I'm trying to do/understand the following exercise:
"Let $E$ be a finite extension of a field $F$. If $[E:F] = 2$, show that $E$ is a splitting field of $F$."*
Background: Just beginning studying fields, I know nothing of Galois theory. Here is my reasoning so far:
$[E:F] = 2$ means that E, when viewed as a vector space over F, has dimension 2. This means that
$\beta \in E \implies \beta = b_0 + b_1 \alpha$, where $\alpha \in E \setminus F$
In other words, $E$ must not only be a finite extension but a simple extension $F(\alpha)$, where the degree of $\alpha$ over F is 2. Did I interpret the definitions correctly here? Does $[E:F]$ imply that $E$ is such a simple extension? [Cleared up definition of a field splitting a field.]
Now, assuming the above is correct, $\alpha \in E$ implies that $\alpha^2 = c+d\alpha$, for some $c$, $d \in F$. Consider this polynomial of degree 2 in $F[x]$:
We see that:
$p(\alpha)=\alpha^2-d\alpha-c = 0$
$p(x) = (x-\alpha)q(x)$
Where $q(x)$ must be of degree one, i.e. a linear factor. So $E$ is a splitting field of $p(x)$. Is my reasoning correct?
*pg. 330 from Judson's Abstract Algebra: Theory and Applications, freely available here http://abstract.ups.edu/download.html.