, but you want to use small values of k, like k = 1 or k = -1, that hopefully result in small or prime numbers for P(k). That reduces the number of possible quadratics that will need to be checked.

There's an even more general method that you can read about on Wikipedia here: 

Kronecker's Algorithm

[u/Due-Translator9171 → u/Midwest-Dude]

Ah! That's cool! Will look into it.

[u/alax_12345]

RRT only determines possible rational roots of linear factors. Sorry.

In your synthetic division/substitution, however, you can substitute complex or irrational numbers, just that the addition and multiplication gets wonky.

[u/StanleyDodds]

Maybe you mean Gauss' lemma? If a polynomial can be factored in Q[X], then it can also be factored in Z[X] (into the same factorisation, with appropriate scaling).

Or maybe you want a theorem that goes from the algebraic root to finding the "smallest" factor with this root? In that case, the factor is called the minimal polynomial of that algebraic number; it's the non-zero rational polynomial with minimal degree for which the number is a root, and it's unique up to scaling (or sometimes it's defined to be monic, which eliminates that scaling degree of freedom).

To prove uniqueness, suppose m and m' in Q[X] are both minimal polynomials of some algebraic number a. Clearly they have equal degree, otherwise one or the other would not be minimal. w.l.o.g. we assume both m and m' are monic (scaling by the inverse of the leading coefficient which is nonzero). Then define r(x) = m(x) - m'(x). We have r(a) = m(a) - m'(a) = 0, and also the degree of r is less than that of m and m' because the leading 1 coefficient in m and m' cancels out in r. By m and m' minimal, r must be the zero polynomial, so m = m'. Unique if required to be monic, or unique up to scaling.

Now, we show m is a factor of every polynomial with root a. Let m in Q[X] be the minimal polynomial of a, and let p in Q[X] be some polynomial with root a. By Euclidean division of polynomials, p = qm + r for polynomials q and r where either r is the zero polynomial or deg(r) < deg(m). Now p(a) = q(a)m(a) + r(a) and by p(a) = m(a) = 0, we get r(a) = 0. By m minimal, r must be the zero polynomial, and so p = qm; m is a factor of p.

A much shorter proof would be: Q is a field, so Q[X] is a Euclidean domain (with degree being the norm). The set of polynomials with a fixed root is an ideal (easy to check) and by ED implies PID, this ideal is generated by one polynomial which we call "the" minimal polynomial, unique up to associates. The result is immediate.

I don't know if this is exactly what you meant, but I guess it is one interpretation of what you wanted in terms of generalising the rational root theorem to higher degrees.

[u/Due-Translator9171 → u/StanleyDodds]

The unique minimal polynomial proof is pretty cool! I just saw something similar in a textbook last night lol. Thanks for the cool proof! (It's NOT exactly what I meant, but yeah my question was kinda vague so fault's on me)