Let $E/\mathbb{Q}$ be a finite dimensional field extension of $\mathbb{Q}$.
Suppose
$dim$ $E/ \mathbb{Q}$ = 6
and
$Q \subset F_1 \subset E$,
$Q \subset F_2 \subset E$
are subextensions with
$dim$ $F_1 / E=2 $,
$dim$ $F_2 / E=3$.
Let $z_1 \in F_1$ and $z_2 \in F_2$ with $z_1,z_2 \not\in \mathbb{Q}$. Must there exist $c_1,c_2 \in \mathbb{Q}$ such that
$\mathbb{Q}(c_1 * z_1 + c_2 * z_2) = E$
I think this is related to the Theorem of the Primitive Element but I can not see how to use it. Can someone help me?
First note that $\mathbb Q(z_1,z_2)=E$. This is because $[F_1:\mathbb Q]$ and $[F_2:\mathbb Q]$ are prime. Spelled out a little bit more, $[\mathbb Q(z_1):\mathbb Q]\neq1$ and $[\mathbb Q(z_1):\mathbb Q]\mid[F_1:\mathbb Q]=2$, so $[\mathbb Q(z_1):\mathbb Q]=2$ and $\mathbb Q(z_1)=F_1$. Similarly, $\mathbb Q(z_2)=F_2$, so $\mathbb Q(z_1,z_2)=F_1F_2=E$. Once you know this, you're done by the Primitive Element theorem since it guarantees a primitive element of the form $z_1+cz_2$ with $c\in\mathbb Q$. If you don't know why your primitive element can to taken to be in this form, read on.
Consider some choice of $c\in\mathbb Q$, and let $z_c=z_1+cz_2$. What does it take to have $E_c:=\mathbb Q(z_c)=E$? Well, one sufficient (& necessary) condition would be that $\sigma(z_c)\neq\sigma'(z_c)$ for any two distinct embeddings $\sigma,\sigma':E\to\overline{\mathbb Q}$. This is because each embedding $E\to\overline{\mathbb Q}$ restricts to an embedding $E_c\to\overline{\mathbb Q}$, and so this would show that $\#\{E\to\overline{\mathbb Q}\}=\#\{E_c\to\overline{\mathbb Q}\}$ but in general for a field extension $F/\mathbb Q$, we have $[F:Q]=\#\{F\to\overline{\mathbb Q}\}$.
With this in mind, we aim to show that some choice of $c$ has this desired property. To show this, consider the polynomial $$f(\lambda)=\prod_{\substack{\sigma,\sigma':E\to\overline{\mathbb Q}\\\sigma\neq\sigma'}}(\sigma(z_\lambda)-\sigma'(z_\lambda)).$$ Note that $f(\lambda)\in\mathbb Q[\lambda]$ since it's fixed by all embeddings $E\to\overline{\mathbb Q}$ (since they just permute its factors). Hence, because polynomials only have finitely many roots while $\mathbb Q$ is infinite, there must exist some $c\in\mathbb Q$ such that $f(c)\neq0$. By construction, this means that $\sigma(z_c)\neq\sigma'(z_c)$ for every pair of distinct embeddings $\sigma,\sigma':E\to\overline{\mathbb Q}$ which is exactly what we wanted in order to show that $\mathbb Q(z_c)=E$.
Edit: Forgot to mention one thing. We know that $f(\lambda)$ is nonzero precisely because $z_1,z_2$ generate $E$. We can write $f$ as $$f(\lambda)=\prod_{\substack{\sigma,\sigma':E\to\overline{\mathbb Q}\\\sigma\neq\sigma'}}\bigg(\big(\sigma(z_1)-\sigma'(z_1)\big)+\lambda\big(\sigma(z_2)-\sigma'(z_2)\big)\bigg).$$ Since $z_1,z_2$ generate $E$, an embedding $E\to\overline{\mathbb Q}$ is determined by its what it does to $z_1$ and $z_2$, so we must have $\sigma(z_1)\neq\sigma'(z_1)$ or $\sigma(z_2)\neq\sigma'(z_2)$ always.