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Vector Spaces - Exercises

Exercise

Find a basis for `QQ(root(4)(2))` 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(root(4)(2))` when it is viewed as a vector space over the base field `QQ` (the rational numbers). We must find the basis set B and the dimension `[QQ(root(4)(2)) : QQ]`.

General Strategy for a Solution:

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

Specific Theory Used:

Field Extension: The smallest field containing both `QQ` and the algebraic number `alpha`.
Minimal Polynomial: The unique monic irreducible polynomial `p(x) in QQ[x]` such that `p(alpha) = 0`.
Dimension of simple Extension: The degree of the field extension , `[QQ(alpha) : QQ]` , is equal to the degree of the minimal polynomial of `alpha` over `QQ`.
Basis for Field Extension: If `[QQ(alpha) : QQ] = n` , then the basis for `QQ(alpha)` as a vector space over `QQ` is `B = {1 , alpha , ... , alpha^{n-1}}`.

Do the Exercise:

Step 1: Determine the Minimal Polynomial and Dimension:
1. The algebraic number is `alpha = root (4)(2)`.
2. We look for the simplest polynomial `p(x) in QQ[x]` such that `p(alpha) = 0 : x = root (4)(2) implies x^4 - 2 = 0`.
3. The polynomial `p(x) = x^4 - 2` is monic and has rational coefficients.
4. To confirm it is the minimal polynomial , we must show it is irreducible over `QQ`. We can use Eisenstein's Criterion with the prime p = 2 :

  • 2 divides the constant term (- 2).

  • 2 divides all intermediate coefficients (which are 0).

  • `2^2 = 4` does not divide the constant term (-2).

  • The leading coefficient 1 is not divisible by 2.

  • Therefore , `p(x) = x^4 - 2` is irreducible over `QQ`.
5. The degree of the minimal polynomial is n = 4.
6. The dimension of the vector space is : `[QQ(root(4)(2)) : QQ] = 4`.
Step 2: Construct the Basis:
1. Since the dimension n = 4 , the basis set B must contain four elements.
2. Using the formula `B = {1 , alpha , alpha^2 , alpha^3}` with `alpha = root (4)(2)` , we get `B = {1 , 2^{1/4} , (2^{1/4})^2 , (2^{1/4})^3}`.
3. Simplifying the powers: `B = {1 , 2^{1/4} , 2^{2/4} , 2^{3/4}}`.

Conclusion: The basis for `QQ(root(4)(2))` over `QQ` is `B = {1 , root(4)(2) , sqrt 2 , root(4)(8)}`. Any element in `QQ(root(4)(2))` is uniquely written as `a + b * root (4)(2) + c * sqrt 2 + d * root(4)(8)` , while `a , b , c , d in QQ`. END

Question

The vector space `QQ(root(4)(2))` over the field `QQ` has a basis `B = {1 , root(4)(2) , sqrt 2 , root(4)(8)}` and a dimension of 4. This dimension is equal to the degree of the minimal polynomial for `alpha = root(4)(2)` over `QQ`.
What critical rquirement must the minimal polynomial p(x) satisfy to correctly determine this dimension n = 4:

A) p(x) must be monic and have a degree equal to n , regardless of its reducibility over Q.

B) p(x) must be the unique monic polynomial of degree n that has `alpha` as a root and is irreducible over the field `QQ` ?