Theory: `"The Finite Basis Theorem & Algebraic Extensions"`.
When we study field extensions , we are essentially looking at how a smaller field `bbb"F"` sits inside a larger field `bbb"E"`. If we take an element `alpha` from the larger field that satisfies a polynomial equation with coefficients in `bbb"F"` , we call `alpha` algebraic.

Theorem: `"Finite Basis Theorem (or the Extension Degree Theorem)"`.
If `alpha` is algebraic over `bbb"F"` and its minimal polynomial has degree `n` , the field generated by `alpha` (written `bbb"F"(alpha)`) is actually a vector space over `bbb"F"`. Specifically:

- The dimension of this vector space is exactly `n`.
- A natural basis for this space is the set `{1 , alpha , alpha^2 , ... \ , alpha^{n-1} }`.
- Because every element in `bbb"F"(alpha)` can be written as a linear combination of these basis elements , every single element in `bbb"F"(alpha)` is also algebraic over `bbb"F"`.

Key Definitions:
1. Algebraic Extension: A field `bbb"E"` is an algebraic extension of `bbb"F"` if every element in `bbb"E"` is algebraic over `bbb"F"`.
2. Finite Extension: If `bbb"E"` has a finite dimension as a vector space over `bbb"F"` , we call it a finite extension.
3. Degree of Extension: We denote this dimension as `[bbb"E" : bbb"F"]`.

`"Note"`: It is important to remember that a "finite extension" refers to the dimension of the vector space , not the number of elements in the field itself.

Exercise: `"Constructing the Basis"`.
Let `bbb"F" = QQ` (the rational numbers) and let `bbb"E" = RR`.
Consider the element `alpha = sqrt 2`.
1. Find the degree of `alpha` over `QQ`.
2. Determine the dimension `[QQ(sqrt 2) : QQ]`.
3. Identify the basis for `QQ(sqrt 2)` as a vector space over `QQ`.

Solution: The element `alpha = sqrt 2` is a root of the polynomial `x^2 - 2 in QQ[x]`. Since this polynomial is irreducible over `QQ` , the degree `n` is `2`. According to the theorem , the dimension of the extension is equal to the degree , so `[QQ(sqrt 2) : QQ] = 2`. The basis is `{1 , alpha}` , which in this case is `{1 , sqrt 2}`. Any element `beta in QQ(sqrt 2)` can be written uniquely in the form `a + b sqrt 2` , where `a , b in QQ`. QED