# On the splitting field of $f(X)$, when $\deg f =2$

Let $$F$$ be a field, $$f(X)\in F[X]$$ with $$\deg f=2$$ and $$E$$ is its splitting field.

We want to show that there are no proper intermediate fields in the extension $$E/F$$.

My thought. We know that $$f$$ has at most $$2$$ roots in its splitting field. Let $$\alpha,\beta$$ be its roots. Then, is is $$E=F(\alpha,\beta)$$. So, we have the Tower of Fields $$F \leq F(\alpha)\leq F(\alpha,\beta)=E$$. So, from the Tower Law, $$[F(\alpha,\beta):F]=[F(\alpha)(\beta):F(\alpha)][F(\alpha):F].$$ Since $$f(\alpha)=0_F$$, we have $$m_{(\alpha,F)}(X)|f(X) \implies \deg m_{(\alpha,F)} \leq \deg f \iff [F(\alpha):F]\leq 2.$$ Now, consider the extension $$F \leq F(\alpha)$$. Then, we have $$m_{(\beta,F(\alpha))}|m_{(\beta,F)}$$.

But like before, since $$f(\beta)=0_F$$, we have $$m_{(\beta,F)}(X)|f(X)$$. So, $$\deg m_{(\beta,F(\alpha))} \leq \deg m_{(\beta,F)} \leq \deg f \implies [F(\alpha,\beta):F(\alpha)]\leq 2.$$ Therefore, $$[E:F]\in \{1,2,3,4\}$$. If it is $$1$$, trivially our claim holds and if $$[E:F]\in \{2,3\}$$, since $$2,3$$ are prime, there are no proper intermediate fields in $$E/F$$. So, we take the case that $$[E:F]=4$$. By above the only possible case is to consider $$[F(\alpha)(\beta):F(\alpha)]=[F(\alpha):F]=2$$

But how can we continue from that? Is this proof right/in the correct direction?

Thank you.

• Doesn't this immediately follow from the fact that for each intermediate $L$, we have $[E:F]=[E:L][L:F]$ and $[E:F]=2$? – russoo Jan 23 at 23:40
• Thank you for your comment. Why $[F(\alpha, \beta) :F] =2$? – Chris Jan 23 at 23:49
• Yes sorry. We have $[E:F]=2$ if $f$ is irreducible and $1$ otherwise. – russoo Jan 24 at 0:00

It seems like the easier approach is to note that if $$x^2-ax+b=(x - \alpha)(x - \beta)$$, then $$a=\alpha+\beta$$ (and $$b=\alpha \beta$$) so $$a \in F \subseteq F[\alpha]$$ and $$\alpha \in F[\alpha] \Rightarrow \beta = a - \alpha \in F[\alpha]$$ and $$E=F[\alpha]$$.
• Thank you for your answer. So, $F(\beta)\subseteq F(\alpha)$. But how do we conclude that there are no intermediate fields from this claim? Is that because $[F(\alpha):F]\leq 2$? – Chris Jan 22 at 2:26
• @Chris If $K\subseteq F\subseteq E$ is a tower of extensions, then $[E:K]=[E:F]\cdot [F:K]$. So we have a factorization of $[E:K]$. If $[E:K]$ is a prime number... – Jyrki Lahtonen Jan 22 at 6:21
• Got it. If $[F(\alpha):F]=1\iff F(\alpha)=F$ and if $[F(\alpha):F]=2$, since it is a prime, there are no proper intermediate fields. Right? – Chris Jan 22 at 19:23