Simple extension

A field extension $𝐿 : 𝐾$ is called simple iff $𝐿$ is generated by the adjunction of a single element, #m/def/field i.e. $𝐿 = 𝐾 (𝜗)$ for some $𝜗 \in 𝐿$ .¹ Such a $𝜗$ is called a primitive element.

Classification

Let $𝐾 (𝜗) : 𝐾$ be a simple extension, and consider the evaluation map

𝜖 : 𝐾 [𝑡] \to 𝐾 [𝜗] \subseteq 𝐾 (𝜗) 𝑓 (𝑡) \mapsto 𝑓 (𝜗)

Then

If $𝜖$ is injective then $𝐾 (𝜗) : 𝐾$ is an infinite extension, whence $𝐾 (𝜗)$ is isomorphic to the field of rational functions $𝐾 (𝑡)$ ;
If $𝜖$ is not injective then $𝐾 (𝜗) : 𝐾$ is an extension of finite degree $𝑛$ , and

𝐾 (𝜗) ≅ 𝖥 𝗅 𝖽 𝐾 [𝑡] ⟨ 𝑚 𝜗 (𝑡) ⟩

where $𝑚 𝜗 (𝑡) \in 𝐾 [𝑡]$ is the minimal polynomial of $𝜗$ .

Proof

By the First isomorphism theorem, the image of $𝜖$ is isomorphic to $𝐾 [𝑡] / k e r 𝜖$ . Since $𝐾 (𝜗)$ is an integral domain, so is $𝐾 [𝜗]$ , and thus by $𝑅 / 𝐼$ for commutative $𝑅$ is an integral domain iff $𝐼$ is prime, $k e r 𝜖$ must be a prime ideal in $𝐾 [𝑡]$ .

First, consider $k e r 𝜖 = 0$ , i.e. $𝜖$ is injective. By the universal property for the Field of fractions, $𝜖$ extends to a unique homomorphism

¯ 𝜖 : 𝐾 (𝑡) \to 𝐾 (𝜗)

where $𝐾 (𝑡) ≅ ¯ 𝜖 (𝐾 (𝑡)) \leq 𝐾 (𝜗)$ is a field containing $𝐾$ and $𝜗$ , whence by definition $¯ 𝜖 (𝐾 (𝑡)) = 𝐾 (𝜗)$ . By injectivity, ${𝜗 𝑖} \infty 𝑖 = 0$ are linearly independent so we have an infinite extension.

Now consider $𝔞 = k e r 𝜖 \neq 0$ . Since $𝐾 [𝑡]$ is a Euclidean domain and thus a PID, it follows $𝔞 = ⟨ 𝑝 (𝑡) ⟩$ for a unique monic irreducible nonconstant polynomial $𝑝 (𝑡) \in 𝐾 [𝑡]$ . Since $⟨ 𝑝 (𝑡) ⟩$ is maximal in $𝐾 [𝑡]$ , the image of $𝜖$ is a subfield containing $𝜗 = 𝜖 (𝑡)$ , and by the same token as above we have $𝜖 (𝐾 (𝑡)) = 𝐾 (𝜗)$ , giving the claimed isomorphism.

Properties

If $𝐾 (𝛼)$ and $𝐾 (𝛽)$ have the same minimal polynomial $𝑓 (𝑥) \in 𝐾 [𝑥]$ , then there exists a unique isomorphism of field extensions $𝜑 : 𝐾 (𝛼) \to 𝐾 (𝛽)$ such that $𝜑 (𝛼) = 𝛽$ .
More generally, an isomorphism of ground fields $𝜓 : 𝐾 1 \to 𝐾 2$ such that $𝜓 (𝑚 𝛼 (𝑥)) = 𝑚 𝛽 (𝑥)$ lifts to an isomorphism of extensions $¯ 𝜓 : 𝐾 1 (𝛼) \to 𝐾 2 (𝛽)$ .

Proof of 1–2

Note that an isomorphism of simple field extensions is completely determined by the image of the primitive element.

Now using the isomorphism described in Classification,

𝐾 (𝛼) ≅ 𝐾 [𝑥] ⟨ 𝑓 (𝑥) ⟩ ≅ 𝐾 (𝛽)

which necessarily fixes $𝐾$ and maps $𝛼 \mapsto 𝛽$ , proving ^P1, whereof ^P2 is a straightforward generalization.

Results

Bound on the automorphism group of a finite simple extension.
Simplicity of an algebraic extension
By the Primitive element theorem, every finite separable extension is simple.

#state/develop | #lang/en | #SemBr

2009. Algebra: Chapter 0, §VII.1.2, pp. 387–388 ↩

Simple extension

Classification

Properties

Results

Footnotes