# Exercise on simple extensions

Let $E/K$ be an extension and $L_1,L_2$ intermediate fields of $E/K$ with $L_i:K$ finite. Then necessarily $L_1L_2:L_2$ $\le$ $L_1:K$. Prove $L_1L_2:L_2$ is not necessarily a factor of $L_1:K$. Hint: Consider $L_1= \mathbb{Q}(2^{\frac{1}{3}})$ and $L_2= \mathbb{Q}(e^{\frac{2\pi i}{3}}2^{\frac{1}{3}})$

My Progress:

Let $K=\mathbb{Q}$. Then $L_1:K=3$

Let $f(X)= X^2+ e^{\frac{2\pi i}{3}}2^{\frac{1}{3}}X+(e^{\frac{2\pi i}{3}}2^{\frac{1}{3}})^{2}$

Then $f(X)$ is an element of $L_2[X]$ and $f(2^{\frac{1}{3}})=0$. Now I want to show that $f(X)$ is the minimal polynomial of $2^{\frac{1}{3}}$ over $L_2$. So I suppose it’s not i.e. there exists a $g(X)$ element of $L_2[X]$ such that

$g$ | $f$, $g\not=0$ and most importantly $g(2^{\frac{1}{3}})=0$.Implying $g$ is of degree 1. And this is where I get stuck, how do I show that $2^{\frac{1}{3}}$ is not an element of $L_2$?

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You could take a few minutes and TeX this. It'd look much nicer and easier to read. –  lhf Sep 19 '12 at 3:22
I'm sorry, trying to learn, but I just started using TeX yesterday. –  Chris Sep 19 '12 at 3:37
$f$ can not be the minimal polynomail since it is not monic –  Belgi Sep 19 '12 at 4:25
Corrected that now, thanks for the spot. –  Chris Sep 19 '12 at 4:36
I added my way of solving this which have less calculations and is more "theory-based" IMO, hope this helps! –  Belgi Sep 19 '12 at 4:45

I would argue that $2^{1/3}$ is not an element of $L_2$ by proving each of the following claims:

1. First, that $L_1$ and $L_2$ are isomorphic as field extensions of $\mathbb{Q}$.
(They are both isomorphic to $\mathbb{Q}[X]/(X^3-2)$.)
2. Second, that $2^{1/3}\in L_2$ iff $e^{2\pi i/3}\in L_2$ iff there exists $\omega\in L_2$ such that $\omega^2+\omega+1=0$.
3. Third, that since there is no real number $\omega$ satisfying $\omega^2+\omega+1=0$, there is no such $\omega$ in $L_1$, and hence no such $\omega$ in $L_2$ by claim 1. Therefore, by claim 2, $2^{1/3}\notin L_2$.

Good luck!

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I understand how to prove claim 2 and claim 3 you essentially proved :) but I'm confused about how to show an isomorphism between $L_1$ and $L_2$. –  Chris Sep 19 '12 at 4:52
Ahhh nevermind, just rereading Falko Lorenz's Algebra, and its a theorem that K[algebraic element] is isomorphic to K[X]/f, where f is the minpolynomial of that element over K. –  Chris Sep 19 '12 at 5:08
New question, how do I know that $x^{3}-2$ is the min polynomial over Q? –  Chris Sep 19 '12 at 5:09
–  Belgi Sep 19 '12 at 5:11
@Belgi Thank you –  Chris Sep 19 '12 at 5:13

$2^{\frac{1}{3}}$ and $e^{\frac{2\pi i}{3}}2^{\frac{1}{3}}$ are both roots of $X^3 - 2$. Since $X^3 - 2$ is irreducible over $\mathbb{Q}$, $[L_1 : \mathbb{Q}] = [L_2 : \mathbb{Q}] = 3$. Since $L_1L_2$ contains $e^{\frac{2\pi i}{3}}$ and $2^{\frac{1}{3}}$, $L_1L_2 = \mathbb{Q}(e^{\frac{2\pi i}{3}}, 2^{\frac{1}{3}})$. Since $e^{\frac{2\pi i}{3}}$ is not real, it is not contained in $\mathbb{Q}(2^{\frac{1}{3}})$. Since $e^{\frac{2\pi i}{3}}$ is a root of $X^2 + X + 1$, $[L_1L_2 : L_1] = 2$. Hence $[L_1L_2 : \mathbb{Q}] = [L_1L_2 : L_1][L_1 : \mathbb{Q}] = 6$. Since $[L_1L_2 : \mathbb{Q}] = [L_1L_2 : L_2][L_2 : \mathbb{Q}] = 6$, $[L_1L_2 : L_2] = 2$. This is not a factor of $[L_1: \mathbb{Q}] = 3$

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Did you deduce the degree is $6$ without using the notaion of splitting fiels as I did ? –  Belgi Sep 19 '12 at 4:59
@Belgi I did not use the notion of splitting field. –  Makoto Kato Sep 19 '12 at 5:07

Note that a splitting field for $f:=X^{3}-2\in\mathbb{Q}[x]$ must be of degree $6$ over $\mathbb{Q}$since it is of degree at most $3!$ and $X^{3}-2$ have non-real roots hence not in $\mathbb{Q}(\sqrt[3]{2})$ so the the latter can not be the splitting field.

Also note that both $L_{1},L_{2}$ are simple extensions over $\mathbb{Q}$ generated by roots of $f$ and that $L_{1}L_{2}$is the splitting field of $f$ since $f$ have two roots $\lambda_{1},\lambda_{2}\in L_{1}L_{2}$ hence in $L_{1}L_{2}[x]$ you have a factorization $$f=(x-\lambda_{1})(x-\lambda_{2})g(x)$$ where $g$ is monic and of degree $1$ thus $g$ is of the form $x-\lambda_{3}$ where $\lambda_{3}\in L_{1}L_{2}$ and must be a root of $f$ .

$[L_{1}L_{2}:\mathbb{Q}]=6$ and you know $[L_{i}:\mathbb{Q}]=3$ since $f$ is irreduciable over $\mathbb{Q}$. Can you see why $[L_{1}L_{2}:L_{2}]$ is not a factor of $[L_{1}:\mathbb{Q}]$ ?

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I haven't been taught what a splitting field is yet. –  Chris Sep 19 '12 at 4:49