Prove that a field is a vector space over a subfield

Question

I am having great difficulty understanding this. Let $(F,+,.)$ is a field with characteristic $p$. Now I have to prove that the order of $F$ is $p^m$. Proving this involves the proof that the field $F$ is, in fact, a vector space over the field of order $p$. I know the definition of a vector space, but I don't see how it can fit in this context. For example, $GF(8)$ contains polynomials and $0,1$. In fact, there is no vector in $GF(8)$. Then how can we say that $GF(8)$ is a vector space over $\{0,1\}$?. Would you please try to explain this to me at the elementary level?

Answer 1

Don't confuse the abstract notion of a field with a specific construction of that field...

Let $F\subseteq K$ be a subfield and a field. Then you have an addition $+$ and a product $\cdot$ that make $K$ into a field, and the restriction of those operations on $F$ make $F$ a field.

We can now view $K$ as a vector space over $F$ by letting "vector addition" be the addition $+$ on $K$. We let "scalar multiplication" be the restriction of the product of $K$ to only product where the first factor is in $F$ and the second factor is in $K$. You can now verify that these definitions satisfy all the axioms of a vector space over $F$, so that makes $K$ a vector space over $F$.

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Now, even if you want to stick to thinking of $K=\mathrm{GF}(8)$ as a set of equivalence classes of polynomials with coefficients in $\{0,1\}$, then it is clear how they form a vector space over $\{0,1\}=\mathbb{F}_2$: addition is the addition in $\mathrm{GF}(8)$, and scalar multiplication is done by letting multiplication by $0$ give $0$, and multiplication by $1$ doing nothing. What is the problem?

Answer 2

Hint. Try to use just the definition of a vector space to show that the real numbers are a vector space over the subfield of rational numbers.

That argument will work pretty much word for word whenever one field is a subfield of another.

Answer 3

I'll try to explain this with a different but very concrete example:

We have that $\Bbb Q$ is a subfield of $\Bbb R$.

That means that $\Bbb R$ is a vector space over $\Bbb Q$. The vector addition between two vectors (two real numbers) is just the regular addition in $\Bbb R$. The scalar multiplication of a scalar (a rational number) and a vector (a real number) is just the regular multiplication in $\Bbb R$. It's straightforward to prove that this two operations satisfies the properties required for a vector space.

In a similar way, one can prove that any field is a vector space over any of its subfields.