Division algebra with only algebraic elements over an algebraically closed field

Let $𝕂$ be an algebraically closed field and $𝐴$ be a division algebra over $𝕂$ such that every $𝑎 \in 𝐴$ is an algebraic element over $𝕂$ .¹ Then $𝐴 = 𝕂$ . #m/thm/falg

Proof

Let $𝑎 \in 𝐴$ and $𝑚 𝑎 (𝑥) \in 𝕂 [𝑥]$ be its minimal polynomial. Since $𝐴$ has no zero divisors, $𝑚 𝑎 (𝑥)$ must be an irreducible polynomial: For if $𝑚 𝑎 (𝑥) = 𝑝 (𝑥) 𝑞 (𝑥)$ then $𝑝 (𝑎) 𝑞 (𝑎) = 0$ and hence either $𝑝 (𝑎) = 0$ or $𝑞 (𝑎) = 0$ , a contradiction. Since $𝑚 𝑎 (𝑥)$ is irreducible it is linear by ^A2, thus $𝑚 𝑎 (𝑥) = 𝑥 - 𝜆$ whence $𝑎 = 𝜆 \in 𝕂$ .

Corollaries

The following situations guarantee every element $𝑎 \in 𝐴$ is algebraic over $𝕂$ .

#state/tidy | #lang/en | #SemBr

Equivalently $𝐴$ is an algebra such that every $𝑎 \in 𝐴$ has a minimal polynomial $𝑚 𝑎 (𝑥) \in 𝕂 [𝑥]$ with a nonzero constant term ↩

Division algebra with only algebraic elements over an algebraically closed field

Corollaries

Footnotes