Yes—and to elaborate, the reason why an algebraic field extension of ℚ cannot contain π is because:
- If it is a field, it contains π, π², π³, … which are linearly independent.
- By definition, an algebraic field extension is finite dimensional.
- If it is a field, it contains π, π², π³, … which are linearly independent.
- By definition, an algebraic field extension is finite dimensional.
A number x is algebraic over ℚ if and only if it generates a finite field extension, i.e. if x, x^2, x^3, etc. have a linear dependence relation.
However, as jopolous pointed out, you can get infinite dimensional algebraic field extensions by adjoining infinitely many algebraic numbers. For example, the set of all numbers which are algebraic over ℚ is a field, and this field is an infinite degree extension of ℚ.
Do we know that? My search doesn’t get more than https://www.encyclopediaofmath.org/index.php/Lindemann_theor..., which proves it for “𝑒, 𝑒², 𝑒³, …“.
If you look at the Lindemann theorem, you can transform the equation so that it uses π instead of e. Multiply all of the exponents by i (which is algebraic!) and then use Euler’s identity. You end up with the same formula, but with π instead of e.
However, if we already know that π is transcendental (which is proven by the Lindemann theorem using the above technique), we can rewrite any linear combination of B = {1, π, π², π³, …} as P(π) where P is a polynomial with coefficients in ℚ. Because π is transcendental, we know that P(π)=0 only if P is the zero polynomial (that is the definition of transcendental number).
In general, one of the big tricks here is that the set of polynomials is a vector space, and the powers B = {1, x, x², x³, …} span the entire vector space.
It's not even a finite extension