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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.

Appreciate your help!

Anton2107
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  • 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. – babu_babu Jun 19 '22 at 11:58
  • Can you explain to me why this solves it? I don't see it rn – Anton2107 Jun 19 '22 at 12:28
  • 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. – babu_babu Jun 19 '22 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). – Jyrki Lahtonen Jun 20 '22 at 05:09

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