Reduce polynomial $x^5-x+2$

Question

The polynomial $p(x) = x^5-x + 2$ is not reducible in $\mathbb {Z}$ (i.e mod $0$). Since it does not have a linear factor, so there is only the case where $p (x) = q (x) r (x)$, where $\deg (q) = 3$ and $\deg (r) = 2$. I have tried to show it by taking the polynomials $p (x) = (x^3+ax^2+bx+c)\cdot (x^2+dx+e)$ to get to the system of equations $$\left\{\begin{matrix} d+a=0\\ e+ad+b=0\\ ae+bd+c=0\\ be+dc=-1\\ ce=2 \end{matrix}\right.$$

but I cannot come to a contradiction, I would appreciate your help

Answer 1

HINT.-A way is to show, as you want to do, that $p (x) = (x^3+ax^2+bx+c)\cdot (x^2+dx+e)$ is not possible in $\mathbb Z[x]$ but avoiding to go to your system of five equations. For this note that $p(0)=p(1)=p(-1)=2$ so you have the system $$\begin{cases}ce=2\\(1+a+b+c)(1+d+e)=2\\(-1+a+b+c)(1-d+e)=2\end{cases}$$ From this you have $(c,e)$ is equal to $(1,2)$ or $(2,1)$ then you get $d=0,\space\space a-b=1$ and $a+b=-2$.

It follows that $a=\dfrac{-1}{2}$ and $b=\dfrac{-3}{2}$.

Thus the coefficients $a$ and $b$ are not rational integers and your factors $q(x)$ and $r(x)$ are not in $\mathbb Z[x]$.

Answer 2

A much more efficient and practical way to prove $x^5 - x + 2$ is irreducible over $\mathbf Z$ is to use the reduction mod $p$ test: if there is a prime $p$ such that $f(x) \bmod p$ is irreducible in $(\mathbf Z/p\mathbf Z)[x]$ then $f(x)$ is irreducible in $\mathbf Z[x]$. While $f(x) \bmod 2$ is reducible, $f(x) \bmod 3$ is irreducible: if $f(x)$ were reducible mod $3$ then it would have a linear or quadratic factor in $(\mathbf Z/3\mathbf Z)[x]$. Since $f(a) \not\equiv 0 \bmod 3$ for $a = 0, 1, 2$, there is no linear factor. For quadratic factors, the only quadratic irreducible polynomials in $(\mathbf Z/3\mathbf Z)[x]$ are $x^2 + x + 2$ and $x^2 + 2x + 2$. You can check in $(\mathbf Z/3\mathbf Z)[x]$ that $$ x^5 - x + 2 \equiv x+2 \bmod x^2 + x + 2, \ \ \ x^5 - x + 2 \equiv x+2 \bmod x^2 + 2x + 2 $$ so $x^5 - x + 2$ has no quadratic irreducible factor in $(\mathbf Z/3\mathbf Z)[x]$. Thus $x^5 - x + 2$ is irreducible in $(\mathbf Z/3\mathbf Z)[x]$, so it is irreducible in $\mathbf Z[x]$.

The nice feature of working in $(\mathbf Z/p\mathbf Z)[x]$ is that there are finitely many polynomials of each degree, so at worst there are only finitely polynomial divisions to make to exhaustively check a polynomial in this ring has no irreducible factor of a specified degree without having to worry about trying to restrict the coefficients in some way first, as you may need to do when you work in $\mathbf Z[x]$, which has infinitely many (irreducible) polynomials of each degree.