I would like to ask the following:
Is $\mathbb{C}$ the Galois extension of $\mathbb{Q}$?
I considered the Galois extension in a finite field, thus I am stuck here.
I think $\mathbb{C}$ is not the finite extension of $\mathbb{Q}$, because it is not an algebraic extension. ($\pi$ is not algebraic over $\mathbb{Q}$, and we have the proposition "finite extension is algebraic extension").
Please help me.
An extension doesn't need to be algebraic in order to be Galois (at least, not if it hasn't been explicitly required in the definition). There are different conventions regarding that, which give different answers.
Suppose a Galois extension is one that is both separable and normal. Then, every field extension over a field of characteristic zero (like the rationals) is separable, so you only need to check normality, which means every irreducible polynomial in $ \mathbf Q[X] $ which has a root in $ \mathbf C $ splits completely over $ \mathbf C $. This is trivially satisfied in this case because $ \mathbf C $ is algebraically closed - every polynomial in $ \mathbf Q[X] $ splits completely in $ \mathbf C[X] $. As such, the extension $ \mathbf C/\mathbf Q $ is Galois, by definition, even though it's not algebraic.