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My attempt:

To construct a field of 27 elements. We need a 3 degree irreducible polynomial over $\mathbb F_{3}$. We know that such a polynomial $x^{3}+2x^{2}+1$ is irreducible over $\mathbb F_{3}$. Then we can construct a field

$\mathbb F_{27} $is isomorphic to $\frac{. \mathbb F_{3}}{x^{3}+2x^{2}+1}$.

Is there is any way to construct irreducible polynomial?

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    $\begingroup$ A cubic is irreducible if it has no roots in $\mathbb F_3.$ $\endgroup$ – Thomas Andrews Dec 2 at 23:35
  • $\begingroup$ The cubic irreducible polynomials over $\mathbb F_{3}$ are the cubic irreducible factors of $x^{27}-x$ mod $3$. Ask WA, $\endgroup$ – lhf Dec 3 at 0:49
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For the second part of the question:

$\mathbb F_{27}$ is a vector space of dimension $3$ over $\mathbb F_{3}$. Therefore, $\mathbb F_{27} \cong \mathbb F_{3} \times \mathbb F_{3} \times \mathbb F_{3}$ as additive groups.

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In a fixed algebraic closure, $\mathbf F_3$ admits a unique extension of degree $3$, which we can denote $\mathbf F_{27}$. Let us determine it using Artin-Schreier theory, which provides a canonical irreducible polynomial, see e.g. math.stackexchange.com/a/3462533/300700. Over a field $k$ of characterisic $p\neq 0$, the Artin-Schreier operator $P$ is defined by $P(x)=x^p-x$. Here the image $P(\mathbf F_3)$ consists only of $0$ and $P(-1)=-1$. Then $\mathbf F_{27}$ is the splitting field over $\mathbf F_3$ of the A.-S. polynomial $P(X)=X^3-X+1=X^3+2X+1$. Note that this is not the polynomial that you gave. As for the structure of the additive group, $\mathbf F_{p^n}$ is obviously a vector space of dimension $n$ over $\mathbf F_p$.

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