Show that if $[E : F] = 2$, then $E$ is a splitting field over $F$?

Question

At present, I know this might involve the Factor Theorem/Division Algorithm in $E[x]$ and finding a root in the splitting field $E$, such that the root is not in $F$. Could anyone please elaborate further? Thank you in advance for the help.

Answer 1

This area is not my strength, so there may be problems with this answer, but I thought I would give my ideas anyway in case they might be helpful:

Let $F \subseteq E$ be a field extension. Suppose $[E:F] = 2$. We want to show $E$ is the splitting field of some polynomial over $F$.

Since $[E:F] = 2$, we have that $\{1, \alpha\}$ is a basis for $E$ as a vector space over $F$, where $\alpha$ is a specific element of $E - F$. Note that this means we can write $E = F(\alpha)$. Then, since $\alpha^{2} \in E$, we have $\alpha^{2} = f_{1} + f_{2} \alpha$ for some $f_{1}, f_{2} \in F$, i.e., $\alpha^{2} - f_{2} \alpha - f_{1} = 0$.

Then since $\alpha$ is a root of $h(x) = x^{2} - f_{2}x - f_{1}$ which is a polynomial in $F[x]$, and $h$ is a degree 2 polynomial, it must be that this polynomial splits in $E$. However, it doesn't split in $F$ since $\alpha \not \in F$.

Answer 2

Let $\{1,\alpha\}$ be a basis of $E$ over $F$. Then, from $F\subset F(\alpha)\subseteq E$, we get $$2=[E:F]=[E:F(\alpha)]\,[F(\alpha):F]$$ and so, since $[F(\alpha):F]>1$, we have $F(\alpha)=2$ and $F(\alpha)=E$.

Thus the minimal polynomial of $\alpha$ has degree $2$ over $F$, let it be $X^2-sX+p$. Since this polynomial factors in $F(\alpha)[X]$ as $(X-\alpha)(X-\beta)$, we have that $\beta=s-\alpha\in F(\alpha)$.

Answer 3

We cannot have $E=F$; for that would imply $|E:F |=1$. Hence choose some $u\in E\backslash F$. Then by Tower's law:

$2=|E:F|=|E:F(u)||F(u):F|$.

Since $|F(u):F|=2$, we must have $|E:F(u)|=1$ and so $E=F(u)$.

The nice thing about the above is that, replacing $2$ by a prime $p$, then if $|E:F|=p$ then $E=F(u)$ for some $u\in E$.

Therefore the degree of the minimal polynomial $m$ of $u$ in $F$ is $2$.

Now just show that $m$ must contain all of its roots in $F(u)$.