Show that $x^5+x^3+x+1$ is irreducible over $\mathbb Q$.

Question

Show that $p(x)=x^5+x^3+x+1$ is irreducible.

My attempt: First, I tried to shifting Eisenstein's up to $10$, i.e., $p(x+a)$, where $a=1,\dots, 10$ using calculator. I did not continue because if I am being honest, you will get bored after a several failures. Also, please, I ask you to do not recommend rational root test for irreducibility.

Actually my original question was $x^5+Ax^3+Ax+1$ where $A \in \mathbb Z$.

I put myself this example to observe my own special version, but even can't show easier version. Or, my question can be

Is $x^7+Ax^5+Bx^3+Ax+1$, with $A,B\in\mathbb Z$, irreducible?

(YES! But how can we show?). I couldn't solve also this example.

I know that one way to show is the following: If polynomial is irreducible for some modulo $p$ then it is irreducible over $\mathbb Z$ and by Gauss Lemma irreducible over $\mathbb Q$.

Any help is appreciated.

Answer 1

Working over $\Bbb{F}_2$ it is clear that $1\in\Bbb{F}_2$ is a root of $x^5+x^3+x+1\in\Bbb{F}_2[x]$, and we find that $$x^5+x^3+x+1\equiv(x+1)(x^4+x^3+1)\pmod{2}.$$ The quartic factor clearly has no roots in $\Bbb{F}_2$, and because the only irredicuble quadratic over $\Bbb{F}_2$ is $x^2+x+1$, it suffices to verify that $$x^4+x^3+1\neq(x^2+x+1)^2,$$ to conclude that $x^4+x^3+1$ is irreducible over $\Bbb{F}_2$. Over $\Bbb{Q}$, this means that $x^5+x^3+x+1$ has an irreducible factor of degree at least $4$. That is to say, it is either irreducible or it has a root in $\Bbb{Q}$. The rational root test tells you that if it has a root in $\Bbb{Q}$, then that root must be $1$ or $-1$. But a quick check shows that these are not roots, and hence the original polynomial is irreducible over $\Bbb{Q}$.

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The proof of the more general case of $x^5+Ax^3+Ax+1$ can be handled in a similar way, albeit with a bit more work. Of course the argument above holds identically for any odd $A\neq1$. But for even $A$ it does not.

If we try the same argument over $\Bbb{F}_3$ we get the following factorizations into irreducible factors: $$x^5+Ax^3+Ax+1\equiv\begin{cases} (x+1)(x^4+2x^3+x^2+2x+1)&\text{ if } A\equiv0\pmod{3}\\ x^5+x^3+x+1&\text{ if } A\equiv1\pmod{3}\\ (x+2)(x^4+x^3+2)&\text{ if } A\equiv2\pmod{3} \end{cases}$$ So for $A\equiv1\pmod{3}$ we immediately get that $x^5+Ax^3+Ax+1$ is irreducible over $\Bbb{Q}$. For $A\equiv0,2\pmod{3}$ we see, as before, that $x^5+Ax^3+Ax+1$ has an irreducible factor of degree at least $4$. If it has an irreducible factor of degree exactly $4$ then it must also have a linear factor, and hence a rational root. By the rational root test this root must be either $1$ or $-1$, which tells us that either $$1^5+A\cdot1^3+A\cdot1+1=0\qquad\text{ or }\qquad (-1)^5+A\cdot(-1)^3+A\cdot(-1)+1=0,$$ that is to say, either $A=-1$ or $A=0$. Indeed for these two values of $A$ the polynomial $x^5+Ax^3+Ax+1$ is clearly reducible over $\Bbb{Q}$, and then for all other values of $A$ it is irreducible.

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As for the septic polynomial $x^7+Ax^5+Bx^3+Ax+1$ with $A,B\in\Bbb{Z}$, it is not irreducible for all $A,B\in\Bbb{Z}$. For example, by the rational root theorem it has a linear factor if and only if $1$ or $-1$ is a root, or equivalently, if and only if $$1^7+A\cdot1^5+B\cdot1^3+A\cdot1+1=0\qquad\text{ or }\qquad (-1)^7+A\cdot(-1)^5+B\cdot(-1)^3+A\cdot(-1)+1=0,$$ which is in turn equivalent to $2A+B=0$ or $2A+B=-2$. There are also values of $A$ and $B$ for which it is reducible but without linear factors. Finding them all is an interesting exercise!

Answer 2

Since the degree of $p(x)$ is low, we can use Rational Root Theorem to solve it.Assume that $p(x)$ is reducible in $\mathbb{Q}[x]$,there must exist $g(x),h(x)$,such that $p(x)=g(x)h(x)$.What's more ,we have $\deg g(x)=3$ and $\deg h(x)=2$(that's because f(x) doesn't have rational root.If $\deg h(x)=1$,then $f(x)$ has the root $h(x)$ has,then contradiction.Thus,we can let $g(x)=x^3+a_1x^2+a_2x+1 $ and $ h(x)=x^2+a_3x+1$(another situation is that $g(x)=x^3+a_1x^2+a_2x-1 $ and $ h(x)=x^2+a_3x-1$.But I think it's probably same).Write down $f(x)=g(x)h(x)$ and compare the number,we have

$ \begin{cases} a_1+a_3=0\\ a_2+a_1a_3+1=1\\ a_1+a_2a_3+1=0\\ a_2+a_3=1 \end{cases} $

But there has no solution for $a_3 $ in $\mathbb{Q}$.

I think this method cannot solve the situation that $\deg p(x)$ is too high.

Answer 3

By Gauss' Lemma, any non-trivial factorization over $\Bbb Q$ with two monic polynomials is already over $\Bbb Z$. Assume that $$ f=x^5 + x^3 + x+1 $$ is reducible over $\Bbb Q$, thus also over $\Bbb Z$, and let us write it in the form $f=gh$. Now take this relation modulo $3$, so we have also the corresponding decomposition over the field $F=\Bbb F_3$ with three elements. $f$ has no root in this field, so the decomposition involves a polynomial of degree two, and a polynomial of degree three. There are only three polynomials of degree two, which are irreducible in $F[x]=\Bbb F_3[x]$, and these are: $$ \begin{aligned} &x^2+1 &&\text{ but }&f&=(x^2+1)\cdot x^3+(x+1)\ ,\\ &x^2+x-1&&\text{ but }&f&=(x^2+x-1)\cdot (x^3-x^2-1)-x\ ,\\ &x^2-x-1&&\text{ but }&f&=(x^2-x-1)\cdot (x^3+x^2+1)-x-1\ , \end{aligned} $$ so there is always a rest. The assumption is false, so our polynomial is irreductible (even seen over $\Bbb F_3[x]$).

$\square$

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Note: For $A=-1$ and $A=0$ the polynomial $X^5+Ax^3+Ax+1$ has a factorization: $$ \begin{aligned} x^5 -x^3-x+1 &=(x-1)(x^4+x^3-1)\ ,\\ x^5 +1 &=(x+1)(x^4-x^3+x^2-x+1)\ , \end{aligned} $$

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Note: There are also values of $A,B$ such that the other polynomial has a factorization... For instance $A=B=0$. But a lot of other cases. One can easily take any $A$ and force $\pm 1$ as a root, which leads to some $B$-values. But also $(x^{3} - x^{2} + 1) \cdot (x^{4} + x^{3} + x^{2} + 1)=(x^7+2x^3+1)$ to have some fancy example.