Find All Irreducible Monic Cubics over the Field with Three Elements

Interpreting the coefficient ring as the field with three elements, the answer is a list of eight polynomials. The key simplification is that a cubic over a field is reducible exactly when it has a linear factor, and a linear factor corresponds to a root in the field.

The notation $\mathbb{Z}_3$ can also mean the 3-adic integers. This article uses the finite-field interpretation $\mathbb{Z}/3\mathbb{Z}$, often written $\mathbb{F}_3$. The distinction matters; see the notation note below if you meant the 3-adic integers.

Why checking roots is enough

Every monic cubic over $\mathbb{F}_3$ has the form

$$f(x)=x^3+ax^2+bx+c,$$

where $a,b,c\in\{0,1,2\}$, and all arithmetic is modulo 3. If the cubic is reducible, its factor degrees must add to 3. At least one factor must therefore have degree 1, which means the polynomial has a root in $\mathbb{F}_3$. Conversely, any root gives a linear factor. Thus, the only possible roots to test are 0, 1, and 2.

Test the possible coefficients

First, $f(0)=c$. So a polynomial with $c=0$ has a root and is reducible. Only $c=1$ or $c=2$ need consideration.

For the other two field elements, the evaluations are

$$f(1)=1+a+b+c,\qquad f(2)=2+a+2b+c \pmod 3.$$

For a candidate to be irreducible, both values must be nonzero. Applying those two conditions gives the following surviving pairs $(a,b)$:

Constant term $c$	$b=0$	$b=1$	$b=2$
1	$a=2$	$a=2$	$a=0,1$
2	$a=1$	$a=1$	$a=0,2$

For example, when $c=1$, the conditions are $a+b\ne1$ and $a+2b\ne0$, modulo 3. When $c=2$, they are $a+b\ne0$ and $a+2b\ne2$. These rules give four pairs for each possible nonzero constant term.

The eight polynomials

Substituting the surviving coefficients gives all the irreducible monic cubics:

$(a,b,c)$	Polynomial in $\mathbb{F}_3[x]$
$(2,0,1)$	$x^3+2x^2+1$
$(2,1,1)$	$x^3+2x^2+x+1$
$(0,2,1)$	$x^3+2x+1$
$(1,2,1)$	$x^3+x^2+2x+1$
$(1,0,2)$	$x^3+x^2+2$
$(1,1,2)$	$x^3+x^2+x+2$
$(0,2,2)$	$x^3+2x+2$
$(2,2,2)$	$x^3+2x^2+2x+2$

The coefficient 2 is the same field element as $-1$. You may therefore see the same polynomials written using minus signs; for instance, $x^3+2x^2+1$ is $x^3-x^2+1$ over this field.

Worked check

Take $f(x)=x^3+2x^2+2x+2$. Its values at the three field elements are

$$\begin{aligned} f(0)&=2,\\ f(1)&=1+2+2+2=7\equiv1\pmod3,\\ f(2)&=2+2+4+2=10\equiv1\pmod3. \end{aligned}$$

None is zero, so the cubic has no linear factor over $\mathbb{F}_3$ and is irreducible there.

Common mistakes

• Forgetting to reduce modulo 3. For example, the value 7 is 1 in $\mathbb{F}_3$, not 7.
• Testing only 0 and 1. The element 2 is also a possible root; missing it can incorrectly label a reducible cubic as irreducible.
• Including constant term 0. Such a polynomial is divisible by $x$, so it is immediately reducible.
• Confusing irreducibility with having no repeated roots. The test here is that there are no roots at all in the field. A cubic can have repeated factors and still be reducible.

As a check on the number of results, the standard count of monic irreducible polynomials of degree 3 over a field with $q$ elements is $(q^3-q)/3$. For $q=3$, this is $(27-3)/3=8$, matching the list.

If $\mathbb{Z}_3$ means the 3-adic integers

The 3-adic integers are not the field with three elements. Over the 3-adic integers, there are infinitely many irreducible monic cubics; for example, $x^3-3u$ is Eisenstein at 3 for any 3-adic unit $u$. The finite list above is for coefficients modulo 3, that is, for $\mathbb{Z}/3\mathbb{Z}$.