Let $x\in \mathbb{R}$ be transcendental over $\mathbb{Q}$. I want to show that $\mathbb{Q}(x) \not \subseteq \mathbb{Q}(x+i)$.
So I assume that $\mathbb{Q}(x) \subseteq \mathbb{Q}(x+i)$ $\Rightarrow $ $i\in \mathbb{Q}(x+i)$
I know that shouldn't be possible but I can't find the right words to express that. Maybe that $i$ could then be expressed as a linear combination of $x+i$ but that isn't possible because $x+i$ is transcendental?
Suppose that $i \in \mathbb{Q}(x + i)$. Then there are polynomials $f(t), g(t) \in \mathbb{Q}[t]$ with $$ i = \frac{f(x+i)}{g(x+i)} $$ in $\mathbb{Q}(x, i)$. Write $f(t) = \sum_{k=0}^m a_kt^k$ and $g(t) = \sum_{k=0}^n b_kt^k$, for $a_k, b_k \in \mathbb{Q}$ with $a_m, b_n$ nonzero. Then $ig(x+i) = f(x+i)$ in $\mathbb{C}[x]$, so $$ i\sum_{k=0}^nb_k(x+i)^n = \sum_{k=0}^ma_k(x+i)^k $$ as polynomials in $\mathbb{C}[x]$. Comparing leading coefficients, we must have $m=n$ and $ib_n = a_n$. But since $b_n$ is nonzero, we have $i = \frac{a_n}{b_n} \in \mathbb{Q}$, which is a contradiction.