3.3 Normal extensions

Normal extensions are a key concept in Galois Theory, bridging the gap between field extensions and polynomial splitting. They're special because every irreducible polynomial with a root in the extension splits completely there.

Understanding normal extensions is crucial for grasping the Fundamental Theorem of Galois Theory. They're closely tied to splitting fields and help us analyze the structure of field extensions and their automorphisms.

Normal Extensions

Definition and Properties

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• An extension $L/K$ is normal if every irreducible polynomial in $K[x]$ that has a root in $L$ splits completely in $L[x]$
  • Example: The extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is normal because the minimal polynomial $x^3-2$ splits completely in $\mathbb{Q}(\sqrt[3]{2})[x]$
• A normal extension is algebraic, meaning every element of $L$ is a root of some polynomial in $K[x]$
• The composition of normal extensions is normal
  • If $L/K$ and $M/L$ are normal, then $M/K$ is normal
  • Example: If $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2},i)/\mathbb{Q}(\sqrt{2})$ are normal, then $\mathbb{Q}(\sqrt{2},i)/\mathbb{Q}$ is normal
• Every finite extension of a finite field is normal
  • Example: The extension $\mathbb{F}_{8}/\mathbb{F}_{2}$ is normal

Equivalence of Definitions

• Prove that an extension $L/K$ is normal if and only if $L$ is the splitting field of a family of polynomials in $K[x]$
• Show that an extension $L/K$ is normal if and only if every $K$-embedding of $L$ into an algebraic closure of $K$ maps $L$ onto itself
  • Example: For the normal extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$, any $\mathbb{Q}$-embedding of $\mathbb{Q}(\sqrt{2})$ into $\mathbb{C}$ maps $\mathbb{Q}(\sqrt{2})$ onto itself
• Demonstrate that an algebraic extension $L/K$ is normal if and only if the fixed field of the group of $K$-automorphisms of $L$ is precisely $K$
  • Example: For the normal extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$, the fixed field of the group of $\mathbb{Q}$-automorphisms of $\mathbb{Q}(\sqrt{2})$ is precisely $\mathbb{Q}$

Normal Extensions and Splitting Fields

Characterization

• A finite extension $L/K$ is normal if and only if $L$ is the splitting field of some polynomial $f(x)$ in $K[x]$
  • Example: The extension $\mathbb{Q}(\sqrt{2},i)/\mathbb{Q}$ is normal because it is the splitting field of the polynomial $x^4-2$
• If $L/K$ is normal, then $L$ is the splitting field of the minimal polynomial of any element $\alpha$ in $L$ over $K$
• Every splitting field is a normal extension
  • Example: The splitting field of $x^3-2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[3]{2},\omega)$, which is a normal extension

Fundamental Theorem of Galois Theory

• Use the Fundamental Theorem of Galois Theory to check if the corresponding Galois group acts transitively on the roots of the minimal polynomial of a primitive element of $L$ over $K$
  • Example: For the normal extension $\mathbb{Q}(\sqrt{2},i)/\mathbb{Q}$, the Galois group acts transitively on the roots of the minimal polynomial $x^4-2$ of the primitive element $\sqrt{2}+i$

Definitions of Normal Extensions

Refer to the content under "Definition and Properties" and "Equivalence of Definitions" in the "Normal Extensions" section above.

Identifying Normal Extensions

Checking Irreducible Polynomials

• Check if every irreducible polynomial in $K[x]$ that has a root in $L$ splits completely in $L[x]$
  • Example: To show that $\mathbb{Q}(\sqrt{2},i)/\mathbb{Q}$ is normal, check that every irreducible polynomial in $\mathbb{Q}[x]$ with a root in $\mathbb{Q}(\sqrt{2},i)$ splits completely in $\mathbb{Q}(\sqrt{2},i)[x]$

Verifying Splitting Fields

• Determine if $L$ is the splitting field of some polynomial $f(x)$ in $K[x]$
  • Example: To show that $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is normal, verify that it is the splitting field of the polynomial $x^3-2$

Examining Embeddings and Automorphisms

• Verify if every $K$-embedding of $L$ into an algebraic closure of $K$ maps $L$ onto itself
• Examine if the fixed field of the group of $K$-automorphisms of $L$ is precisely $K$, assuming $L/K$ is algebraic
  • Example: To show that $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ is normal, verify that every $\mathbb{Q}$-embedding of $\mathbb{Q}(\sqrt{2})$ into $\mathbb{C}$ maps $\mathbb{Q}(\sqrt{2})$ onto itself and that the fixed field of the group of $\mathbb{Q}$-automorphisms of $\mathbb{Q}(\sqrt{2})$ is precisely $\mathbb{Q}$