Let us start with an exercise that bridges the concepts of ideals , simple extension fields, and introduces vector spaces. Exercise 1: OK

Exercise 2:Field extensions as Vector Spaces.
Let F = `QQ` be the field of rational numbers. Consider the polynomial `p(x) = x^2 - 2` `in QQ[x]`.

Problem statement:
Consider the field extension `L = QQ(sqrt 2)` , which is the smallest field containing both the rational numbers and the square root of 2.
1. Demonstrate that L is a vector space over the field of rational numbers `QQ`.
2. Determine the basis for L as a vector space over `QQ`.
3. Calculate the dimension of L over `QQ` , denoted as `[L : QQ ]`.

Strategy and repetition of Vector Spaces:
A vector space (or linear space) V over a field F is a set V equipped with two operations:

1. Vector Addition: For any u , v `bb in V` , there is a unique element `bb {u + v in V}`.
2. Scalar Multiplication: For any c `in F` and u `bb {in V}` , there is a unique element `c bb {u in V}`.

Solution :
Part 1: Demonstrating that `QQ(sqrt 2)` is a Vector Space over `QQ`: QED
Part 2: Finding a Basis for `QQ(sqrt 2)` over `QQ` :
We need to find a set of elements in `QQ(sqrt 2)` that is both spanning and linearly independent.
Spanning: Any element of `QQ(sqrt 2)` is of the form `a + b sqrt 2` , where `a , b in QQ`. We can rewrite this as a linear combination with rational coefficients: `bb {a * 1 + b * sqrt 2}`. This shows that the set `{ 1 , sqrt 2 }` spans the vector space.
Linear Independence: We must show that the only solution to the equation `c_1 * 1 + c_2 * sqrt 2 = 0` is `c_1 = 0` and `c_2 = 0` , where `c_1 , c_2 in QQ`. If `c_2 ne 0` we could rearrange the equation to get
`sqrt 2 = - c_1 / c_2 `. Since `c_1` and `c_2` are rational , their quotient `- c_1 / c_2` would also be rational. This would imply that `sqrt 2` is a rational number , which is a known contradiction. Therefore , our assumption that `c_2 ne 0` must be false. So , `c_2 = 0`. Substituting this back into the original equation , we get `c_1 + 0 * sqrt 2 = 0` , which simplifies to `c_1 = 0`. Since the only solution is `c_1 = 0` and `c_2 = 0` , the set `{ 1 , sqrt 2}` is linearly independent. Since `{ 1 , sqrt 2 }` is both spanning and linearly independent , it is a basis for `QQ(sqrt 2)` over `QQ`. QED
Part 3: Calculating the Dimension :
The dimension of a vector space is the number of elements in its basis. The basis we found for `QQ(sqrt 2)` over `QQ` is `{ 1 , sqrt 2 }` , which contains two elements.
Therefore , the dimension of `QQ(sqrt 2)` is 2 : `bb {[QQ(sqrt 2) : QQ ] = 2}`. QED
This exercise was a fundamental step in linking field theory to linear algebra. The next logical step is to see how this concept extends to other field extensions. END of exercise