Check if $f(x)=x^6+x+1$ is irreducible in $\mathbb{Z}[x]$

Question

Let $f(x)=x^6+x+1.$ I want to check if this one is irreducible in $\mathbb{Z}[x]$

I’m kinda stuck in this one not gonna lie, neither the Einsenstein criterion works, neither can I see any simple solution to this. Could someone give me a small help on this one.

Edit: I want to check if it is irreducible.

Answer 1

Hint If $$f(x) = x^6 + x + 1$$ factors over $\Bbb Z$ then it factors modulo $2$, and in particular one factor modulo $2$ must have degree $\leq 3$.

Since $f$ has an odd number of terms, neither $0$ nor $1$ is a root of $f$ and hence $f$ has no linear factor. Now, the only quadratic and cubic polynomials irreducible modulo $2$ are

$$x^2 + x + 1,$$ $$x^3 + x + 1, \qquad x^3 + x^2 + 1 ,$$

and dividing each of these into $f$ shows that none of these are factors of $f$ either.

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Edit When I first wrote an answer, I'd misread the polynomial as $x^6 + x^2 + 1$. For posterity, here is the hint I posted for that case:

Hint There are only $1$ and $2$ irreducible polynomials modulo $2$ of degrees $2$ and $3$, respectively, so we can deduce quickly that the polynomial $$f(x) = x^6 + x^2 + 1$$ factors modulo $2$ into irreducible polynomials as $$f(x) \equiv (x^3 + x + 1)^2 \pmod 2 .$$ So, if $f$ factors over $\Bbb Z$ into a product of polynomials of smaller degree, it factors as a product of $2$ cubic polynomials with constant terms $\pm 1$. Moreover, since $f$ is even, the substitution $x \mapsto -x$ must either fix both cubic factors or interchange them (up to an overall sign).

Answer 2

Brute force check cases:

$f(x)=(x-a)\times(\text{quintic})$ easily ruled out.

$f(x)=(x^2+mx+1)(x^4+ax^3+bx^2+cx+1)$.

$\ \ \ \ x^5$ coeff is $0\Rightarrow a+m=0$

$\ \ \ \ x^4$ coeff is $0\Rightarrow a^2=b+1$

$\ \ \ \ x^3$ coeff is $0\Rightarrow a-ab+c=0$

$\ \ \ \ x^2$ coeff is $0\Rightarrow 1-ac+b=0$

$\ \ \ \ x$ coeff is $1\Rightarrow c=1+a$

$\ \ \ \ $ go find a contradiction among these conditions.

$f(x)=(x^2+mx-1)(x^4+ax^3+bx^2+cx-1)$ case similar to above.

$f(x)=(x^3+ax^2+bx+1)(x^3+mx^2+nx+1)$

$\ \ \ \ x^5$ coeff is $0\Rightarrow a=-m$

$\ \ \ \ x^4$ coeff is $0\Rightarrow n+am+b=0$

$\ \ \ \ x^3$ coeff is $0\Rightarrow 1+an+bm+1=0$

$\ \ \ \ x^2$ coeff is $0\Rightarrow m+bn+a=0\Rightarrow bn=0$

$\ \ \ \ x$ coeff is $1\Rightarrow n+b=1\Rightarrow am=-1$

$\ \ \ \ \Rightarrow a=\pm 1,m=\mp 1$ and either $n=0,b=1$ or $n=1,b=0$,

$\ \ \ \ $but these contradict $1+an+bm+1=0$.

$f(x)=(x^3+ax^2+bx-1)(x^3+mx^2+nx-1)$ case is similar.