elements of $E\setminus\mathbb{Q}$ are transcendental over $\mathbb{Q}$

59 Views Asked by At

Let $u$ be a real transcendental number over $\mathbb{Q}$ and $\alpha$ a root of $X^2 + u^2 + 1$ in $\mathbb{C}$. We note $K=\mathbb{Q}(u)$ and $E=K(\alpha)$.

I have to show that

elements of $E\setminus\mathbb{Q}$ are transcendental over $\mathbb{Q}$

I considered $x\in E\setminus\mathbb{Q}$ and supposed that $x$ was algebraic over $\mathbb{Q}$.

We get that $\mathbb{Q}(x)$ is a finite extension of $\mathbb{Q}$ but now I feel really stuck.

I tried to find a contradiction by considering $\mathbb{Q}(x)(u)$, $\mathbb{Q}(x)(\alpha)$ and even the fact that $\mathbb{Q}(x)$ is countable but couldn't find anything.

I also tried to find something like $[\mathbb{Q}(x):\mathbb{Q}]=[\mathbb{Q}(x):\mathbb{Q}(u)][\mathbb{Q}(u):\mathbb{Q}]$ so I could have a contradiction since $\mathbb{Q}(u)$ is an $\infty$-dimensional $\mathbb{Q}$ vector space but didn't really succeed.

Can I have an hint :)?