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

Exercise:Exploring a field E as a Vector Space.
Let F = `QQ` be the field of rational numbers. Consider the polynomial `p(x) = x^2 - 2` `in QQ[x]`.
1. Show that the ideal `I = langle p(x) rangle` generated by p(x) in `QQ[x]` is a maximal ideal.
2. Deduce that the quotient ring `K = QQ[x]"/" I` is a field. This is a simple extension field of `QQ`.
3. Demonstrate that K is a vector space over `QQ`.
4. Find a basis for K as a vector space over `QQ` and determine its dimension.

Detailed Solution (including explanation of Vector Space theory used): Before we dive into the solution , let us briefly revisit the definition of a Vector Space.

Introduction to 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}`.

Continuing:
Solution:
Let us adress each part of the exercise:
1. Show that the ideal `I = langle p(x) rangle` generated by p(x) in `QQ[x]` is a maximal ideal.QED
2. Deduce that the quotient ring `bb {K = QQ[x]"/"I}` is a field. QED

Connecting `K = QQ[x]"/"langle x^2 - 2 rangle to QQ(sqrt 2)`:
The field `bb {QQ(sqrt 2)}` is defined as the smallest field containing `QQ` and `sqrt 2` , its elements are of the form `bb {a + b sqrt 2}` , where `a , b in QQ`. Let us see how the elements of K map directly to elements of `QQ(sqrt 2)`.
Consider the element x + I in K. What does it represent? Since we are working modulo `x^2 - 2` , the relation `x^2 = 2 (mod x^2 - 2)` means that the square of the element is 2 + I. So `(x + I)(x + I) = x^2 + I = 2 + I`. Meaning the element x + I in K behaves exactly like `sqrt 2 in QQ(sqrt 2)`. It's an element whose square is 2.

Let us denote `alpha = x + I`. Then `alpha^2 = 2`. Any element (ax + b) + I in K can be written as
a(x + I) + b(1 + I). If we identify 1 + I with the rational number 1 (which is consistent , as q + I behaves like q for q in `QQ`) , then we can write `a(x + I) + b(1 + I) = a alpha + b`. So , the elements of K are of the form `a alpha + b` where `alpha` is an abstract element such that `alpha^2 = 2`. If we map `alpha to sqrt 2` , then `a alpha + b to a sqrt 2 + b`.