Let $K$ be a field extension of $F$ and let $a \in K$. Show that $[F(a):F(a^3)] \leq 3$. Find examples to illustrate that $[F(a):F(a^3)]$ can be $1,2$ or $3$.
Attempt: $F \subset F(a^3) \subseteq F(a)$
The minimal polynomial for $a^3$ over $F$ is $ x-a^3=0$
I, unfortunately, don't have much idea than this on this problem. Could you please tell me how to move ahead?
Let $K$ be an extension of $F$. Suppose that $E_1$ and $E_2$ are contained in $K$ and are extensions of $F$. If $[E_1:F]$ and $[E_2:F]$ are both prime, show that $E_1 = E_2$ or $E_1 \bigcap E_2 = F $
Attempt: $[K:F] = [K:E_1][E_1:F] = [K:E_2][E_2:F]$
Since, $[E_1:F]$ and $[E_2:F]$ are both prime $\implies [E_2:F]$ divides $[K:E_1]$ and $[E_1:F]$ divides $[K:E_2]$
How do i move ahead?
Thank you for your help.
Answer
Consider $x^3-a^3$, where here $a^3$ is the number which generates $F(a^3)/F$. Then $a$ is a root of this polynomial, and so we will let $m_a(x)$ denote the minimal polynomial for $a$ over $F(a^3)$. We know
$$\begin{cases}m_a(x)|(x^3-a^3)\\ \deg m_a(x)\le \deg (x^3-a^3)=3\end{cases}.$$
But since
$$F(a)\cong F(a^3)[x]/(m_a(x))$$
we know
$$[F(a):F(a^3)]=\text{deg }_{F(a^3)} F(a)=\text{deg } m_a(x)\le 3$$
For the second question we proceed similarly.
Let $F\subseteq E'\subseteq E_1$, then by the tower law $[E':F]\big|[E_1:F]$ since $[E_1:F]$ is prime, we have $[E':F]\in \{1,[E_1:F]\}$, so either $E'=F$ or $E'=E_1$, and similarly for any subfield of $E_2$.
But then
$$F\subseteq E_1\cap E_2\subseteq E_1\implies E_1\cap E_2= F\text{ or } E_1$$
If it's $F$, we're done, if not then by using the same condition on $E_2$ we see that $E_1\cap E_2=E_2$ as well (since it's not $F$). Hence $E_1=E_2$.
Note: Why don't we see $K$ in this computation? What is it there for? The answer is that $E_1\cap E_2$ doesn't make sense unless they are both subsets of a common set, $K$ is there because of a set theory restriction, but isn't really essential to the proof other than that, unless we're being very pedantic.
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