Find a Basis for `QQ(i)` over `QQ`.

Reformulate the given Exercise: The task is to find a set of elements B that forms a basis for the field extension `QQ(i)` (the Gaussian rationals) when it is viewed as a vector space over the base field `QQ` (the rational numbers). We aim to find B and the dimension `[QQ(i) : QQ]`.

General Strategy for a Solution:

1. Identify the Element: Let `alpha = i` (the imaginary unit).
2. Find the Minimal Polynomial: Determine the non-zero polynomial p(x) with coefficients in `QQ` of the lowest degree such that i is a root.
3. Determine the Dimension: The degree of the minimal polynomial of i over `QQ` is the dimension of the vector space , `bb{n = [QQ(i) : QQ]}`.
4. Construct the basis: The basis B will be the set of powers of i up to n - 1 : `bb{B = {1 , i , i^ 2 , ... , i^ {n-1}}}`.

Specific Theory Used:

Field Extension `QQ(i)`: This field consists of all elements of the form a + bi , where a and b are rational numbers (scalars from `QQ`).
Minimal Polynomial for i over `QQ`: The unique monic , irreducible polynomial `p(x) in QQ[x]` such that `p(i) = 0`. Since `i^2 = -1` , the polynomial is `bb{x^2 + 1}`. This polynomial is irreducible over `QQ` because it has no rational roots.
Dimension of simple extension: The degree of the field extension `bb{[QQ(i) : QQ]}` , is equal to the minimal polynomial of i over `QQ` : `bb{[QQ(i) : QQ] = deg(x^2 + 1) = 2}`.
Basis for Field Extension: If `[QQ(alpha) : QQ] = n` , the basis is `{1 , alpha , ... , alpha^{n-1}}`.

Do the Exercise:

Step 1: Determine the Minimal Polynomial and Dimension:
1. The base field is `bb{F = QQ}`. The extension element is `bb{alpha = i}`.
2. The minimal polynomial is `p(x) = x^2 + 1` , which has degree n = 2.
3. The dimension of the vector space is `bb{[QQ(i) : QQ] = 2}`.
Step 2: Construct the Basis:
1. Since the dimension n = 2 , the basis set B must contain two elements.
2. Using the form `B = {1 , alpha}` with `alpha = i` , we get `bb{B = {1 , i}}`.

Verification:

Spanning: Any element `z in QQ(i)` has the form `z = a + bi` , where `a , b in QQ`. This is a linear combination of the basis elements 1 and i with scalars a and b from the field `QQ` : `bb{z = a * 1 + b * i}`.
Linear Independence: We showed in the previous exercise that `c_1 * 1 + c_2 * i = 0` implies `c_1 = 0` and `c_2 = 0` for real numbers. Since `QQ subset RR` , this holds true for rational numbers as well. Thus , `{1 , i}` is linear independent over `QQ`.

Conclusion: The basis for `QQ(i)` over `QQ` is `bb{B = {1 , i}}`. END