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.
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.