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) = x^2 - 2 in QQ[x]` is a maximal ideal.
In a polynomial ring F[x] over a field F , an ideal `langle f(x) rangle` is maximal if and only if f(x) is an irreducible polynomial over F. So we need to show that `p(x) = x^2 - 2` is irreducible over `QQ`.

Irreducibility Test: For a polynomial of degree 2 or 3 , it is irreducible if and only if
it has a root in `QQ`. Let us check if `x^2 - 2` has any rational roots. If r is a rational root , then `r^2 = 2`. Suppose `r = a/b` where a , b `in ZZ` , `b ne 0` and gcd(a , b) = 1. Then `(a/b)^2 = 2 rArr a^2/b^2 = 2 rArr a^2 = 2b^2`. This implies that `a^2` is an even number , which means a is an even number. Let a = 2k for some integer k. Substituting a = 2k into the equation: `(2k)^2 = 2b^2 rArr 4k^2 = 2b^2 rArr 2k^2 = b^2`. This implies that `b^2` is an even number , which means b must be an even number.However , if both a and b are even , then gcd(a , b)`le` 2 , which contradicts our assumption that gcd(a , b) = 1. Therefore , `x^2 - 2` has no rational roots. Since `x^2 - 2` is a polynomial of degree 2 and has no rational roots , it is irredicible over `QQ`. Thus , the ideal `bb {I = langle x^2 - 2rangle}` is a maximal ideal in `QQ[x]`. QED