4.3 Perfect fields and separable closure

Perfect fields and separable closures are crucial concepts in Galois Theory. They help us understand when all algebraic extensions are separable and how to construct the largest separable extension of a field. These ideas are key to studying field extensions and their properties.

Understanding perfect fields and separable closures allows us to analyze the structure of field extensions more deeply. We can determine when all polynomials have distinct roots and explore the relationship between a field and its largest separable extension, shedding light on important algebraic properties.

Perfect fields

Properties of perfect fields

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• A field $F$ is perfect if every irreducible polynomial over $F$ is separable, meaning it has distinct roots in an algebraic closure of $F$
• In a perfect field, every algebraic extension is separable
  • This implies that for a perfect field $F$, any polynomial $f(x) \in F[x]$ that factors into linear terms in an algebraic closure of $F$ already factors into linear terms in $F$ itself
• A field of characteristic $0$ is always perfect (examples: $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$)
• A field of characteristic $p > 0$ is perfect if and only if every element of the field is a $p$-th power
  • In other words, the Frobenius endomorphism $x \mapsto x^p$ is surjective for a perfect field of characteristic $p$

Examples of perfect fields

• The prime field $\mathbb{F}_p$ and all its finite extensions are perfect fields
  • For instance, $\mathbb{F}_2$, $\mathbb{F}_3$, $\mathbb{F}_4$, $\mathbb{F}_5$, etc. are all perfect
• If $F$ is perfect, then any algebraic extension of $F$ is also perfect
  • For example, if $F = \mathbb{Q}$ is perfect, then $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(i)$, and any other algebraic extension of $\mathbb{Q}$ is also perfect
• The algebraic closure of a perfect field is also perfect
  • So, the algebraic closures $\overline{\mathbb{Q}}$, $\overline{\mathbb{F}_p}$, etc. are all perfect fields

Separable closure of a field

Existence and uniqueness of separable closure

• The separable closure $F_s$ of a field $F$ is the largest separable extension of $F$ inside an algebraic closure of $F$
• To prove existence, consider the composite of all separable extensions of $F$ inside an algebraic closure
  • This composite is a separable extension of $F$ and contains all other separable extensions
  • Thus, the composite is the separable closure $F_s$
• To prove uniqueness, suppose $F_s$ and $F_s'$ are two separable closures of $F$
  • Then there exists an $F$-isomorphism between $F_s$ and $F_s'$ by the universal property of the separable closure
  • This means the separable closure is unique up to isomorphism over $F$

Properties of separable closure

• The separable closure is the smallest separably closed extension of a field
  • A field $K$ is separably closed if every separable polynomial over $K$ has a root in $K$
• Any finite separable extension of $F$ is contained in the separable closure of $F$
  • For example, if $F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$, then $K \subseteq \mathbb{Q}_s$
• If $F \subseteq K \subseteq F_s$, where $F_s$ is the separable closure of $F$, then $K$ is separable over $F$
  • This follows from the fact that $F_s$ is separable over $F$, and any intermediate field is also separable over $F$

Constructing separable closures

Construction process

• To construct the separable closure of a field $F$, first construct an algebraic closure $F_a$ of $F$
• Inside $F_a$, consider the set $S$ of all elements that are separable over $F$
  • An element $\alpha \in F_a$ is separable over $F$ if its minimal polynomial over $F$ is separable
• Show that $S$ is a subfield of $F_a$ containing $F$
  • This involves proving that $S$ is closed under addition, multiplication, and taking inverses
• Prove that $S$ is the separable closure of $F$ by showing that it is separable over $F$ and contains all separable extensions of $F$
  • To show $S$ is separable over $F$, use the fact that every element in $S$ is separable over $F$
  • To show $S$ contains all separable extensions, use the definition of $S$ and the properties of the algebraic closure

Special cases

• In characteristic $0$, the separable closure coincides with the algebraic closure
  • This is because every irreducible polynomial over a field of characteristic $0$ is separable
  • So, $\mathbb{Q}_s = \overline{\mathbb{Q}}$, $\mathbb{R}_s = \overline{\mathbb{R}}$, etc.
• In characteristic $p > 0$, the separable closure is obtained by adjoining all $p$-power roots of elements in $F$
  • This means that $F_s = F(x_1^{1/p^{\infty}}, x_2^{1/p^{\infty}}, \ldots)$ for all $x_i \in F$
  • For example, if $F = \mathbb{F}_p(t)$, then $F_s = \mathbb{F}_p(t^{1/p^{\infty}})$

Separable closure in field extensions

Applications of separable closure

• Use the fact that the separable closure is the smallest separably closed extension of a field
  • This can help determine if a given field extension is separably closed or not
• Any finite separable extension of $F$ is contained in the separable closure of $F$
  • This can be used to study the structure of finite separable extensions and their relation to the separable closure
• If $F \subseteq K \subseteq F_s$, where $F_s$ is the separable closure of $F$, then $K$ is separable over $F$
  • This property can be used to prove that certain field extensions are separable

Absolute Galois group

• The Galois group of the separable closure over $F$ is called the absolute Galois group of $F$, denoted $\text{Gal}(F_s/F)$
• The absolute Galois group acts on the set of algebraic extensions of $F$ and the set of separable extensions of $F$
  • This action is defined by the restriction of automorphisms in $\text{Gal}(F_s/F)$ to the subfields of $F_s$
• Use the separable closure to study the structure of the absolute Galois group and its relation to the arithmetic properties of the base field
  • For example, the absolute Galois group of $\mathbb{Q}$ is closely related to the arithmetic of the rational numbers
  • The absolute Galois group of a finite field $\mathbb{F}_q$ is isomorphic to the profinite completion of $\mathbb{Z}$, denoted $\hat{\mathbb{Z}}$