Investigate whether the polynomial $q(x) = 2x^5 - 78x^3 + 39x + 21$ is irreducible in $\mathbb{F}_{13}[x]$.

Question

Investigate whether the polynomial $q(x) = 2x^5 - 78x^3 + 39x + 21$ is irreducible in $\mathbb{F}_{13}[x]$.

Solution: In $\mathbb{F}_{13}[x]$, $q(x) = 2x^5 + 8 = 2(x^5 + 4)$. This polynomial has a root in $\mathbb{F}_{13}: p(a) = 0 \in \mathbb{F}_{13} $ if and only if $a^5 = -4 \in \mathbb{F}_{13}$. Since $\mathbb{F}_{13}^*$ is cyclic of order $13 - 1 = 12$, and $\text{gcd}(12, 5) = 1$, every congruence of the form $x^5 \equiv b \mod 13$ has a solution.

I have an old math exam question with the solution included, but there are certain steps of the solution I don't understand.

Questions:

• Since the polynomial has a zero point it means that the polynomial is reducible, since the polynomial can be written as a product of factors although how do I know that one of the factors must be a unit?
• Why do we look at the order of $\mathbb{F}_{13}^*$ instead of at $\mathbb{F}_{13}$ and how does knowing the order of $\mathbb{F}_{13}^*$ and that the $\text{gcd}(12,5)=1$ lead to the conclusion that $x^5 \equiv b \mod 13$ must have a solution?

Some definitions:

• Let R be an integral domain. An element $ p \in R $ is called irreducible if $ p $ is neither the zero element nor a unit, and if the following holds: If $ p = ab $ with $ a, b \in R $, then either $ a $ is a unit or $ b $ is a unit.
• The set $\mathbb{F}_{13}^*$ consists of the residue classes that are coprime to $13$.

Thanks in advance!

Answer 1

• It is a classical result that for any field $K$ and any $P\in K[X]$, we have $$ P(a)=0\iff X-a\mid P. $$ In particular, an irreducible polynomial in $K$ of degree $\geqslant 2$ has no roots in $K$.
• We're looking at $\mathbf{F}_{13}^*$ because we are interested in the multiplicative law modulo $13$. For the other part of the question, you know that $\mathbf{F}_{13}^*$ is cyclic, a generator being $2$ here. Since $12\wedge 5=1$ there exists $u,v\in\mathbf{Z}$ such that $12u+5v=1$ so $$ b\equiv b^{12u+5v}\equiv (b^u)^{12}(b^v)^5\equiv (b^v)^5\pmod{13} $$ because of Fermat's little theorem, so the equation $x^5\equiv b\pmod{13}$ has a solution.

Answer 2

The claim is that it decomposes into two smaller polynomials neither of which is a unit, as a factor $x - a$ is not a unit in $\mathbf{F}_{13}[x]$. To answer your second question, think about what happens when we take $k$-th powers of elements in $\mathbf{F}_{13}^*$ with $\text{gcd}(k, 12) = 1$. Hint: when do wo have that $\left(a^k\right)^n = 1$ (take maybe some generator of $\mathbf{F}_{13}^*$)?

Answer 3

$q(x) = 2x^5 - 78x^3 + 39x + 21$ is equivalent to $f(x)=2x^5+8=2(x^5+4)$ in $\Bbb F_{13}[x]$ and if $x^5=-4=9$ modulo $13$ then $x=3$ because $3^3=1$ and $3^2=9$ modulo $13$. Then $q(x)$ is not irreducible in $\Bbb F_{13}[x]$. On the other hand $$\dfrac{x^5+4}{x-3}=x^4+3x^3+9x^2+x+3+\dfrac{247}{x-3}$$ so, because $247\equiv0\pmod{13}$, the quotient in $\Bbb F_{13}[x]$ is $$h(x)=x^4+3x^3+9x^2+x+3$$ We can verify that $h(x)\ne0$ for all the elements of $\Bbb F_{13}$ so $h(x)$ has not linear factor.I don't look at the possibility of two quadratic factors because the question was about the irreducibility of $q(x)$ and the answer is not, $q(x)$ is reducible and we have $$q(x)=2(x-3)(x^4+3x^3+9x^2+x+3)\in\Bbb F_{13}[x]$$