# If $R$ is a finitely generated $D$-algebra, $D$ a division ring, then $R$ has finite basis over $D$?

Let $k \subset D \subset R$, where $k$ is a field of characteristic zero, $D$ is a division $k$-algebra, and $R$ is affine over $D$ (= $R$ is a finitely generated $D$-algebra).

$R$ is a free $D$-module, since $D$ is a division ring (see this question), so $R$ has a basis, finite or infinite, over $D$.

Is the assumption that $R$ is affine over $D$ implies that $R$ has a finite basis as a $D$-module?

Is the fact that $D$ is Noetherian helps? (a division ring has no left ideals except the zero ideal, so it is trivially Noetherian).

My motivation is a similar result in field theory that says that an algebraic field extension which is finitely generated is finite dimensional. Perhaps here I should further assume that $R$ is left algebraic over $D$, and only then get a positive answer?

Edit: The following is true: If $R$ is affine over a division ring $D$ (=finitely generated as a $D$-algebra) and also $R$ is left algebraic over $D$, then $R$ is finitely generated as a $D$-module. The proof is as in the commutative case, with one exception: We need to be careful on which side we multiply; more specifically, if $d_m r^m+d_{m-1}r^{m-1}+\cdots+d_1r+d_0=0$ with minimal $m$, then $r^m= -d_m^{-1}(d_{m-1}r^{m-1}+\cdots+d_1r+d_0)$, from which it follows that $1,r,\ldots,r^{m-1}$ are generators for $D[r]$ over $D$ (indeed, higher powers of $r$ are $D$-linear combinations of $1,r,\ldots,r^{m-1}$-- multiply by $r$ on the right both sides etc.). Finally, it is not difficult to see that the tower of finitely generated modules $D \subset D[r_1] \subset D[r_1,r_2] \ldots \subseteq D[r_1,\ldots,r_l]=R$ (not necessarily free modules) is also a finitely generated module.
• No, it is not even true if $D=k$, e.g. a polynomial ring. See this question. – Kimball Nov 30 '17 at 3:18
• @Kimball, thanks, but what if $R$ is left algebraic over $D$? Is the proof for fields can be adjusted to the non-commutative setting ? – user237522 Nov 30 '17 at 8:37
• What does left algebraic over $D$ mean? – Kimball Nov 30 '17 at 13:31
• It means that every element of $R$ is left algebraic over $D$. An element $r \in R$ is left algebraic over $D$ if there exist $d_m,\ldots,d_1,d_0 \in D$ such that $d_mr^m+\cdots+d_1r+d_0=0$ (the coefficients are on the left. Similarly, right algebraic is when the coefficients are on the right: $r^md_m+\cdots+rd_1+d_0=0$). – user237522 Nov 30 '17 at 17:19