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. OK
Question 5: What is the characteristic polynomial of this matrix ? OK

Question 6: What is the minimal polynomial of `alpha = 1 + i` over `QQ` ?
Here is the key: The minimal polynomial of an element `alpha` in a field extension is a divisor of the characteristic polynomial. In this case , your characteristic polynomial is `bb{p(x) = x^2 -2x + 2}` , which is a quadratic polynomial of degree 2. We can find the roots using the quadratic formula:
`x = {-(-2)+- sqrt {(-2)^2 - 4 * 1 * 2}}/{2 * 1} = { 2 +- sqrt {4 - 8}}/2 = {2 +- sqrt {-4}}/2 = {2 +- 2i}/2 = 1 +- i`.
The roots are `bb {1 + i}` and 1 - i . Since these roots are not in `QQ` , the polynomial `x^2 -2x + 2` cannot be factored into linear polynomials with rational coefficients. Therefore , the polynomial is irreducible over `QQ`. This is also confirmed by using the Eisenstein Criterion. The direct observation that it has no roots in `QQ` and is of degree 2 is sufficient to show it is irreducible. Now let us tie this back to the definition of the minimal polynomial:
- The minimal polynomial must be monic. The polynomial p(x) is monic.
- It must have `bb {alpha}` as a root. The polynomial p(x) has `bb {alpha = 1 + i}` as a root.
- It must be of the least degree possible. Since p(x) is irreducible and has `alpha` as a root , it cannot be factored into polynomials of lower degree over `QQ`. This means it must be the minimal polynomial. Therefore , the minimal polynomial of `alpha = 1 + i` over `QQ` is indeed `bb {x^2 -2x + 2}`.
- The minimal polynomial is unique. The characteristic polynomial , on the other hand , is not always the minimal polynomial , but is a multiple of the minimal polynomial. In the special case where the characteristic polynomial is irreducible , as it is here , it must be the minimal polynomial. OK