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.

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    $\begingroup$ As Adam didn't give you the examples sought after in the last sentence of your first question I guess it is ok to remark that part three of your question from today gives an example of a situation where $[F(a):F(a^3)]=2$ for a certain $a$ and $F$. I think that it is easier to find examples for the other two degrees. $\endgroup$ – Jyrki Lahtonen Aug 16 '14 at 17:20
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    $\begingroup$ yeah :-). If $F= \mathbb Q, f(x) = x^3-1 , g(x) = x^2+x+1$. Let $1,\omega, \omega^2$ denote the cube roots of unity, then : $\omega^3 = 1$ and $[\mathbb Q(\omega) : \mathbb Q(\omega^3)] = [\mathbb Q(\omega): \mathbb Q] = 2$ as the minimal polynomial $x^2+x+1$ for $\omega$ over $\mathbb Q$ is of degree $2$. Hope I inferred it the right way .. $\endgroup$ – MathMan Aug 16 '14 at 19:34
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    $\begingroup$ Correct. Well done, again! $\endgroup$ – Jyrki Lahtonen Aug 16 '14 at 19:59
  • $\begingroup$ Thank you for the example :-) $\endgroup$ – MathMan Aug 16 '14 at 20:05

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.

  • $\begingroup$ Thank you Adam for the answer :-). Should have thought in this direction. May be I am saturated and not able to think :-( $\endgroup$ – MathMan Aug 14 '14 at 18:06
  • $\begingroup$ @VHP no worries, it happens to the best of us. I'm glad I found something you could understand and accept. $\endgroup$ – Adam Hughes Aug 14 '14 at 18:09
  • $\begingroup$ Thanks and could you give me any hint for the second question as well? $\endgroup$ – MathMan Aug 14 '14 at 18:11
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    $\begingroup$ @VHP my answer is updated. $\endgroup$ – Adam Hughes Aug 14 '14 at 18:16

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