0

An affine domain $A$ over $k$ is a finite dimensional $k$-algebra which is also an integral domain as a ring. Here's my thought.

Fix $a \in A$ and $\varphi:k[x] \rightarrow A $ be defined as $f \rightarrow f(a)$. Then since it's $k$-algebra homomorphism, it's $k$-linear and not injective. So, $\ker\varphi$ is non trivial principal, prime ideal since $k[x]$ is ED and $A$ is domain. Writing $\ker\varphi =(x-b)$ for some $b\in k$ (here's where algebraically closedness is used), $a=b\in k$. Hence $A=k$.

But when I looked up for Wedderburn-Artin theorem for algebraically closed fields, it seems most people use additional requirement that $A$ is a division algebra. But I don't know why it's needed. Is this something related to the implicit use of unity in $A$ in my thought? Or is my thought wrong somewhere?

user26857
  • 52,094
Chogun
  • 45
  • Maybe you can supply a reference? This doesn''t sound like Wedderburn's theorem (which says every finite domain is a field, or perhaps every finite division ring is a field) – D_S Apr 24 '21 at 02:47
  • Oh it was Weddernburn Artin theorem which states that 'every semi simple k algebra is isomorphic to product of matrix algebras over division algberas over k'. Corollary states that 'if k is algebraically closed, then the division algebras are exactly k'. I googled and found some proofs that finite dimensional division algebras over k is k. That's the curious part. Why needed division algebra? – Chogun Apr 24 '21 at 03:11
  • One reference https://math.stackexchange.com/questions/405199/division-algebra-over-a-an-algebraically-closed-field – Chogun Apr 24 '21 at 03:11
  • 4
    One usually says that a $k$-algebra is affine if it is finitely generated as an algebra over $k$, not necessarily finite dimensional over $k$. For example, the polynomial ring $k[x]$ is an affine $k$-algebra.

    Your argument is correct (for the finite dimensional case) and is not really related to the Wedderburn-Artin theorem.

     – Allen Bell Apr 24 '21 at 03:24
  • Okay I got it! I was totally thinking wrong. Then it must be true that finitely 'generated' division algebra over an algebraically closed field is the field ifself. Can offer any sketch of rather elementary proof for this? I'm at elementary level. – Chogun Apr 24 '21 at 03:47
  • I think the question of whether a finitely generated $k$-algebra that is also a division algebra has to be finite dimensional over $k$ is open. For example, see Section 3 of this paper by Smoktunowicz. – Allen Bell Apr 24 '21 at 16:39
  • A finite dimensional domain is a field (by some finite dimension linear algebra argument). – FiMePr Apr 24 '21 at 16:44

0 Answers0