# 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 $E \supset L \supset F$

1. $E \supset L$ is algebraic and $L \supset F$ is algebraic $\Rightarrow E \supset F$ is algebraic
2. $E \supset F$ is algebraic $\Rightarrow E \supset L$ is algebraic and $L \supset F$ is algebraic
3. $E \supset L$ is separable and $L \supset F$ is separable $\Rightarrow E \supset F$ is separable
4. $E \supset F$ is separable $\Rightarrow E \supset L$ is separable and $L \supset F$ is separable
5. $E \supset L$ is splitting and $L \supset F$ is splitting $\Rightarrow E \supset F$ is splitting
6. $E \supset F$ 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

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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 $\mathbf Q$.

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A good reference, which I'll have to defend up and down, that organizes all of this is Lang's Algebra. The presentation there is supposedly lifted from Artin's Notre Dame lectures. – Dylan Moreland May 8 '12 at 0:42
And note that normality does do a few things right: for example, if $E/F$ is normal then so is $E/L$. – Dylan Moreland May 8 '12 at 0:59