Wednesday, September 13, 2017

abstract algebra - 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.)


Answer



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 $\;\left(\sqrt[4]2\right)^2=\sqrt2\mapsto\sqrt2\;$


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