Show that $x^3-a$ is irreducible in $\mathbb{Z}_7$ unless $a=0$ or $\pm1$

Question

Show that $x^3-a$ is irreducible in $\mathbb{Z}_7$ unless $a=0$ or $\pm1$

My Idea:

Suppose $x^3-a=(Ax+b)(Bx^2+cx+d).$

Then $A=B=1$ or $A=B=-1$ WLOG $A=B=1.$

Then $x^3-a=x^3+x^2(b+c)+x(bc+d)+bd\\ \Rightarrow c+b=0,d+bc=0,bd=-a.$

I can't go further from here. Help...Thank You!

Answer 1

Let $F$ be any field, and let

$p(x) = x^3 + ax^2 + bx + c \in F[x] \tag 1$

be any cubic polynomial over $F$. Then we have the following

Fact: $p(x)$ is reducible in $F[x]$ if and only if it has a zero in $F$.

Proof of Fact: Clearly if $p(x)$ has a zero $z \in F$, then $p(x) = (x - z)q(x)$ where $q(x) \in F[x]$ is of degree $2$; thus $p(x)$ is reducible. Now if $p(x)$ is assumed reducible, we may write $p(x) = r(x)s(x)$ where precisely one of $r(x), s(x) \in F[x]$ is of degree one, since $\deg r + \deg s = \deg p = 3$. But a factor $\alpha x + \beta$ of degree $1$ yields a root $-\beta/\alpha$, an thus we are done. End of Proof of Fact.

So the question becomes, for which $a \in \Bbb Z_7$ does $x^3 - a$ have no zeroes. Now it is just a matter of simple arithmetic to discover which $a \in \Bbb Z_7$ are not perfect cubes. We have

$0^3 = 0, \; 1^3 = 1, \; 2^3 = 1, \; 3^3 = 6 = -1, \; 4^3 = 1, \; 5^3 = 6 = - 1, \; 6^3 = (-1)^3 = -1; \tag 2$

we see that $0, 1, -1$ are perfect cubes, but that

$a = 2, 3, 4, 5, 6 \tag 3$

are not cubes in $\Bbb Z_7$; thus $x^3 - a$ is irreducible for $a$ given by (3).

By the way, $x^3 = xx^2$ is trivially reducible, and

$x^3 - 1 = (x - 1)(x^2 + x + 1), \tag 4$

$x^3 + 1 = (x + 1)(x^2 - x + 1). \tag 5$

Answer 2

It's a cubic, so you need not work so hard. Indeed, you can just check if it has a root.

By exhaustion: $x^3-a\cong 0\implies x^3\cong a$, so
$$ 0^3=0=a\\ 1^3=1=a\\ 2^3=1=a\\ 3^3=-1=a\\ (-3)^3=1=a\\ (-2)^3=-1=a\\ (-1)^3=-1=a $$ where equality is in the the mod $7$ sense.