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Proposition 8.8 from Atiyah Macdonald states:

Let $A$ be an Artin local ring. Then the following are equivalent:

  1. every ideal in $A$ is principal;
  2. the maximal ideal $\mathfrak{m}$ of $A$ is principal;
  3. $\dim_k(\mathfrak{m} / \mathfrak{m}^2) \leq 1$.

Okay, the implication from 1 to 2 is clear of course, but I don’t get the implication from 2 to 3, why is it clear? In my lecture we didn’t discuss the dimension a lot. So $k$ is the residue field given by $A / \mathfrak{m} = k$ and we know that $\mathfrak{m} / \mathfrak{m}^2$ becomes a $k$-vector space, but why does it have dimension less or equal $1$? I struggle to understand it at the moment, maybe someone can enlight me! Thank you

Jendrik Stelzner
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Zanjo
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1 Answers1

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Say the maximal ideal is generated as an $A$-module by $x$. Then, $m/m^2$ is generated by $x+m^2$ as an $A$-module. But, $m/m^2$ is clearly killed by $m$, so it is also a $k=A/m$ vector space. Thus, it suffices to see that $m/m^2$ is spanned by $x+m^2$ as a $k$-vector space.

How does the vector space structure work on $m/m^2$? Well, a scalar $\lambda + m$ acts on an arbitrary element of $m/m^2$, say $rx+m^2$, by $(\lambda+m)(rx+m^2) = \lambda rx + m^2$. Either $r$ is in $m$, in which case $rx \in m^2$ so $rx+m^2=m^2$ is the zero vector, or $r \not \in m$ in which case $r$ is a unit of $A$, so $r+m$ is a nonzero element of $k$. Hence every element $rx+m^2$ in $m/m^2$ is in the $k$-span of $x+m^2$. Since we have a spanning set of size $1$, $\dim_k m/m^2 \le 1$.

In fact, this proof shows for any local ring $A$ with maximal ideal $m$, if $m=(x_1,\cdots,x_d)$ then $m/m^2$ is spanned by $x_1+m^2,\cdots,x_d+m^2$ as an $A/m$-vector space.

walkar
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