Subalgebra generated by an algebraic element

Let $𝐴$ be a K-monoid over a field $𝕂$ and $𝑎 \in 𝐴$ be an algebraic element with minimal polynomial $𝑚 𝑎 (𝑥)$ . Then the unital subalgebra generated by $𝑎$ is¹ falg

𝕂 [𝑎] = ⟨ 𝑎 ⟩ \leq 𝖴 𝖠 𝗌 𝖠 𝗅 𝗀 𝕂 𝐴 = {𝑝 (𝑎) : 𝑝 (𝑥) \in 𝕂 [𝑥], d e g 𝑝 < d e g 𝑚 𝑎}

and is isomorphic to

⟨ 𝑎 ⟩ \leq 𝖴 𝖠 𝗌 𝖠 𝗅 𝗀 𝕂 𝐴 ≅ 𝕂 [𝑥] ⟨ 𝑚 𝑎 (𝑥) ⟩ ⊴ 𝕂 [𝑥]

Proof

Let $𝑀 = d e g 𝑚 𝑎$ First we will show ^eq1. First note that the RHS is clearly a vector subspace, so it suffices to show that $𝑎 𝑛 \in R H S$ for all $𝑛 \in ℕ 0$ . Applying the division algorithm for polynomials
$𝑥 𝑛 = 𝑞 (𝑥) 𝑚 𝑎 (𝑥) + 𝑟 (𝑥)$
where $d e g 𝑟 < 𝑁$ . But
$𝑎 𝑛 = 𝑞 (𝑎) 𝑚 𝑎 (𝑎) + 𝑟 (𝑎) = 𝑟 (𝑎)$
so $𝑎 𝑛 \in R H S$ .

For the second statement, let $𝐼 = ⟨ 𝑚 𝑎 (𝑥) ⟩ ⊴ 𝕂 [𝑥]$ . It follows from above that to every $𝑏 \in ⟨ 𝑎 ⟩ \leq 𝑎$ there corresponds a unique $𝑟 𝑏 (𝑥) \in 𝕂 [𝑥]$ with $d e g 𝑟 𝑏 < 𝑀$ such that $𝑟 𝑏 (𝑎) = 𝑏$ . Let $𝜑$ be the map
$𝜑 : ⟨ 𝑎 ⟩ \leq 𝐴 \to 𝕂 [𝑥] / 𝐼 𝑏 \mapsto 𝑟 𝑏 (𝑥) + 𝐼$
Now $𝜑$ is a ring isomorphism, since for any $𝑏, 𝑐 \in ⟨ 𝑎 ⟩ \leq 𝐴$
$𝑟 𝑏 + 𝑐 (𝑥) + 𝐼 = 𝑟 𝑐 (𝑥) + 𝑟 𝑏 (𝑥) + 𝐼 𝑟 𝑏 𝑐 (𝑥) = 𝑟 𝑏 (𝑥) 𝑟 𝑐 (𝑥) + 𝐼 𝑟 1 (𝑥) = 1$
with an inverse by the evaluation map $𝜂 (𝑎)$ .

tidy | en | SemBr

Stated without proof in 2008. Advanced Linear Algebra, §18, p. 259. ↩

Subalgebra generated by an algebraic element

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Subalgebra generated by an algebraic element

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