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

Understanding the Elements of `K = QQ[x]"/" langle x^2 - 2 rangle`:
The elements of K are cosets of the form f(x) + I , where `f(x) in QQ[x]`. Two polynomials f(x) and h(x) are in the same coset , i.e. f(x) + I = h(x) + I , if and only if their difference is in the ideal I : `f(x)- h(x)in I`. This means f(x) = h(x) (mod `(x^2 - 2)`). The crucial property of working modulo `I = langle x^2 - 2 rangle` is that any multiple of `x^2 - 2` is equivalent to zero in K. In particular:
`x^2 - 2 = 0` (mod `(x^2 - 2)`). This implies `bb {x^2 = 2}` (mod `(x^2 - 2)`). This congruence is what allows us to simplify polynomials in K. Any polynomial f(x) can be written as `f(x) = q(x)(x^2 - 2) + r(x)` , where r(x) is the remainder after division by `x^2 - 2`. Since `q(x)( x^2 - 2) in I`, we have
f(x) = r(x) (mod `(x^2 - 2)`). Because deg `(x^2 - 2) = 2` , the remainder r(x) must have a degree less than 2. So r(x) must be of the form ax + b , where `a , b in QQ`.
Therefore , every element in K can be uniquely represented by a coset of the form (ax + b) + I , where a , b `in QQ`. For simplicity and common practise , we often drop the "+ I" and just write the element as ax + b , implicity understanding that `x^2 = 2` in this context.
How the elements in K look like:
They look like polynomials of degree at most 1: ax + b , where `a , b in QQ`.