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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.

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

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  • $\begingroup$ 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 $[F(\alpha):F]\leq 2$? $\endgroup$ – Chris Jan 22 at 2:26
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    $\begingroup$ @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... $\endgroup$ – Jyrki Lahtonen Jan 22 at 6:21
  • $\begingroup$ 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? $\endgroup$ – Chris Jan 22 at 19:23

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