Extensions of degree $1$.

Question

My doubt is very simple:

Let $F|K$ be a field extension, if $[F:K]=1$, what can we say about $F$ and $K$? can I say $F=K$?

I'm trying to prove the equality without success.

Thanks in advance

Answer 1

Suppose $\exists a \in K \setminus F$. Then $1, a$ are $F$-linearly independent. Proof: If $f_1, f_2 \in F$ such that $f_1 1 + f_2 a = 0$, it follows that $f_2 a = -f_1$. If $f_2 \neq 0$, then $a = - \frac{f_1}{f_2} \in F$, a contradiction. Otherwise, we have $f_1 = 0$ as well, i.e. the only solution is $f_1 = f_2 = 0$. Thus $1, a$ are $F$-linearly independent.

Of course, there cannot exist two $F$-linearly independent elements in $K$ if $[K:F]=1$.

Edit: Sorry for the change in notation, I assumed $F \subset K$.

Answer 2

$F$ is a $K$-vector space and $\{1\} \subseteq F$ is linearly independent. Hence, if $F$ is $1$-dimensional over $K$, $\{1\}$ is a basis and then $F = K \cdot 1 = K$.

Answer 3

The fact that $[F:K] = 1$ means that as a $K$-vector space, $F$ has a basis of cardinality $1$. Let us suppose this basis is $B = \{k_0\}$.

Since a basis is a spanning set, this means that for ANY $\alpha \in F$, we have: $\alpha = c\cdot k_0$ for some $c \in K$. But the $K$-action (scalar multiplication) of $F$ as an extension field of $K$ is just the $F$-multiplication.

So $\alpha = ck_0$. Now as a subring of $F$, $K$ is closed under multiplication, so we conclude $F \subseteq K$.

Since $F$ is an extension of $K$, we have a priori, $K \subseteq F$, thus the two are equal.