Suppose $F$ is a field extension of $\Bbb{Z}_2$ of degree $3$, prove $F$ is finite, what is the size of $F$?

Question

Suppose $F$ is a field extension of $\Bbb{Z}_2$ of degree $3$, prove $F$ is finite, what is the size of $F$?

Okay so what we have $[F:\Bbb{Z}_2]=3$ so if we view $F$ as a vector space over $\Bbb{Z}_2$ it is $3$ dimensional that is it has a basis consisting of $3$ vectors.

What I don't understand though is how we relate viewing $F$ as a vector space over the field $\Bbb{Z}_2$ to draw conclusions about the field $F$ itself.

It seems like it might be useful to try and compare $F$ and $\Bbb{Z}_2$ and use the fact that we can write any element of $F$ as $au_1+bu_2+cu_3$ where $a,b,c \in \Bbb{Z}_2$ and $u_1,u_2,u_3$ are the basis vectors of $F$.

Since $|\Bbb{Z}_2|=2$ would this mean that there are a total of $2^3=8$ different elements of $F$ that is $|F|=8$ is this correct?

Answer 1

Yes, exactly.

An $n$ dimensional vector space over a field $K$ is isomorphic to $K^n$, i.e. (after having a basis chosen) each vector can be identified with an $n$-tuple of elements of $K$.

About your question: actually, this exercise is a very good example. A field extension of a base field $K$ holds a $K$-vector space structure, and for example it gives pretty good information about the number of elements.
Likewise, anything you want to know about the addition of the extension field will already hold in the underlying vector space.

Answer 2

This is the right idea. Indeed, since the extension is degree $3$, we can find a basis of $3$ linearly independent vectors; we'll call them $\{1, v, w \}$. As you said, elements of this extension will necessarily look like $c_1 + c_2v + c_3w$ for $c_j \in \mathbb{Z}_2$.

To show that each unique tuple of $c_j$'s produces a unique element in the extension field, suppose we have $c_1 + c_2v + c_3w = d_1 + d_2v + d_3w$ where $c_j \neq d_j$ for some $j$. Suppose WLOG that $j=2$; rearranging yields $(c_2-d_2)v = (d_1-c_1) + (d_3-c_3)w$. After multiplying through by the inverse of $(c_2-d_2)$, we arrive at a contradiction since we have now written $v$ as a linear combination of $1$ and $w$.

Therefore, we simply count every possible combination of choices for $c_1, c_2, c_3$. Two choices for each of the three $c_j$'s $\implies 8$ combinations $\implies$ 8 elements in the extension field.