Describe in detail how to construct a field of order 16 from a field $F$ of order 2.

Question

From $F$ we want to construct a field $K$ of order $16 = 2^4.$ Since $F$ is order two then, we want $|K:F| = 4.$ So we would want to find an irreducible polynomial of degree $4$ in $F.$ Then $\frac{F[x]}{p(x)} \cong F[\alpha]$ for a root $\alpha$ of $p(x)$ (in an extension field of $F$ since $p(x)$ is irreducible in $F$). Furthermore, this field will be order $16,$ since it looks like $\{a + b\alpha + c \alpha^2 + d\alpha^3 : a, b, c, d \in F \}.$ So define $K = F[\alpha].$

I am not sure if I can be more specific than this. There is also the possibility that the irreducible polynomial of degree $4$ does not exist? What would one do in this case?

See https://math.stackexchange.com/questions/4143115/finite-field-gf16 — Wuestenfux, Jan 15 '22 at 16:42

score 3 · Accepted Answer · answered Jan 15 '22 at 16:44

3

Just take $p(x)=x^4+x+1$. Since it has no roots in $\Bbb Z_2$, the only way it could be reducible was if it could be written as the product of two quadratic polynomials with no roots in $\Bbb Z_2$. But there is only one such quadratic polynomial, which is $x^2+x+1$, and its square is $x^4+x^2+1$.

answered Jan 15 '22 at 16:44

José Carlos Santos

427,504

Why can we just consider $\mathbb{Z}_2$ and roots over $\mathbb{Z}_2$ when we are looking at a general field? – Mike Jan 15 '22 at 16:47
1

Because $\Bbb Z_2$ is the only field of order $2$. – José Carlos Santos Jan 15 '22 at 16:50

Describe in detail how to construct a field of order 16 from a field $F$ of order 2.

1 Answers1