0

Let $K$ be a field and $k$ be a subfield. Let $k'$ be a finite algebraic extension of $k$. Then it's said that $K\otimes_k k'$ is a $K$-algebra of finite rank, and hence it's a direct product of artinian local rings:

$$K\otimes_kk'=A_1\times A_2\times \cdots\times A_r.$$

I don't know why it has such a decomposition and each component $A_i$ is artinian? I also saw a statement that: if ($A,m$) is a local ring containing a field $k$, and if $B$ is a finite $A$-algebra, then $B/mB$ is a finite $A/m$-algebra, and hence artinian. So, does it means finite algebra is artinian? Why? Hope someone could help. Thanks!

user26857
  • 52,094
Yuyi Zhang
  • 1,442
  • Have you tried the case when $k'$ is a simple extension? – user26857 Jan 09 '20 at 15:09
  • $A/m$ is a field, and every finite algebra over a field is artinian (since it is a finitely dimensional vector space). For more information see, for instance, https://math.stackexchange.com/questions/46654/atiyah-macdonald-exercise-8-3-artinian-iff-finite-k-algebra – user26857 Jan 09 '20 at 15:11

0 Answers0