Division algebra

Division algebra with only algebraic elements over an algebraically closed field

Let 𝕂 be an algebraically closed field and 𝐴 be a division algebra such that every 𝑎 ∈𝐴 is an algebraic element over 𝕂.¹ Then 𝐴 =𝕂. falg

Proof

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

Corollaries

The following situations guarantee every element 𝑎 ∈𝐴 is algebraic over 𝕂.

1. All elements of a finite-dimensional unital associative algebra are algebraic.
2. Dixmier’s lemma
3. Quillen’s lemma

---

tidy | en | SemBr

Footnotes

1. Equivalently 𝐴 is an algebra such that every 𝑎 ∈𝐴 has a minimal polynomial 𝑚𝑎(𝑥) ∈𝕂[𝑥] with a nonzero constant term ↩