4.1 Separable polynomials and extensions

Separable polynomials and extensions are key concepts in Galois Theory. They help us understand the structure of field extensions and their automorphisms. Separable polynomials have distinct roots, while separable extensions are built from these polynomials.

These ideas are crucial for the Fundamental Theorem of Galois Theory. They allow us to connect field extensions with their Galois groups, providing a powerful tool for solving polynomial equations and understanding field theory.

Separable Polynomials

Properties of Separable Polynomials

• A polynomial $f(x)$ over a field $F$ is separable if it has distinct roots in some extension field of $F$
• The multiplicity of a root $\alpha$ of $f(x)$ is the largest positive integer $m$ such that $(x-\alpha)^m$ divides $f(x)$
  • A root is simple if it has multiplicity 1
• A polynomial is separable if and only if all its roots are simple
• The derivative $f'(x)$ of a polynomial $f(x)$ is the polynomial obtained by differentiating each term of $f(x)$ with respect to $x$
  • A polynomial $f(x)$ is separable if and only if $f(x)$ and $f'(x)$ are relatively prime ($\gcd(f(x), f'(x)) = 1$)
• The product of separable polynomials is separable ($f(x)$ and $g(x)$ separable implies $f(x)g(x)$ separable)
• If $f(x)$ is a separable polynomial over $F$ and $E$ is an extension of $F$, then $f(x)$ is also separable over $E$

Examples of Separable Polynomials

• The polynomial $f(x) = x^2 - 2$ over $\mathbb{Q}$ is separable because it has distinct roots $\pm\sqrt{2}$ in the extension field $\mathbb{Q}(\sqrt{2})$
• The polynomial $g(x) = x^3 - 3x + 1$ over $\mathbb{Q}$ is separable because $\gcd(g(x), g'(x)) = \gcd(x^3 - 3x + 1, 3x^2 - 3) = 1$
• The polynomial $h(x) = (x^2 - 2)(x^2 - 3)$ over $\mathbb{Q}$ is separable because it is the product of separable polynomials $(x^2 - 2)$ and $(x^2 - 3)$

Separable Extensions

Characterization of Separable Extensions

• An algebraic extension $E/F$ is separable if every element of $E$ is the root of a separable polynomial over $F$
• For a finite extension $E/F$, the following are equivalent:
  1. $E/F$ is separable
  2. There exists a primitive element $\alpha \in E$ such that $E = F(\alpha)$ and the minimal polynomial of $\alpha$ over $F$ is separable
  3. Every irreducible polynomial in $F[x]$ that has a root in $E$ is separable
• Every splitting field is a separable extension
• If $E/F$ is a finite separable extension, then $E$ is the splitting field of some separable polynomial over $F$

Examples of Separable Extensions

• Finite fields over their prime subfields ($\mathbb{F}_{p^n}/\mathbb{F}_p$ is separable for any prime $p$ and positive integer $n$)
• $\mathbb{Q}(\sqrt[n]{2})/\mathbb{Q}$ is separable for any positive integer $n$ because the minimal polynomial of $\sqrt[n]{2}$ over $\mathbb{Q}$ is $x^n - 2$, which is separable
• The splitting field of $x^4 - 2$ over $\mathbb{Q}$ is a separable extension of $\mathbb{Q}$ because $x^4 - 2$ is a separable polynomial

Definitions of Separable Extensions

Equivalent Definitions of Separable Extensions

• Theorem: Let $E/F$ be a finite extension. The following are equivalent:
  1. $E/F$ is separable
  2. There exists a primitive element $\alpha \in E$ such that $E = F(\alpha)$ and the minimal polynomial of $\alpha$ over $F$ is separable
  3. Every element of $E$ is the root of a separable polynomial over $F$
  4. Every irreducible polynomial in $F[x]$ that has a root in $E$ is separable

Proof of Equivalence

• The proof involves showing the implications (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (4) $\Rightarrow$ (1) using properties of minimal polynomials, primitive elements, and the separability of irreducible factors
  • (1) $\Rightarrow$ (2): If $E/F$ is separable, then there exists a primitive element $\alpha \in E$ such that $E = F(\alpha)$ and the minimal polynomial of $\alpha$ over $F$ is separable (Primitive Element Theorem)
  • (2) $\Rightarrow$ (3): If $\alpha$ is a primitive element of $E/F$ with a separable minimal polynomial, then every element of $E$ can be expressed as a polynomial in $\alpha$ and is thus the root of a separable polynomial over $F$
  • (3) $\Rightarrow$ (4): If every element of $E$ is the root of a separable polynomial over $F$, then every irreducible polynomial in $F[x]$ that has a root in $E$ must be separable (as it divides a separable polynomial)
  • (4) $\Rightarrow$ (1): If every irreducible polynomial in $F[x]$ that has a root in $E$ is separable, then $E/F$ is separable by definition

Separability of Polynomials and Extensions

Determining Separability of Polynomials

• To determine if a polynomial $f(x)$ over $F$ is separable, check if $f(x)$ and $f'(x)$ are relatively prime using the Euclidean algorithm
  • Example: $f(x) = x^3 - 2$ over $\mathbb{Q}$ is separable because $\gcd(f(x), f'(x)) = \gcd(x^3 - 2, 3x^2) = 1$
• Alternatively, factor $f(x)$ into irreducible factors over $F$ and check if all factors have multiplicity 1
  • Example: $g(x) = (x^2 - 2)(x - 1)^2$ over $\mathbb{Q}$ is not separable because $(x - 1)$ has multiplicity 2

Determining Separability of Extensions

• For a finite extension $E/F$, find a primitive element $\alpha$ such that $E = F(\alpha)$ and check if the minimal polynomial of $\alpha$ over $F$ is separable
  • Example: $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is separable because the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$ is $x^3 - 2$, which is separable
• Alternatively, factor the minimal polynomials of elements in $E$ over $F$ and check if all irreducible factors are separable
  • Example: $\mathbb{F}_4/\mathbb{F}_2$ is separable because the minimal polynomial of any element in $\mathbb{F}_4$ over $\mathbb{F}_2$ is either $x$ or $x^2 + x + 1$, both of which are separable
• Use the properties of separable extensions, such as the fact that every splitting field is separable, to determine the separability of a given extension
  • Example: The splitting field of $x^4 - 2$ over $\mathbb{Q}$ is a separable extension of $\mathbb{Q}$ because $x^4 - 2$ is a separable polynomial

Examples of Inseparable Extensions

• $\mathbb{F}_p(x^{1/p})/\mathbb{F}_p(x)$ is inseparable for any prime $p$ because the minimal polynomial of $x^{1/p}$ over $\mathbb{F}_p(x)$ is $y^p - x$, which is not separable
• $\mathbb{F}_p(t^{1/p})/\mathbb{F}_p(t^p)$ is inseparable for any prime $p$ because the minimal polynomial of $t^{1/p}$ over $\mathbb{F}_p(t^p)$ is $y^p - t$, which is not separable