# Corollary 6.11 of Atiyah's "Introduction to commutative algebra"

On page 78, the corollary 6.11 says "let $$A$$ be a ring in which the zero ideal is a product $$m_1...m_n$$ of (not necessarily distinct) maximal ideals. Then $$A$$ is Noetherian iff $$A$$ is Artinian".

The proof: "Consider the chain of ideals $$A \supset m_1 \supseteq m_1m_2 \supseteq ... \supseteq m_1...m_n = 0$$. Each factor $$m_1... m_{i-1}/m_1... m_i$$ is a vector space over the field $$A/m_i$$. Hence a.c.c. $$\Longleftrightarrow$$ d.c.c. for each factor. But a.c.c.(resp. d.c.c.) for each factor $$\Longleftrightarrow$$ a.c.c.(resp. d.c.c.) for $$A$$, by repeated application of (6.3). Hence a.c.c. $$\Longleftrightarrow$$ d.c.c. for $$A$$."

I've got issues with the statement "Each factor $$m_1... m_{i-1}/m_1... m_i$$ is a vector space over the field $$A/m_i$$".

Let $$m_1... m_{i-1}/m_1... m_i$$ be denoted by $$I$$. It suffices to prove that $$I/m_iI$$ is sub $$A$$-module of $$A/m_i$$. We know that $$(A\cap I)/(m_i \cap I)$$ is a subgroup of $$A/m_i$$ and $$A\cap I = I$$, then it suffices to prove that $$m_i \cap I = m_iI$$.

If $$I\nsubseteq m_i$$, then $$I + m_i = A$$,for $$m_i$$ is a maximal ideal of $$A$$, hence $$m_i \cap I = m_iI$$.

But if $$I\subseteq m_i$$, in fact $$m_i \cap I \neq m_iI$$. Now I'm stuck. Does it mean that I spot an error?

You over complicate it. In general, if $$M$$ is a module over $$R$$ and $$I\subseteq R$$ is an ideal then the quotient $$M/IM$$ (which is obviously a module over $$R$$) is a module over $$R/I$$ with respect to the operation $$(r+I)(m+IM)=rm+IM$$. This is well defined, because if $$r-r'\in I$$ and $$m-m'\in IM$$ then:
$$rm-r'm'=rm-rm'+rm'-r'm'=r(m-m')+(r-r')m'\in IM+IM=IM$$
So in the theorem, take $$M=m_1...m_{i-1}$$ and $$I=m_i$$.