# Is the polynomial $(x^3 + 2x + 1)$ irreducible in $\mathbb{Z}_{3}[x]$?

Let $$F$$ be a field, $$f \in F[x]$$ of degree 2 or 3. Theorem: If $$f$$ has no roots, then $$f$$ is irreducible.

Using this theorem, I am trying to prove if $$(x^3 + 2x + 1)$$ is irreducible in $$\mathbb{Z}_{3}[x]$$. I am having a hard time finding roots for $$(x^3 + 2x + 1)$$ in $$\mathbb{Z}_{3}[x]$$ as the $$x^3$$ is throwing me off. When I have a degree 2 polynomial such as $$(x^2 + 8)$$ or $$(x^2 + 1)$$, it is a lot easier as degree-2 polynomials can be factored into $$(x + n) \cdot (x + n)$$. Is there an easier way to go about trying to find the roots of a degree-3 polynomial like $$(x^3 + 2x + 1)$$? Any feedback would be appreciated.

• Is $f(1), f(2), f(3) \equiv 0 \pmod{3}$? Dec 2, 2019 at 5:34
• What do you mean by $\mathbb Z_3$? If you mean $\mathbb Z/3\mathbb Z$, then you may just test the three values one by one. If you mean the ring of $3$-adic integers, then it's the same argument, just modulo $3$. Dec 2, 2019 at 5:34
• You really do not need to solve the equations.
– Keon
Dec 2, 2019 at 5:35
• are you sure? $f(1) = 1 + 2 + 1$ right? Dec 2, 2019 at 5:38
• @CalvinLin I'm sorry. I thought $f(1)$ = 3. Since $f(1), f(2)$, and $f(3)$ are $\not\equiv$ 0 (mod 3), does this mean that $f$ has no roots? Dec 2, 2019 at 5:41

Let $$f=X^3+2X+1\in (\mathbf{Z}/3\mathbf{Z})[X]$$.

Let us check if $$f$$ has roots in $$\mathbf{Z}/3\mathbf{Z}=\{\overline0,\overline1,\overline2\}$$. (For $$f\in(\mathbf{Z}/3\mathbf{Z})[X]$$ to have a root in $$\mathbf{Z}/3\mathbf{Z}$$ means that there exists $$x\in\mathbf{Z}/3\mathbf{Z}$$ such that $$f(x)\equiv0\bmod{3}$$.)

We see that $$f(\overline0),f(\overline1),f(\overline2)\not\equiv0\bmod{3}$$, so $$f$$ has no roots in $$\mathbf{Z}/3\mathbf{Z}$$. Since the degree of $$f$$ is $$\leqslant 3$$, we can conclude that $$f$$ is irreducible over $$(\mathbf{Z}/3\mathbf{Z})[X]$$ by your previous question.

Yes: it doesn't have a root in $$\Bbb Z_3$$.

A cubic polynomial is reducible if and only if it has a factor of degree $$1$$, for if

$$\deg c(x) = 3, \tag 1$$

and

$$c(x) = a(x)b(x) \tag 2$$

with

$$\deg a(x), \deg b(x) \ge 1, \tag 3$$

and also

$$\deg a(x) + \deg b(x) = \deg c(x) = 3, \tag 4$$

elementary arithmetic forces precisely one of $$\deg a(x)$$, $$\deg b(x)$$ to take the value $$1$$, and the other $$2$$; without loss of generality we may take

$$\deg a(x) = 1, \; \deg b(x) = 2; \tag 5$$

with

$$\deg a(x) = 1 \tag 6$$

we may write

$$a(x) = a_1 x + a_0, \; a_1 \ne 0; \tag 7$$

then

$$x_0 = -\dfrac{a_0}{a_1} \tag 8$$

is clearly a root of $$c(x)$$:

$$c(x_0) = a(x_0)b(x_0) = 0. \tag 9$$

We have thus shown that a reducible cubic over $$\Bbb Z_3$$ has a root in that field; in fact, careful scrutiny of what we have done reveals this result holds over ant field $$\Bbb F$$; there is nothing intrinsic to $$\Bbb Z_3$$ in our argument. In any event, the question at hand may be simply resolved by evaluating $$x^3 + 2x + 1$$ at every element of $$\Bbb Z_3$$:

$$f(0) = 0^3 + 2 \cdot 0 + 1 = 1, \tag{10}$$

$$f(1) = 1^3 + 2 \cdot 1 + 1 = 1, \tag{11}$$

$$f(2) = 2^3 + 2 \cdot 2 + 1 = 8 + 4 + 1 = 1, \tag{12}$$

where all arithmetic is of course done in $$\Bbb Z_3$$; we see that $$f(x) = x^3 + 2x + 1$$ has no zeroes in the field $$\Bbb Z_3$$ and hence is irreducible.

In closing, I think it is only fair to point out that the basic result of this answer, that a reducible cubic polynomial must have a zero, is not always so easy to apply in practice; our advantage here is that $$\Bbb Z_3$$ has only three elements, ensuring that the amount of calculation is relatively small; over a field of large though finite cardinality, or over an infinite field, the amount of arithmetic required may become intracibly large. Nevertheless, the number of situations in which the present test may be efficiently applied is probably large enough to make it worth keeping in one's toolbox of polynomial "hammers".