Prove that $\{x\}$ is transcendental base of $\mathbb{Q}(x,i)$

Question

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.

Answer 1

If $y$ is transcendental, then the only elements of $\mathbb Q(y)$ which are algebraic over $\mathbb Q$ already lie in $\mathbb Q$. For, suppose $\alpha\in\mathbb Q(y)-\mathbb Q$ satisfies $p(\alpha)=0$ for some polynomial $p$ of degree $d$. Writing $\alpha=f(y)/g(y)$ we have $0=g(y)^dp(\alpha)=q(y)$ for some (nonzero) polynomial $q$, a contradiction.

In your question, suppose $\mathbb Q(x)$ is contained in $\mathbb Q(x+i)$. Then $i\in\mathbb Q(x+i)$, but $i$ is algebraic and $x+i$ is transcendental. Contradiction.

Answer 2

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 $K(x)$.