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 (Theory):
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}`.

These operations must satisfy the following ten axioms for all u , v , w `in` V and all c , d `in` F:
Axioms for Vector addition:
1. Closure under addition : `bb {u + v in V}`.
2. Commutativity under addition: `bb {u + v = v + u}`.
3. Associativity under addition: `bb {(u + v) + w = u + (v + w)}`.
4. Zero vector (additive identity): There exists an element `bb {0 in V`
    that u + 0 = u for all `bb {u in V}`.
5. Inverse under addition: For every `bb {u in V}` , there exists an element `bb {-u in V}`
    such that u + (-u) = 0.
Axioms 1-5 mean that (V , +) is an abelian group.

Axioms for Scalar multiplication:
6. Closure under scalar multiplication: `c bb u in bb V`.
7. Distributivity (scalar over vector addition): `c * bb {(u + v)}` = cu + cv.
8. Distributivity (vector over scalar addition): `(c + d) * bb {u}` = cu + du.
9. Associativity (scalar multiplication): c(du) +(cd)u
10. Identity (multiplication): 1u = u (where 1 is multiplicative identiy in F).

Note: There is no such thing as a "multiplicative identity vector" in a vector space. Instead, the concept of a multiplicative identity in vector spaces comes from the field of scalars—the scalar 1 in the field F , not from a particular vector in V.