The smallest field containing all roots of a polynomial -- where symmetry lives
The splitting field of a polynomial f(x) over F is the smallest extension of F that contains all the roots of f. It is where Galois theory lives: the splitting field is always a normal extension, meaning that every irreducible polynomial that has one root in the field has all its roots there.
In this lesson, watch polynomials split root by root, compare normal and non-normal extensions, and compute splitting fields step by step.
Pick a polynomial and watch it factor one root at a time. As each root is adjoined, the field tower grows and the polynomial factorization updates. Notice how some roots "come for free" -- adjoining one root may force others into the field automatically.
Polynomial: x^2 - 2
Key insight: For x^2 - 2, adjoining sqrt(2) gives -sqrt(2) for free (it equals -1 times sqrt(2)). For x^3 - 2, adjoining the real cube root does NOT give the complex roots for free -- you need to also adjoin a primitive cube root of unity omega. This is why the splitting field of x^3 - 2 has degree 6, not 3.
An extension E/F is normal if every irreducible polynomial over F that has one root in E splits completely in E. Equivalently, E contains all conjugates of each of its elements. Splitting fields are always normal; arbitrary extensions need not be.
Minimal polynomial: x^2 - 2
Q(sqrt(2))
2 / 2 roots present in field
Both roots of x^2 - 2 lie in Q(sqrt(2)). Since all roots of the minimal polynomial are present, this is a normal extension.
Key insight: Q(sqrt(2))/Q is normal because x^2 - 2 has both roots (+/-sqrt(2)) in Q(sqrt(2)). But Q(cbrt(2))/Q is NOT normal: x^3 - 2 has only one root (the real cube root) in Q(cbrt(2)), while the other two roots omega*cbrt(2) and omega^2*cbrt(2) are complex and missing.
Select a polynomial and compute its splitting field step by step. The factorization tree shows how the polynomial breaks apart over successive extensions. The final result displays the splitting field, its degree over Q, a basis, and all roots with symmetry lines connecting conjugates.
Degree 2 polynomial -- select to compute its splitting field step by step
Key insight: The degree of the splitting field divides n! (where n is the degree of the polynomial), but is often much smaller. For x^4 - 5x^2 + 6 = (x^2-2)(x^2-3), the splitting field Q(sqrt(2), sqrt(3)) has degree 4, not 4! = 24, because the roots are related by simple algebraic identities.