Artin local ring

Question

I have studied "structure theorem for Artin rings" which states

"An Artin ring $A$ is unique a finite direct product of Artin local rings".

Let $A$ be Artin ring. By Chinese remainder theorem, $A \cong \prod (A/m_i^k)$, in which $m_i$ is a maximal ideal in $A$ and $R^k= (m_1 \cap m_2 \cap... \cap m_n)^k =0 $.

Since $A$ is Artin then $A/m_i^k$ is also Artin.

However, I do not understand why $A/m_i^k$ is a local ring.

I am beginner then could you help me explain in detail about it?

I guess the maximal ideal of the ring $A/m_i^k$ is $m_i/m_i^k$.

Thank you in advance.

score 7 · Answer 1 · answered Dec 01 '14 at 12:02

7

A maximal ideal of $A/m^k_i$ should be of the form $M/m^k_i$, where $M$ is a maximal ideal of $A$ containing $m^k_i$. But now, $$M \supset m^k_i \Rightarrow M=\sqrt{M} \supset \sqrt{m^k_i} = m_i$$ and by maximality of $m_i$ we have equality.

So $m_i/m_i^k$ is the unique maximal ideal of $A/m^k_i$.

Note that this holds without assuming $A$ is artinian.

answered Dec 01 '14 at 12:02

Crostul

37,715

I think this is the shortest way to show that we can ignore the power $k$. Do you have better ideas? – Crostul Dec 01 '14 at 12:06
2

But prime ideals (and so maximal as well) are equal to thier radical, so everything works. – Crostul Dec 01 '14 at 12:32
Thank you so much. Now I understand. :) – user69833 Dec 02 '14 at 01:11

Artin local ring

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