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I want to construct a field with $8$ elements and a field with $27$ elements for an ungraded exercise.

For $\bf 8$ elements: So we can't just have $\Bbb Z/8\Bbb Z$ since this is not even an integral domain. But rather we can construct $\Bbb F_2 \oplus \Bbb F_2 \oplus\Bbb F_2 \oplus \Bbb F_2 = \{0,1,\alpha,\alpha+1,\beta,\beta+1,\gamma,\gamma+1\}$.

This line of thinking seems to break from what I tried. Is there a better way to construct these things?

I saw this answer: Construct a finite field of order 27

We pick a polynomial irreducible polynomial and take the quotient of $\Bbb Z_3[x]$ but this wasn't helpful in me understanding the general ideal/method.

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Galois in the Field
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    That's precisely the general method. To construct a field of order $p^{n}$, you find an irreducible polynomial $p(x) \in \mathbb{F}{p}[x]$ of degree $n$. The quotient field $\mathbb{F}{p}[x]/p(x) \cong \mathbb{F}_{p}^{n}$. Also, a product of fields is not a field. In particular, your example is not isomorphic to the field with $8$ elements. – Alex Wertheim Oct 11 '15 at 06:59
  • Note that if $\mathbb{F}$ is a field, the polynomial ring $\mathbb{F}[x]$ is a principal ideal domain (actually a Euclidean domain). So if $p(x)$ is an irreducible polynomial in that ring, the principal ideal generated by $p(x)$ is a maximal ideal (and $\mathbb{F}[x]/p(x)$ is a field). – hardmath Oct 11 '15 at 07:13
  • In the second section of this answer you see a copy of $\Bbb{F}_8$ together with examples of its arithmetic. I'm not sure it answers your question well, but it gives you something to chew. As an additive group it is a direct sum of three copies of $\Bbb{F}_2$ - not four as in your attempt. That 4-way direct sum has 16 elements. You left out guys like $\alpha+\beta$, $\alpha+\gamma+1$ et cetera. – Jyrki Lahtonen Oct 11 '15 at 07:40
  • I’ve found that it seems rather easy to find a degree-$n$ irreducible over a finite field just by blind trial and error. – Lubin Oct 11 '15 at 18:48

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There are two ways you might want to represent a field of order $q^k$ (where $q\in \mathbb{N}$ is a prime and $k \gt 0$ a positive integer).

One is to imagine a simple algebraic field extension of $\mathbb{Z}_q$ which has a basis $\{1,\alpha,\ldots,\alpha^{k-1}\}$, namely $\mathbb{Z}_q[\alpha]$ where $\alpha$ satisfies an irreducible monic polynomial of degree $k$ over $\mathbb{Z}_q$:

$$ \alpha^k + c_1 \alpha^{k-1} + \ldots + c_{k-1}\alpha + c_k = 0 $$

Any element of this field extension can be represented in terms of the basis, and the addition is just the same as the addition of the $k$-dimensional vector space over $\mathbb{Z}_q$. But when we multiply, we have to perform substitutions to eliminate powers $\alpha^k$ and higher by using:

$$ \alpha^k = - c_1 \alpha^{k-1} - \ldots - c_{k-1}\alpha - c_k $$

The other way to think about it is to start with an irreducible polynomial $p(x)$ of degree $k$ over $\mathbb{Z}_q$ and construct $\mathbb{Z}_q[x]/p(x)$ as a quotient ring. Since the ideal generated by $p(x)$ is maximal, the quotient ring is a field and has dimension $k$ over $\mathbb{Z}_q$ as a vector space.

These two constructions are equivalent, with $\alpha$ being a root of:

$$ p(x) = x^k + c_1 x^{k-1} + \ldots + c_{k-1}x + c_k $$

and identified with $x \bmod{p(x)}$ in $\mathbb{Z}_q[x]/p(x)$.

It turns out that all field extensions of degree $k$ over $\mathbb{Z}_q$ are isomorphic, so as a computational convenience the irreducible polynomial $p(x)$ may be chosen to be simple in some way (e.g. having as few nonzero coefficients as possible).

It's not hard to come up with a monic irreducible polynomial of degree $3$ over $\mathbb{Z}_2$ in order to construct a field of $8$ elements. You only need to check for divisibility by linear (first degree) factors to be sure of irreducibility.

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hardmath
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Let $p$ be some prime and $k$ some natural exponent.

The field $\mathbb F_{p^k}$ can be thought of as a splitting field of $χ = X^{p^k} - X$ over $\mathbb F_p$.

There is a general way to construct splitting fields for any set of polynomials, in your case you just repeatedly construct $F[X]/(f)$ for some irreducible non-linear factor $f$ of $χ$ and some field $F$ you wish to extend. You start with $F = \mathbb F_p$, see here.

k.stm
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  • Do we ever need more than a single irreducible factor $f$ of degree $k$? – hardmath Oct 12 '15 at 00:03
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    @hardmath No, but the existence of such a factor is not obvious. The only way of proving it that I see at the moment is to already use the existence of a field with $p^k$ elements: Take a cyclic generator of its multiplicative group. Then it also generates the whole field, so its minimal polynomial has degree $k$, which therefore is an irreducible factor of $χ$ of degree $k$. But this would be kind of putting the cart before the horse. – k.stm Oct 12 '15 at 07:04
  • The nym of the OP (Galois in the Field) makes me wonder how much they already know about such things. – hardmath Oct 12 '15 at 14:51
  • @hardmath That’s also why I left that out. – k.stm Oct 12 '15 at 15:05
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Use Zech logarithms. For primitive a, compute 1+a^k = a^Z(k). See Collected Algorithms of the ACM 1973 Lam and McKay. Arithmetic over a finite field. Number 469.

John Mckay
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  • Given your enthusiasm for this approach, perhaps you would care to elaborate on what is meant by Zech logarithms? I believe a more detailed discussion would be critical to most Readers for proper appreciation of the suggestion. – hardmath Oct 12 '15 at 00:01