If the extension F(s) is finite, s is algebraic (with respect to F), else s is transcendental. If a base field is not given, the rationals are assumed. An algebraic number, adjoined to the rationals, produces a finite extension. Equivalently, an algebraic number is the root of an irreducible polynomial in the rationals. The square root of 2 and the cube root of 39, for example. A transcendental is not the root of any polynomial over the rationals. There are countably many polynomials over the rationals, and each polynomial has finitely many roots, hence there are contably many algebraic numbers. The reals are uncountable, so there are lots of transcendental numbers. E and π are two examples. (Limerick)

Remember that a finitely generated extension, also called an extension of finite type, could be an infinite extension over F. This happens when at least one of the adjoined elements is transcendental. If all the adjoined elements are algebraic, bring them in one at a time to produce a finite extension.

The extension field E over F is algebraic (not necessarily finite) if every s in E is algebraic over F. Failing this, E is a transcendental extension of F.

Adjoining a finite set of algebraic elements produces a finite extension, and every element in a finite extension is algebraic, hence the extension is algebraic. If a and b are algebraic, F(a,b) is finite, and algebraic, hence a+b, a-b, a*b, and a/b are all algebraic. The square root of 2 divided by the cube root of 17 is the root of some polynomial somewhere.

Two sequential extensions are algebraic iff the composite extension is algebraic.
If b is the root of p(x) over the intermediate field,
and p has coefficients a_{1} a_{2} … a_{n},
the extension of F by these coefficients, and then by b, is finite,
and algebraic, hence b is algebraic over F.
This holds for all b, so the composite extension is algebraic.

Here is an example of an infinite algebraic extension.
Start with the rationals
and adjoin the p^{th} root of 2 for each prime p.
Each polynomial x^{p}-2 is irreducible by
eisenstein's criterion.
Each extension has dimension p,
and if the composite extension were finite,
there would not be room for a p dimensional extension for large p.
Thus our algebraic extension is infinite.

Let E/F be algebraic with a countably infinite dimension. In other words, it has a countable basis. Use these basis elements as generators, and the extension is countably generated. A finite set of generators will not suffice, since that would produce a finite extension. We really need an infinite set of generators.

Conversely, adjoin a countable set of algebraic elements to a base field. each generator multiplies the dimension by a finite number. The basis elements can be enumerated, and the dimension of the extension is countable.