Let $a$ be a complex zero of $x^2+x+1$ over $\mathbb{Q}$. Prove that $\mathbb{Q}(\sqrt{a})=\mathbb{Q}(a)$.

Question

Let $a$ be a complex zero of $x^2+x+1$ over $\mathbb{Q}$. Prove that $\mathbb{Q}(\sqrt{a})=\mathbb{Q}(a)$.

Let $f(x)=x^2+x+1$. Then $f(a)=a^2+a+1=0$.

To show equality of the two fields, we need to show containment of each one in the other.

For the forward containment, I guess I could use $\sqrt{a}\cdot \sqrt{a}=a\implies \mathbb{Q}(\sqrt{a})\subset\mathbb{Q}(a)$

For backward containment, I'm not sure. I might be doing this the wrong way, since I don't understand where $a$ being a complex zero of $f$ affects anything.

Answer 1

Your first implication is not correct it should be: $$a=\sqrt{a}\cdot \sqrt{a}\implies a\in \mathbb{Q}(\sqrt{a})\implies \mathbb{Q}(a)\subset \mathbb{Q}(\sqrt{a})\ $$

in the second inclusion: $$\sqrt{a}=1+a\implies \sqrt a\in \mathbb{Q}(a)\implies \mathbb{Q}(\sqrt a)\subset \mathbb{Q}(a)\ $$

Answer 2

$(a+1)^2 = a^2+2a+1 = a \,\,\,$