We know the algebraic closure of rational numbers $\mathbb{Q}$ is the field of algebraic numbers, and the algebraic closure of real numbers $\mathbb{R}$ is the field of complex numbers $\mathbb{C}$. I want to know
What is the algebraic closure of $\mathbb{Q}(\pi)$, where $\mathbb{Q}(\pi)$ is the smallest field containing $\mathbb{Q}$ and $\pi$? Is it $\mathbb{C}$?
The field $ \mathbb{Q}(\pi) $ consists of all rational functions over $ \mathbb{Q} $, evaluated at $ \pi $. Specifically, the elements of $ \mathbb{Q}(\pi) $ can be written as $ \frac{f(\pi)}{g(\pi)} $, where $ f(x) $ and $ g(x) $ are polynomials with rational coefficients. The algebraic closure $ \overline{\mathbb{Q}(\pi)} $ is the set of all algebraic numbers over $ \mathbb{Q}(\pi) $, which means it contains the roots of all polynomials with coefficients from $ \mathbb{Q}(\pi) $. Since $ \mathbb{Q}(\pi) $ is countable, the set of polynomials with coefficients in $ \mathbb{Q}(\pi) $ is also countable, and therefore the algebraic closure is countable.
For example, since $ x^2 + 1 = 0 $ is a polynomial with coefficients in $ \mathbb{Q}(\pi) $, its roots, $ \pm i \notin \mathbb{Q}(\pi)\subset \mathbb{R}$, must be included in $ \overline{\mathbb{Q}(\pi)} $. However, the algebraic closure is still countable and cannot contain all complex numbers, since $ \mathbb{C} $ is uncountable. For instance, the number $ e^\pi $ is transcendental over $ \mathbb{Q}(\pi) $ and cannot be included in the algebraic closure. Thus, $ \overline{\mathbb{Q}(\pi)} $ is a proper subset of $ \mathbb{C} $ and does not include numbers like $ e^\pi $. This constant is known as Gelfond's Constant and it's nontrivial that it isn't a member of $ \overline{\mathbb{Q}(\pi)} $. But seemed like a nice example.
