# Basis of finite Galois extension generated by Galois group acting on single element.

Let $$E\supset F$$ be a finite Galois extension of fields. Show that there exists an element $$a\in E$$ such that $$\{\sigma(a):\sigma\in\mathrm{Gal}(E/F)\}$$ forms a basis of $$E$$ over $$F$$.

I guess that $$a$$ may be the primitive element of $$E/F$$ such that $$E=F(a)$$. Then $$\sigma(a)$$ are all the roots of the minimal polynomial of $$a$$. Only need to prove roots of this irreducible polynomial are linearly independent over $$F$$ and expand $$E$$. Am I in the right direction?

• Careful: The primitive element might not always work. When we say that $a$ is a primitive element for $E/F$, we mean that $a$ generates $E$ as an algebra over $F$, i.e., the powers of $a$ generate $E$ as an $F$-vector space. But in this problem, we are asking for an $a$ such that $\{\sigma(a) : \sigma \in \text{Gal}(E/F)\}$ generates $E$ as an $F$-vector space. For example: Taking $F = \mathbb{Q}$ and $E = \mathbb{Q(i)}$, you'll notice that $a = i$ is a primitive element for $E/F$, but $\{a, \sigma(a)\}$ isn't a basis for $E/F$. – user263190 Oct 10 '18 at 21:39
• This is the so called Normal basis theorem. Locally see here or here. Google gives hits to various lecture notes. For some strange reason the Wikipedia article titled Normal Basis only handles finite fields (when the proof of existence is simpler due to cyclicity of the Galois group). Go figure? – Jyrki Lahtonen Oct 11 '18 at 5:39
• @JyrkiLahtonen thanks – user498029 Oct 11 '18 at 14:10