Determine if $x^5-4x^3+x^2+1$ is irreducible in $\Bbb Z_7[x]$

Question

I'm tasked with determining if $x^5-4x^3+x^2+1$ is irreducible in $\Bbb Z_7[x]$.

What I've Tried

So far, I have plugged in every number 0 through 6 to see if the polynomial has a root in $\Bbb Z_7[x]$. Unfortunately, $f(a) \neq 0$ (mod 7) for $a = 0$ through $a = 6$. So f(x) is not a product of a degree 1 and degree 4 polynomial.

So the only other possibility is to set this up as a quadratic times a cubic as follows:

$$x^5 - 4x^3+x^2+1=(ax^2+bx+c)(dx^3+ex^2+fx+g)$$

Then I can generate a system of equations for the variables: $$ad=1$$ $$ae+bd=0$$ $$af+be+cd=-4=3$$ $$ag+bf+ce=1$$ $$bg+cf=0$$ $$cg=1$$

But this looks miserable. For starters, there are 6 equations for 7 unknowns, so I'm not sure I could even solve such a system (or find a contradiction). Does anyone have any suggestions?

Answer 1

It's not fun, but you could do polynomial long decision with your three coeficients $a,b,c$ into the original equation.

Doing so I get 7 variables $a,b,c,d,e,f,g$ and 6 equations

$$ad=1$$ $$ae=-db$$ $$af=-4-dc-eb$$ $$ag=1-ec-fb$$ $$bg=fc$$ $$gc=1$$

In $\Bbb Z_7$.

As one of the commenters said we can always turn a non monic polynomial into a monic by distributing the leading coefficients inverse through.

If $ax^2+bx+c$ is a factor, then so is $a(x^2+a^{-1}bx+a^{-1}c)$.

Taking then, $a=1$ we have $d=1$ as well giving us

$$e=-b$$ $$f=-4-c-eb$$ $$g=1-ec-fb$$ $$bg=fc$$ $$gc=1$$

Which is still formidable. Yielding

$$f=-4-c+f^2c^2$$ $$1=c+fc^4-f^2c^4$$

Maybe a better approach is to long divide the quadratics which are irreducible in $\Bbb Z_7$. To see if any have zero remainder.

As it turns out $x^5-4x^3+x^2+1$ is an irreducible polynomial in $\Bbb Z_7$.

Answer 2

The product of the monic, irreducible polynomials over $\mathbb{F}_7$ with degree $1$ or $2$ is given by $x^{7^2}-x$.
If we prove that $p(x)=x^5+3x^3+x^2+1$ and $x^{48}-1$ are coprime we get that $p(x)$ has no irreducible factor with degree $\leq 2$, so it is an irreducible polynomial over $\mathbb{F}_7$.

To compute $(x^{48}-1)\pmod{x^5+3x^3+x^2+1}$ in $\mathbb{F}_7$ is very tedious but completely straightforward$^{(*)}$. It leads to

$$ \gcd(x^{48}-1,p(x)) = \gcd(x^4+2x^3-2x^2-3x-2,\,p(x)) = 1$$ then to the irreducibility of $p(x)$ over $\mathbb{F}_7$.

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$(*)$ This can be done by computing $M^{48}$ in $\mathbb{F}_7$ by repeat-squaring, where $M$ is the $5\times 5$ companion matrix of the polynomial $p(x)$. This is equivalent to performing the following chain of computations

$$x^1\to x^2\to x^3\to x^6\to x^{12}\to x^{24}\to x^{48}\pmod{ p(x)}$$ requiring at most $100$ multiplications in $\mathbb{F}_7$: not that much, after all.

Answer 3

Not sure if there's an easier way, but there are only 49 - 28 = 21 monic irreducible degree 2 polynomials modulo 7.