Proving a Field Extension to be Galois Using Polynomial Splitting

In the realm of abstract algebra, a Galois extension is a field extension that enables the application of Galois theory to study the roots of a polynomial and the symmetries of its solutions. This article delves into the process of proving that a field extension ( mathbb{F} ) is Galois over base field ( mathbb{E} ) by showing that it is the splitting field of a polynomial ( f(X) ) over ( mathbb{E}[X] ). We will cover the necessary conditions, key concepts, and illustrative examples to provide a comprehensive understanding of the topic.

Introduction to Galois Theory and Splitting Fields

Galois theory is a branch of mathematics that combines algebra, particularly polynomial equations, with field theory. Its primary purpose is to investigate the symmetry of polynomial equations and their solutions. A field extension ( mathbb{F} ) over ( mathbb{E} ) is considered Galois when it satisfies specific criteria, one of which involves it being a splitting field of a particular polynomial.

Necessary Conditions for Galois Extensions

To prove that a field extension ( mathbb{F} ) is Galois over ( mathbb{E} ), it must satisfy two main conditions:

Splitting Field: The extension ( mathbb{F} ) must be the splitting field of some polynomial ( f(X) ) over ( mathbb{E}[X] ). This means that ( mathbb{F} ) contains all roots of the polynomial ( f(X) ). Separability: The polynomial ( f(X) ) must be separable over the base field ( mathbb{E} ). This condition ensures that the roots of the polynomial are distinct and does not contribute any multiple roots. In fields of characteristic ( p > 0 ), this is a crucial requirement, though in characteristic 0, like the field of rational numbers ( mathbb{Q} ), it is automatically satisfied.

Concrete Example: The Extension (mathbb{Q}(sqrt{2}, i)) over (mathbb{Q})

To demonstrate the application of these concepts, consider the field extension ( mathbb{Q}(sqrt{2}, i) ) over ( mathbb{Q} ).

Step 1: Identifying the Polynomial

The extension ( mathbb{Q}(sqrt{2}, i) ) does not itself provide an obvious polynomial. However, it can be related to the polynomial whose roots are contained within this field. Specifically, we need to find a polynomial that, when split, includes both ( sqrt{2} ) and ( i ).

Consider the polynomial ( f(X) X^4 - 4 ). The roots of this polynomial are ( pm sqrt{2}, pm isqrt{2} ). As we are interested in the roots ( sqrt{2} ) and ( i ), we can see that ( mathbb{Q}(sqrt{2}, i) ) is the splitting field of another polynomial, namely:

[ g(X) X^2 - 2 text{ and } h(X) X^2 1 ]

Thus, the polynomial ( f(X) X^4 - 4 (X^2 - 2)(X^2 1) ) suffices for our purposes.

Step 2: Verifying the Splitting Field Condition

The splitting field of ( f(X) X^4 - 4 ) over ( mathbb{Q} ) is indeed ( mathbb{Q}(sqrt{2}, i) ). This is because ( sqrt{2} ) and ( i ) are the roots of ( f(X) ), and no smaller extension of ( mathbb{Q} ) containing these roots exists.

Step 3: Checking Separability

Both ( g(X) X^2 - 2 ) and ( h(X) X^2 1 ) are separable polynomials over ( mathbb{Q} ). The discriminants of these polynomials are ( -8 ) and ( 4 ), respectively, which are non-zero, confirming that they have distinct roots.

Conclusion: The Galois Extension (mathbb{Q}(sqrt{2}, i)) over (mathbb{Q})

We have now proven that ( mathbb{Q}(sqrt{2}, i) ) is a Galois extension of ( mathbb{Q} ) by showing that it is the splitting field of the polynomial ( f(X) X^4 - 4 ) and it contains no multiple roots. This example illustrates the key steps in proving a field extension to be Galois using polynomial splitting techniques.

Additional Insights

Understanding the role of splitting fields in Galois theory is crucial for a deeper appreciation of polynomial equations and their symmetries. By focusing on specific polynomials and their roots, we can explore the structure of field extensions and the automorphisms that permute these roots. This serves as a foundation for more advanced topics in algebra and number theory.