# Construct a field of 27 elements and find the structure of its additive group.

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?

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

$$\mathbb F_{27}$$ 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.
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 $$\mathbf F_p$$.