is $f(x)=x^4 +x^3 +x^2 +x +1$ irreducible in $R=(\mathbb Z/7\mathbb Z)[x]$?

Question

is $f(x)=x^4 +x^3 +x^2 +x +1$ irreducible in $R=(\mathbb Z/7\mathbb Z)[x]$?
I know that $f(x)$ cannot be expressed as a product of degree 1 polynomial and degree 3 polynomial since it has no roots in $R$ but what about two degree 2 polynimials?

Answer 1

Since $(X-1)(X^4 + X^3 + X^2 + X + 1) = X^5 - 1$, if $f$ wasn’t irreducible, you’d have a degree two factor $(X - α)(X - β) = X^2 - (α+β) X + αβ~$ of $X^5 - 1$ of $f$ in $ℤ/7ℤ[X]$, where $α$, $β$ and therefore $αβ$ ($∈ ℤ/7ℤ$) had to be $5$-th roots of unity. What $5$-th roots of unity are there in $ℤ/7ℤ$? What possibilities are there for $α$ and $β$? (Note that $α$ and $β$ don’t have to be elements of $ℤ/7ℤ$ a priori, but elements of some extension of $ℤ/7ℤ$.)

Answer 2

If $f=(ax^2+bx+c)(dx^2+ex+f)$ with $a,d \neq 0$ mod $7$ one can divide the first factor by $a$ to make it monic, then remultiply by $a$ to compensate. So resetting the coefficients at this point $f=(x^2+bx+c)(dx^2+ex+f).$ [these are "reset" $d,e,f$] Then the second factor is also monic to make the coefficient of $x^4$ in the product match that of $f.$ So we have $$f=(x^2+bx+c)(x^2+ex+f).$$ Now we have $cf=1$ so that $c,f$ make one of the pairs (in either order) $(1,1),\ (2,4),\ (3,5),\ (6,6).$ We also have $b+e=1$ so that $b,e$ make up one of the pairs $(0,1),\ (2,6),\ (3,5),\ (4,4).$ Now the coefficient of $x^2$ in the product is $(c+f)+(be).$ The values of $c+f$ are $1,2,5,6,$ while the values of $be$ are $0,1,2,5.$ So to get the coefficient of $x^2$ to be $1$ we need a sum of $1$ mod $7$ using one term from the first list and one from the second. The only choice is $c+f=6,\ be=2.$ This leads to $(c,f)=(2,4)$ [without loss] and to $(b,e)=(4,4).$ But then the product of the two factors becomes $x^4+8x^3+22x^2+24x+8,$ which has all coeffients $1$ mod $7$ except that of $x^1$ which happens to come out $3$ mod $7.$

Added note to explain a comment below: The expansion of $(x^2+bx+c)(x^2+ex+f)$ is $$x^4+(b+e)x^3+(c+f+be)x^2+(bf+ce)x+cf.$$ So although true that $bf+ce=1,$ it is also true that the simpler relation $b+e=1$ must hold, in order that this product should match $f$ which has all coefficients $1$ mod $7.$ The above argument works by assuming the cubic and constant coefficients are $1$ to get relations separately between $b,e$ versus $c,f$ and then putting these into the $x^2$ coefficient, since that has a simple expression in terms of sum/products from $b,e$ and $c,f.$ The last thing one would try to match would be the coefficient $bf+ce$ of the power $x^1,$ and that gives what goes wrong in this case since it cannot be made $1$ after the other coefficients of $f$ have been matched.