# Extension of automorphism to finite algebraic extension

Let $L_2/L_1/K$ be a tower of field extensions where $L_2/K$ is finite algebraic [so $L_1/K$ is also finite]. Prove or disprove: For every $\sigma_1\in Aut_K(L_1)$ there is a $\sigma_2\in Aut_K(L_2)$ such that $\sigma_2|L_1=\sigma_1$.

The claim is clearly true if $L_2/K$ is normal, since we can extend $\sigma_1$ to a homomorphism into an algebraic closure and then use normality. Other than that, I'm completely lost. Do I need to induct on the degree of the extension? (Hints are appreciated.)

It is false. Take for example

$$\Bbb Q\subset\Bbb Q(\sqrt2)\subset\Bbb Q(\sqrt[4]2)$$

We have the automorphism

$$\sigma_1\in\text{Aut}_{\Bbb Q}(\Bbb Q(\sqrt2))\;,\;\;\sigma_1(a+b\sqrt2):=a-b\sqrt2\;,\;\;a,b\in\Bbb Q$$

Now, the only elements in $\;\text{Aut}_{\Bbb Q}(\Bbb Q(\sqrt[4]2))\;$ are the ones defined by (i.e, the real ones)

$$\text{Id.}:\begin{cases}1&\mapsto&1\\{}\\\sqrt[4]2&\mapsto&\sqrt[4]2\end{cases}\;\;\;\;,\;\;\;\;\tau:\begin{cases}&1&\mapsto&1\\{}\\&\sqrt[4]2&\mapsto&-\sqrt[4]2\end{cases}$$

yet observe that in both cases $\;\left(\sqrt[4]2\right)^2=\sqrt2\mapsto\sqrt2\;$

• Thanks, I should've put more effort into finding a counterexample (especially since I've seen this extension before). – user228680 Jul 31 '16 at 11:12
• Hah was just in the process of writing this one out :) +1 – Mariah Jul 31 '16 at 11:16