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If $K$ is a commutative ring which is a finite dimensional vector space over $\mathbb C$ what can we say about the maximal ideals of $K$? What can we say if instead of $\mathbb C$ we have some arbitrary field?

If $x\in K$ then $1,x,x^2,...,x^k$ must be linearly dependent for some $k \in \mathbb N$. So $x$ satisfies a polynomial over $\Bbb C$, say $f$. How is $\Bbb C[X]/\langle f\rangle$ related to $K$ if $f$ is the minimal degree polynomial?

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  • $\begingroup$ A key point is that the complex numbers are algebraically closed. That means there are not many options for what the residue field must be... $\endgroup$ – Zhen Lin Feb 16 '14 at 12:15
  • $\begingroup$ Yes, and if $\Bbb C/\langle f \rangle \subseteq K$ then we can study $K$ more easily. So is it true? $\endgroup$ – viplov_jain Feb 16 '14 at 14:44
  • $\begingroup$ Is $K$ commutative? $\endgroup$ – Qiaochu Yuan Feb 17 '14 at 2:51
  • $\begingroup$ Yes,Thanks for pointing out. $\endgroup$ – viplov_jain Feb 17 '14 at 8:15
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In general, if $A$ is a finite $k$-algebra, then $A$ is necessarily $A$ is Artinian. This implies that $\text{Spec}(A)=\text{MaxSpec}(A)$ and that $A$ decomposes as a finite product of local artinian rings.

In fact, let $A$ be a finite type $k$-algebra. Then, the following are equivalent:

  1. $\text{Spec}(A)$ is finite.
  2. $\text{Spec}(A)$ is discrete.
  3. $\text{MaxSpec}(A)$ is finite.
  4. $A$ is a finite $k$-algebra.
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  • $\begingroup$ I'm looking for a more elementary approach.Is there a simple way to state and prove these. $\endgroup$ – viplov_jain Feb 18 '14 at 9:33

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