Let us start with exercises that bridges the concepts of ideals , simple extension fields, and introduces vector spaces. Exercise 1: OK , Exercise 2: OK

Exercise 3:The Extension `QQ(i)`:
Let us consider the field extension `QQ(i)` , which is the smallest field containing both the both the rational numbers `QQ` and the imaginary unit `i = sqrt {-1}`.

Part 2: The linear Transformation:
Consider the element `alpha = 1 + i` in `QQ(i)`. We define a linear transformation `bb {L_alpha : QQ(i) rarr QQ(i)}` by multiplication by `alpha:bb {L_alpha (x) = alpha x}`. Remember our basis for `QQ(i)` is B = { 1 , i }.

Question 4: Using this basis , represent the linear transformation `L_alpha` as a matrix.
Solution question 4:

Let `alpha = 1 + i` and `x = a + bi`. This will give `L_alpha (x) = alpha x = (1 + i)(a + bi)= a + bi + ai + bi^2 = a + bi + ai - b = (a - b) + (a + b)i = `

`((a - b), (0)) + ((0) , (a + b)) = ((a - b) , (a + b)) = bb {[ [1 , -1] , [ 1 , 1]]}((a), (b))`. Which shows our matrix.

Alternative method: We have our basis B = { 1 , i }. We need to see where the transformation `L_alpha` sends each of the basis vectors:
`L_alpha (1) = alpha * 1 = 1 + i`. When we express this in terms of our basis { 1 , i } , we get `1 * 1 + 1 * i`. The coordinat vector is `bb {((1) , (1))}`. This forms the first first column of our matrix.
`L_alpha (i) = alpha * i = ( 1 + i)i = i + i^2 = i - 1 = -1 + i `. When we express this in terms of our basis
{ 1 + i } , we get `-1 * 1 + 1 * i`. The coordinat vector is `bb{((-1), (1))}`. This forms the second column of our matrix. Putting these columns together , we get the matrix:

`bb{[[1 , -1] , [1 , 1]]}`.