# Infinite dimensional simple algebras of finite degree

Let $F$ be a field and let $B$ be an $F$-algebra. The degree of $B$ over $F$ is the smallest positive integer $\deg_F B = d \geq 1$ such that every element of $B$ satisfies a (monic) polynomial of degree $d$ over $F$ if such an $m$ exists, otherwise $\deg_F B = \infty$.

An algebra of finite degree over $F$ need not be finite-dimensional over $F$: for example, any Boolean ring is an $\mathbb{F}_2$-algebra of degree $2$.

Suppose, however, that $F$ has characteristic $0$ and $D$ is a division $F$-algebra of degree $d \in \mathbb{Z}_{>0}$. Is $D$ finite-dimensional?

By the primitive element theorem, the center $Z(D)$ of such a division algebra $D$ is finite-dimensional over $F$, so one can assume without loss of generality that $D$ is central.

I can prove that the answer is yes when $d=2$: in fact, a division $F$-algebra of degree $2$ is either a quadratic field extension of $F$ or a quaternion algebra over $F$. But the argument is a bit special.

(One can make examples of degree $p$ in characteristic $p$ by considering inseparable field extensions; but the question above is still interesting when $\mathrm{char} F = p > d$.)

Yes $D$ must be finite-dimensional over $F$. This follows from a Theorem of Kaplansky's that I found in Herstein's monograph "Noncommutative Rings". The first step is to show that an algebraic algebra of bounded degree satisfies a polynomial identity (see Lemma 6.2.3 in Herstein), i.e., is a P.I. algebra. Note, however, that the degree of this polynomial identity may be larger than the degree of the algebra, as defined in the question. Then since a division algebra $D$ is primitive, one can use:
Theorem 6.3.1 (Kaplansky) If $A$ is a primitive algebra satisfying a polynomial identity of degree $d$ then $A$ is a finite dimensional simple algebra over its center, of dimension at most $\lfloor d/2 \rfloor^2$.
The theorem would still apply in the characteristic $p$ case, but as suggested in the question, it would not follow that the center of $D$ is finite-dimensional over $F$.