Intermediate fields Separable, Algebraic, or Splitting I just took my algebra final, and I had a few questions about intermediate fields and the properties of their extensions
True or False...For $E \supset L \supset F$


*

*$E \supset L$ is algebraic and $L \supset F$ is algebraic $\Rightarrow E \supset F$ is algebraic

*$E \supset F$ is algebraic $\Rightarrow E \supset L$ is algebraic
and $L \supset F$ is algebraic

*$E \supset L$ is separable and $L \supset F$ is separable $\Rightarrow E \supset F$ is separable

*$E \supset F$ is separable $\Rightarrow E \supset L$ is separable
and $L \supset F$ is separable

*$E \supset L$ is splitting and $L \supset F$ is splitting $\Rightarrow E \supset F$ is splitting 

*$E \supset F$ is splitting $\Rightarrow E \supset L$ is splitting 
and $L \supset F$ is splitting 


My guess is that 1,3,5 are true, but I can't work out a proof for 2,4, or 6. My guess is that they are false.
Could anyone point me in the right direction? Thanks in advance
 A: There are many things to verify here; I'll refer to other sources whenever possible. My guess is that your "splitting" is what most people refer to as "normal".
Again, let $E \supset L \supset F$ be a tower of field extensions.
If $E/L$ and $L/F$ are algebraic, then so is $E/F$ [Milne, 1.31]. The converse is easier: if $\alpha \in E$, then there exists a polynomial with coefficients in $F$ having $\alpha$ as a root. Since the coefficients lie in $F \subset L$, $E/L$ is algebraic. Since, in particular, we can find such a polynomial when $\alpha \in L \subset E$, $L/F$ is algebraic.
Both implications hold for separability [Conrad 3.13].
Everything breaks down for normality, in general. Here's the example I always trot out: any extension of degree $2$ is normal, so in particular both steps of $\mathbf Q(\sqrt[4]{2}) \supset \mathbf Q(\sqrt{2}) \supset \mathbf Q$ are normal. And yet $\mathbf Q(\sqrt[4]{2})$ is not normal over $\mathbf Q$. On the other hand, this field is contained in $\mathbf Q(\sqrt[4]{2}, i)$, which is normal over $\mathbf Q$.
