We have the field extension $\mathbb{C}/\mathbb{Q}$. Let $x\in \mathbb{R}\subset \mathbb{C}$ be transcendental over $\mathbb{Q}$. Show, that $\{x\}$ is a transcendental basis of $L = \mathbb{Q}(x,i)$.
I've argued that $\mathbb{Q}(x)$ is a transcendental extension because $x$ is transcendental and $\mathbb{Q}(x,i)/\mathbb{Q}(x)$ is algebraic since $i$ vanishes in $X^2+1$. $\Rightarrow$ $[\mathbb{Q}(x,i):\mathbb{Q}(x)]=2$
Is that enough to prove that $\{x\}$ is a transcendental basis? ($\ast$)
I was also asked to find another transcendental base $B$ so that $\mathbb{Q}(x)$ and $\mathbb{Q}(B)$ aren't subsets of each other.
Can I work with $B= \{x+i\}$? It's pretty clear that it's a transc. basis (if ($\ast$) is sufficient) and $\mathbb{Q}(x+i)\not\subset \mathbb{Q}(x)$ is also obvious. I'm not sure how to elegantly show '$\not \supset$' though.
Any help is appreciated.
Best Answer
We have $\Bbb Q(x,i) = K(x)$, where $K = \Bbb Q(i)$ is algebraic over $\Bbb Q$. Its clear that $\{x\}$ is a transcendental basis of $K(x)$.