3.1 Splitting fields and their construction

Splitting fields are crucial in understanding polynomial roots and field extensions. They're the smallest fields containing all roots of a given polynomial, unique up to isomorphism. This concept bridges basic field theory with more advanced Galois theory.

Constructing splitting fields involves adjoining roots to the base field step-by-step. The process reveals important properties like field degree and Galois group structure, connecting polynomial behavior to abstract algebra and number theory.

Splitting Fields for Polynomials

Definition and Properties

• A splitting field of a polynomial $f(x)$ over a field $F$ is the smallest field extension of $F$ that contains all the roots of $f(x)$
• The splitting field is obtained by adjoining all the roots of $f(x)$ to the base field $F$ (adjoining means adding elements to the field to create a larger field)
• The splitting field is unique up to isomorphism, meaning that any two splitting fields of a polynomial over the same base field are isomorphic (have the same structure and properties)
• The degree of the splitting field over the base field is equal to the order of the Galois group of the polynomial (the Galois group is the group of field automorphisms that fix the base field)

Examples

• The splitting field of $x^2 - 2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{2})$, obtained by adjoining $\sqrt{2}$ to $\mathbb{Q}$
• The splitting field of $x^3 - 2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[3]{2}, \omega)$, where $\omega$ is a primitive cubic root of unity

Constructing Splitting Fields

Step-by-Step Process

• To construct a splitting field, first factor the polynomial completely over the base field
• If all the roots are already in the base field, then the base field itself is the splitting field
• If some roots are not in the base field, adjoin one root at a time to the base field until all roots are included. The resulting field is the splitting field
• The process of adjoining a root $\alpha$ to a field $F$ is denoted by $F(\alpha)$ and is the smallest field containing both $F$ and $\alpha$
• When adjoining multiple roots, the order in which they are adjoined does not matter; the resulting splitting field will be the same up to isomorphism

Examples

• To construct the splitting field of $x^4 - 1$ over $\mathbb{Q}$:
  • Factor: $x^4 - 1 = (x - 1)(x + 1)(x - i)(x + i)$
  • Adjoin $i$ to $\mathbb{Q}$: $\mathbb{Q}(i)$
  • All roots are in $\mathbb{Q}(i)$, so this is the splitting field
• To construct the splitting field of $x^3 - 2$ over $\mathbb{Q}$:
  • Adjoin $\sqrt[3]{2}$ to $\mathbb{Q}$: $\mathbb{Q}(\sqrt[3]{2})$
  • Adjoin $\omega$ (a primitive cubic root of unity) to $\mathbb{Q}(\sqrt[3]{2})$: $\mathbb{Q}(\sqrt[3]{2}, \omega)$
  • All roots are in $\mathbb{Q}(\sqrt[3]{2}, \omega)$, so this is the splitting field

Uniqueness of Splitting Fields

Proving Uniqueness up to Isomorphism

• Let $E$ and $E'$ be two splitting fields of a polynomial $f(x)$ over a field $F$. To prove uniqueness, we need to show that $E$ and $E'$ are isomorphic
• Define a homomorphism $\phi: F[x]/(f(x)) \to E$ by sending $x$ to a root $\alpha$ of $f(x)$ in $E$. This homomorphism is surjective because $E$ is generated by $\alpha$ over $F$
• Similarly, define another homomorphism $\phi': F[x]/(f(x)) \to E'$ by sending $x$ to a root $\alpha'$ of $f(x)$ in $E'$. This homomorphism is also surjective
• By the first isomorphism theorem, $F[x]/(f(x)) \cong E$ and $F[x]/(f(x)) \cong E'$. Therefore, $E \cong E'$, proving the uniqueness of splitting fields up to isomorphism

Examples

• The splitting field of $x^2 - 2$ over $\mathbb{Q}$ is unique up to isomorphism, whether constructed as $\mathbb{Q}(\sqrt{2})$ or $\mathbb{Q}(-\sqrt{2})$
• The splitting field of $x^4 - 2$ over $\mathbb{Q}$ is unique up to isomorphism, whether constructed as $\mathbb{Q}(\sqrt[4]{2}, i)$ or $\mathbb{Q}(\sqrt[4]{2}, -i)$

Degree of Splitting Fields

Determining the Degree

• The degree of a splitting field $E$ over its base field $F$ is equal to the order of the Galois group $G$ of the polynomial $f(x)$ over $F$, denoted as $[E:F] = |G|$
• The Galois group $G$ is the group of all automorphisms of $E$ that fix $F$ elementwise
• To find the degree of the splitting field, determine the order of the Galois group by examining the permutations of the roots of $f(x)$ that preserve the field operations
• If the polynomial $f(x)$ factors into linear factors over $F$, then the degree of the splitting field is equal to the degree of the polynomial
• If $f(x)$ is irreducible over $F$ and has degree $n$, then the degree of the splitting field is a divisor of $n!$

Examples

• The splitting field of $x^2 - 2$ over $\mathbb{Q}$ has degree 2 over $\mathbb{Q}$ because the Galois group has order 2 (the identity and the automorphism that swaps $\sqrt{2}$ and $-\sqrt{2}$)
• The splitting field of $x^4 - 2$ over $\mathbb{Q}$ has degree 8 over $\mathbb{Q}$ because the Galois group has order 8 (the dihedral group $D_4$)