Prove there are infinitely many fields in which the polynomial is reducable

Question

Polynomial: $$ t^7+t^2+1$$

I have the solution, but I don't understand the thinking behind it. How are they coming up with the factorization, and specific values of $t$ (shown in solution).

The solution says: Note that in complex field for $t=e^{i\frac{2\pi}{3}}, t^7=t$. That means that the polynomial is divisible by $t+t^2+1$ and $$ t^7+t^2+1=(t+t^2+1)(t^5-t^4+t^2-t+1)$$ This factorization obviously holds for any polynomials considered under any finite field $(\{0, 1, ..., p-1\}, +, \cdot)$ where $p$ is prime, and $-1=p-1$. Obviously this factorization holds for any field.

Could anyone explain it to me in more details?

Answer 1

I guess you are saying that you do not understand the solution you were given.

Take a cube root of one, $$ \omega = - \frac{1}{2} + i \frac{\sqrt 3}{2}=e^{i\frac{2\pi}{3}}. $$ Then

$$ (x - \omega)(x - \omega^2) = x^2 + x + 1 $$

Both $\omega$ and $\omega^2$ are roots of your polynomial, since $$ 7 \equiv 1 \pmod 3. $$ The polynomials over a field form a Euclidean ring, the two linear factors have gcd 1, therefore your polynomial is divisible by $x^2 + x + 1.$

Suppose the problem were $$ \color{blue}{ g(x) = x^{2017} + x^{1066} + x^{153} + x^{19} + 1.} $$ Then the exponents would be $$ 2, 1, 3, 4, 0 \pmod 5. $$ As a result, we take a non-real fifth root of $1,$ $$ \omega =e^{i\frac{2\pi}{5}}. $$ Since $$ \omega^5 = 1, $$ we get $$ g(\omega) = \omega^2 + \omega + \omega^3 + \omega^4 + 1. $$ However, since $x^5 - 1 = (x-1)(x^4 + x^3 + x^2 + x + 1)$ and $\omega \neq 1,$ we find $$ \omega^4 + \omega^3 + \omega^2 + \omega + 1 = 0, $$ so $$ g( \omega) = 0, $$ $$ g( \omega^2) = 0, $$ $$ g( \omega^3) = 0, $$ $$ g( \omega^4) = 0. $$ Therefore $$ (x^4 + x^3 + x^2 + x + 1) \; | \; g(x) = x^{2017} + x^{1066} + x^{153} + x^{19} + 1. $$

Answer 2

$t^2+t+1$ divides $t^7+t^2+1$, hence $t^7+t^2+1$ cannot be an irreducible polynomial.

In order to check that $t^2+t+1$ is actually a divisor of $t^7+t^2+1$, it is enough to check that $t^7+t^2+1$ vanishes at the roots of $t^2+t+1$, that are the primitive third roots of unity since $t^2+t+1=\frac{t^3-1}{t-1}=\Phi_3(t)$. But is $\omega$ is a primitive third root of unity, $\omega^7=\omega^1$, so $t^7+t^2+1$ evaluated at $t=\omega$ equals $\omega^7+\omega^2+1 = 1+\omega+\omega^2 = 0.$ At last, the identity

$$t^7+t^2+1 = (t^2+t+1)(t^5-t^4+t^2-t+1) $$ is straightforward to check.

Answer 3

All you need is to understand how the factor $t^2+t+1$ is deduced (say directly if $t=e^{i\frac{2\pi}{3}}$ then $t^7=t$ is having too much lucidity). The writer wants to reduce the degree and the ideal reduction he thought is the second degree. Then he saw that $$t^7=t\iff t^7-t=t(t^3-1)(t^3+1)=0$$ hence the non real root of $1$ might agree. This lead him to the factor $t^2+t+1=\frac{t^3-1}{t-1}$ wich besides being good it gives also the other factor after division.