Prove that $x^5 + 4x^4 + 2x^3 + x^2 + 4x + 54$ is irreducible over $\mathbb{Z} / 7\mathbb{Z}$

Question

I wish to show that $$f(x) = x^5 + 4x^4 + 2x^3 + x^2 + 4x + 54$$ is irreducible over $\mathbb{Z} / 7\mathbb{Z}$. Simply testing every element, it is easy to see that there are no roots and thus no linear factors, but I'm not sure how to proceed from there.

Answer 1

You can try case by base: 5 = 1+4 = 2+3 and $F = \mathbb{Z}/7\mathbb{Z}$ is a field so, there are three options: $f = x^5 + 4x^4 + 2x^3 +x^2 + 4x + 54$ is irreducible, it decompose int terms of degree 1 and 4 or terms of degree 2 and 3. As you checked, the second case is impossible because it has no root in $F$. Let's try the other case:

$$ f = (x^2 + b x + c)(x^3 + dx^2 + ex + f) = x^5 + (d+b)x^4 + (e+bd+c)x^3 + (f+be+cd)x^2 + (bf+ce)x + cf $$

So

$$ b+d = 4\\ e+bd+c = 2\\ f+be+cd = 1\\ bf+ce = 4\\ cf=54 $$

Or

$$ \left[ \begin{array}{l} 1& 0 &0\\ b &1 &0\\ c &b &1\\ 0 &c &b\\ 0 &0 &c \end{array} \right] \left[ \begin{array}{l} d\\ e\\ f \end{array} \right] = \left[ \begin{array}{l} 4-b\\ 2-c\\ 1\\ 4\\ 54 \end{array} \right] $$ So, we need to verify if this linear system have a solution;

$$ \left[ \begin{array}{l} 1& 0 &0\\ b &1 &0\\ c &b &1\\ 0 &c &b\\ 0 &0 &c \end{array} \right] \begin{array}{l} 4-b\\ 2-c\\ 1\\ 4\\ 54 \end{array} \\ \simeq \left[ \begin{array}{l} 1& 0 &0\\ 0 &1 &0\\ 0 &0 &1\\ 0 &0 &0\\ 0 &0 &0 \end{array} \right] \begin{array}{l} 4-b\\ 2-c - b(4-b)\\ 1 - c(4-b) - b(2-c - b(4-b))\\ 4 - c(2-c - b(4-b)) - b(1 - c(4-b) - b(2-c - b(4-b)))\\ 54 - c(1 - c(4-b) - b(2-c - b(4-b))) \end{array} $$ We need $$ 4 - c(2-c - b(4-b)) - b(1 - c(4-b) - b(2-c - b(4-b))) = 0\\ 54 - c(1 - c(4-b) - b(2-c - b(4-b))) = 0 $$ And you can use a computer to check all 49 cases in (b,c) and see that this system has no solution in $F$.

Then, this case of decomposition is impossible too and $f$ is irreducible.

Note, this is a completely computacional way to do that. Essentially, I'm using that $F$ is a finite field and we can really check all it's elements.

Answer 2

Showing a "standard" way of settling questions like this.

The key well known fact is that the polynomial $p(x)=x^{49}-x$ is the product of all the irreducible linear and quadratic polynomials over $\Bbb{F}_7$. This is because all those polynomials are minimal polynomials of some elements of $\Bbb{F}_{49}$. But OTOH all the elements of $\Bbb{F}_{49}$ are zeros of $p(x)$.

A consequence of this is that if $f(x)$ has no common factors with $p(x)$ then it has no linear or quadratic factors. That is enough to conclude that $f(x)$ must be irreducible in $\Bbb{F}_7[x]$.

That gcd can be efficiently calculated with the Euclidean algorithm. This is trivial for a suitable CAS, but can actually be done with paper & pencil work. In this particular case there are certain coincidences making it quite fast. Let me demonstrate!

We calculate certain remainders modulo $f(x)$. The most taxing step turns out to be (we are extraordinary lucky here): $$ x^6-(x+3)f(x)\equiv 6+4x\pmod 7.\tag{1} $$ The rest follows from this. Multiplying $(1)$ by $x$ gives the congruence $$ x^7\equiv 6x+4x^2\pmod{f(x)}.\tag{2} $$ Squaring $(2)$ gives then $$ x^{14}\equiv (6x+4x^2)^2\equiv x^2-x^3+2x^4\pmod{f(x)}.\tag{3} $$ On the other hand we are in characteristic seven, so we have the so called freshman's dream $$ (a+b)^7=a^7+b^7 $$ that holds for all polynomials $a,b\in\Bbb{F}_7[x]$. Let's raise the congruence $(2)$ to the seventh power. We first get that (recall Little Fermat telling us that $c^7=c$ for all constants $x\in\Bbb{F}_7$) $$ x^{49}\equiv 6x^7+4x^{14}\pmod{f(x)}.\tag{4} $$ But we already calculated the remainders of both $x^7$ and $x^{14}$ modulo $f(x)$, so we arrive at the congruence $$ \begin{aligned} x^{49}&\equiv 6x^7+4x^{14}\\ &\equiv 6(6x+4x^2)+4(x^2-x^3+2x^4)\\ &\equiv x+3x^2+4x^2+3x^3+x^4\\ &\equiv x+3x^3+x^4. \end{aligned} $$ So our calculations have showed that $$ x^{49}-x\equiv 3x^3+x^4=x^3(x+3)\pmod{f(x)}.\tag{5} $$ This means that any eventual common factor of $p(x)$ and $f(x)$ must also be a factor of the remainder $r(x)=x^3(x+3)$. But $r(x)$ is manifestly a product of linear factors. OTOH you had already checked that $f(x)$ has no linear factors, so we are done!