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

Formally , there is an isomorphism `phi`:
`phi: QQ[x]"/"langle x^2 - 2 rangle to QQ(sqrt 2)` defined by `bb {phi((ax + b) + I ) = a sqrt 2 + b}`. Let us quickly check this:
Well defined:
If (ax + b) + I = (cx + d) + I , then (a - c)x + (b - d) `in I`. Since deg ((a - c)x + (b - d)) < 2 , this means (a - c)x + (b - d) must be the zero polynomial , so a = c and b = d. Thus , `a sqrt 2 + b = c sqrt 2 + d` and the map is well defined.
Homomorphism:
Addition:OK
Multiplication:OK
Surjective: For any `a sqrt 2 + b in QQ(sqrt 2)` , the element `(ax + b) + I in K` maps to it.
Injective: If `phi((ax + I)) = 0` , then `a sqrt 2 + b = 0`. Since `sqrt 2` is irrational , this linear combination can only be zero if a = 0 and b = 0. Thus (0x + 0) + I = 0 + I , meaning the kernel is trivial. Since `phi` is an isomorphic homomorphism , `K = QQ[x]"/" langle x^2 + 2 rangle` is isomorphic to `QQ(sqrt 2)`.
In summary: The ideal `I = langle x^2 - 2 rangle` consists of all polynomials in `QQ[x]` that have `bb {sqrt 2}` as a root.
The elements of `K = QQ[x]"/"I` are cosets (ax + b) + I. The key relation `bb {x^2 = 2}` (mod I) makes the element x + I behave exactly like `sqrt 2`. By identifying x + I with `sqrt 2` , we see that (ax + b) + I in K directly corresponds to an element `a sqrt 2 + b in QQ(sqrt 2)`. This isomorphism confirms that the abstract quotient field K is indeed the familiar simple field extension `bb{QQ(sqrt 2)`.