# Polynomial of degree 3 with coefficients over $\mathbb{F}_3$ has always a root in GF(27) [duplicate]

Prove or give a counterexample:

Every polynomial of degree 3 having coefficients over $$\mathbb{F}_3$$ has always a root in $$\mathbb{F}_{27}$$.

I noticed that $$\mathbb{F}_{27} = \mathbb{F}_{3}[x]/(f)$$, for f irreducible in $$\mathbb{F}_{3}[x]$$. I wrote down the form of an arbitrary polynomial of degree 3 over GF(27) and tried to find an irreducible polynomial (i.e. no roots), but this did not work out.

• One way (if $f$ is irreducible) is to use the uniqueness (up to isomorphism) of the field of order 27. I do not consider the uniqueness to be trivial, but you can find it on this website by searching. Commented Jun 19, 2022 at 11:58
• I don't get it, in fact haven't you already answered the question inside your question ? You have written $\mathbf{F}_{27} \cong \mathbf{F}_3[x]/(f)$. This is the nontrivial statement (it is what I would justify using uniqueness of finite field of order $27$). Once you have this, you already know $f$ has a root inside $\mathbf{F}_{27}$: it is the image of the symbol $x$'' under the isomorphism that you said exists in your question. Commented Jun 19, 2022 at 17:08
• If $f$ is irreducible over $\Bbb{F}_3$ then it has a zero in $\Bbb{F}_{27}=\Bbb{F}_3[x]/\langle f\rangle$. If it isn't irreducible, it has a linear factor and hence ___ (you fill in the blank). Commented Jun 20, 2022 at 5:09