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

Question

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?

Answer 1

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 $I=m_i$.