For which values of $a\in\mathbb ℤ/3\mathbb ℤ$ is the quotient $\mathbb ℤ/3\mathbb ℤ[x]/(x^3+x^2+ax+1)$ a field?

Question

I'm trying to solve the following problem:

Determine for which values of $a\in\mathbb{Z}/3\mathbb{Z}$ the quotient $Q_a=(\mathbb{Z}/3\mathbb{Z})[x]/(x^3+x^2+ax+1)$ is a field.

I see two options:

• Show that ($x^3+x^2+ax+1$) is maximal, or
• show that every element of $Q_a \backslash \{0\}$ is invertible.

Any help would be appreciated.

Answer 1

The first option looks good, and recall that here "maximal" means that the generating element, that is the polynomial, is irreducible over $\mathbb{Z}/3\mathbb{Z}$.

Now, you are left with the task of deciding which of the polynomials are irreducible.

I cannot know which means you have for this, but if nothing else you can note that since the polynomials have degree $3$ they are reducible if and only if they have a root. (If this is not clear, try to prove it.)

Then, you can just check which polynomial has a root, for instance by plugging in the three possible values.

Answer 2

Here is my solution, thanks for your help.

$Q_a$ is a field if and only if $I:=(x^3+x^2+ax+1)$ is maximal.

$I$ is maximal if and only if $I=(p(x))$ for $p(x)\in(\mathbb{Z}/3\mathbb{Z})[x]$ an irreducible polynomial.

If a polynomial of degree 2 or 3 has no roots in $\mathbb{Z}/3\mathbb{Z}$, then it is irreducible in $(\mathbb{Z}/3\mathbb{Z})[x]$.

$f(x):= x^3+x^2+ax+1,\quad f(0)=1,\quad f(1)=a,\quad f(2)=2a+1$

f is irreducible only for $a=2$ so only $Q_2$ is a field.

Tell me if something is wrong.