Prove that $\alpha:=\sqrt{\pi}+\sqrt 2 \in \Bbb{C} $ is trancendental over $\Bbb{Q}$. 
We want to prove that $\alpha:=\sqrt{\pi}+\sqrt 2 \in \Bbb{C} $ is
  trancendental over $\Bbb{Q}$.

Attempt. We use proof by contradiction and so assume that $\alpha \in \Bbb{C}$ is algebraic over $\Bbb{Q}$. Then we have that $$\left[\Bbb{Q}\left(\sqrt{\pi}+\sqrt 2\right):\Bbb{Q}\right]<\infty.$$
It is true that $[K(\sqrt \pi ):K]=\infty$, where $K$ is a field and the proof is the same with this post. So, one can think to use the following relations:


*

*$\sqrt \pi,\sqrt2 \in \Bbb{Q}(\sqrt \pi, \sqrt 2) \implies \sqrt \pi+\sqrt2 \ \in \Bbb{Q}(\sqrt \pi, \sqrt 2) \implies \Bbb{Q}(\sqrt \pi+ \sqrt 2)\subseteq \Bbb{Q}(\sqrt \pi, \sqrt 2) $

*$\Bbb{Q} \leq \Bbb{Q}(\sqrt2)\leq  \Bbb{Q}(\sqrt \pi+ \sqrt 2)\leq \Bbb{Q}(\sqrt \pi, \sqrt 2)$ and one logical thought is to apply the Tower Law, but I m not sure if this helps. 


Any ideas please?
Thank you.
 A: If it was algebraic then, since the algebraic numbers form a field and $-\sqrt2$ is algebraic,$$\left(\sqrt\pi+\sqrt2\right)-\sqrt2$$would be algebraic too. But this number is $\sqrt\pi$, which is transcendental, since $\pi$ is transcendental.
A: You must use the fact that $\pi$ (and so $\sqrt{\pi}$) is transcendental.                               
Before proving your proposition, I would like to note the fact that if $R$ is a subring of some ring $P$ then the integral elements form a ring (for a proof, go on my post: Proof that the algebraic integers form a subring of $\mathbb{C}$  and replace $\mathbb{Z}$ with $R$ and $\mathbb{C}$ with $P$).                   
Consider now the evaluation homomorphism $\psi_{\alpha}: \mathbb{Q}[T] \longmapsto \mathbb{Q}(\sqrt{\pi}; \sqrt{2})$, then \begin{equation} \frac{\mathbb{Q}[T]}{<P_{\alpha}>} \cong \mathbb{Q}(\alpha) \end{equation}, where $P_{\alpha}$ is the minimal polynominal of $\alpha$.                                                                           
From my proof that on the link above you can notice that \begin{equation}dim_{\mathbb{Q}}(\mathbb{Q}(\alpha)) \leq dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi};\sqrt{2})) \leq dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi}) \otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2}))\end{equation} but since \begin{equation} \mathbb{Q}(\sqrt{\pi};\sqrt{2})= \mathbb{Q}(\sqrt{\pi})\mathbb{Q}(\sqrt{2}) \end{equation}, where by some fields $K$ and $F$, by $KF$ we mean the composite extension                                                                                                                                                         also, \begin{equation} dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi}) \otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2}))= dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi})\mathbb{Q}(\sqrt{2})) \Rightarrow dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi}) \otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2}))=dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi};\sqrt{2})) \end{equation}.
Since $\sqrt{{\pi}}$ is transcendental, \begin{equation}dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi};\sqrt{2}))= \infty \end{equation}.
Since $\mathbb{Q}(\sqrt{\pi}) \subseteq \mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\sqrt{\pi};\sqrt{2})$ and $dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi}))= \infty$ (because $\sqrt{\pi}$ is transcendental), obviously: \begin{equation} dim_{\mathbb{Q}}(\mathbb{Q}(\alpha))= \infty \end{equation} and hence $P_{\alpha}=0$ and hence $\alpha$ is transcendental. $\,\,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,   \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,  \,                                                              \square$       
A: Continuing your argument:
$$
\pi \in \Bbb{Q}(\sqrt \pi) \subseteq \Bbb{Q}(\sqrt \pi, \sqrt 2) = \Bbb{Q}(\alpha)
\implies 
\Bbb{Q}(\pi) \subseteq \Bbb{Q}(\alpha)
\implies 
[\Bbb{Q}(\alpha ):\Bbb{Q}] \ge [\Bbb{Q}(\pi):\Bbb{Q}]=\infty
$$
