Posted by Jason Polak on 13. October 2016 · Write a comment · Categories: commutative-algebra

Nakayama’s lemma probably comes in as many flavours as ice cream.

In this post we’ll review some forms of them, and deduce some consequences. Before we continue, recall that in an associative ring $R$ with unity, the Jacobson radical is the intersection of all the left maximal ideals of $R$. It is easy to show that this ideal coincides with the intersection of all the right maximal ideals of $R$; thus, the Jacobson radical is a two-sided ideal.

The form of Nakayama’s lemma I like best is:

Nakayama’s Lemma #1. Let $R$ be a ring, $I\subset R$ an ideal contained in the Jacobson radical of $R$, and $M$ a finitely-generated left $R$-module. If $IM = M$ then $M = 0$.

Proof. Suppose $M$ is not zero. Then it has a nonempty minimal generating set $m_1,\dots,m_k$. We can write $m_1 = r_1m_1 + \cdots + r_km_k$ with $r_i\in I$, since we assumed $IM = M$. Therefore, $(1 – r_1)m_1 = r_2m_2 + \cdots + r_km_k$. Because $I$ is in the Jacobson radical, $1 – r_1$ is invertible, and hence $m_2,\cdots,m_k$ also generates $M$, a contradiction. QED

I like this version of the theorem the most because an immediate consequence in the local commutative case is the following: if $M$ is a finitely-generated $R$-module then minimal generating sets for $M$ correspond to bases of $M/\mfr$ as an $R/\mfr$-vector space where $\mfr$ is the maximal ideal of $R$.

There is another very useful form of Nakayama’s lemma. To state it, let’s recall that the if $M$ is a left $R$-module, the annihilator of $M$ in $R$, denoted by ${\rm ann}_R(M)$, is the set $\{ r\in R : rm = 0\forall_{m\in M}\}$. In words, it is the set of elements of $R$ that send every element of $M$ to zero. Exercise: the annihilator of $M$ is a two-sided ideal of $R$. Now we can state:

Nakayama’s Lemma #2.Let $R$ be a commutative ring, $I\subseteq R$ and ideal, and $M$ a finitely-generated $R$-module. Then $IM = M$ if and only if $I + {\rm ann}_R(M) = R$.
Proof. “If Direction”. If $I + {\rm ann}_R(M) = R$ then there exists an $i\in I$ and a $x\in {\rm ann}_R(M)$ such that $1 = i + x.$. Hence $1M = (i + x)M = iM$ and so $IM = M$.

“Only If Direction”. Suppose that $IM = M$. If $I + {\rm ann}_R(M)\not = R$ then it is contained in some maximal ideal $\mfr$ of $R$. In the localization $R_\mfr$, we have $IM_\mfr = M_\mfr$. Therefore, by Nakayama’s Lemma #1, we have $M_\mfr = 0$. Thus, we can choose for each $m\in M$ an $x\in R – \mfr$ such that $xm = 0$. Since $M$ is finitely generated, we can choose an $x$ such that $xm = 0$ for all $m\in M$. In other words, $x\in{\rm ann}_R(M)$, a contradiction. QED

Let us remark that the ideal $I$ no longer has to be contained in the Jacobson radical, nor does it even have to be proper: the conclusion holds regardless. The good thing about this form is you don’t even have to remember what the Jacobson radical is to use it! (But, who would want to forget the Jacobson radical?!)

Also, I’m not sure if the above form can be proved in the noncommutative case, since the proof I gave does not work the same way there (localizations work in the noncommutative world but they don’t have all the nice properties of commutative localization).

Some Consequences

Let’s look at some consequences of Nakayama’s lemma. As already mentioned, we have:

Theorem. Let $R$ be a commutative local ring with maximal ideal $\mfr$. For a finitely generated $R$-module $M$, the quotient $M\to M/\mfr$ induces a bijective correspondence between the minimal generating sets of $M$ and the $R/\mfr$-bases of $M/\mfr$.

I’ll leave this as an exercise. Onto more pressing matters. Recall that an $R$-module $P$ is projective if the functor ${\rm Hom}_R(P,-)$ from $R$-modules to abelian groups is exact. This is the case if and only if there exists a “dual basis”; that is, elements $f_i\in {\rm Hom}_R(P,R)$ and $p_i\in P$ (indexed by some set $I$) such that for all $p\in P$, we have $f_i(p) = 0$ for almost all $i$ and $p = \sum f_i(p)p_i$.

If $P$ is a finitely-generated projective $R$-module over a commutative ring $R$, then ${\rm ann}_R(P)$ is a projective ideal of $R$.
Let $f_i,p_i$ be a dual basis of $P$. Let $I$ be the ideal generated by elements of the form $f(p)$ where $f\in{\rm Hom}_R(P,R)$ and $p\in P$. Since a dual basis exists for $P$ we have $IP = P$. Since $P$ is finitely generated, we have $I + {\rm ann}_R(P) = R$. All that remains is to show that this sum is a direct sum. Observe that if $r\in {\rm ann}_R(P)$ then $ri = 0$ for $i\in I$. Write $1 = a + b$ where $a\in I$ and $b\in{\rm ann}_R(P)$. If $x\in I\cap {\rm ann}_R(P)$ then $x = 1x= ax + bx$. By our observation this is zero, so the sum is indeed direct. QED

Is the converse of this theorem true: in other words, if $P$ is a finitely generated module and its annihilator is projective, then is $P$ itself projective? This is obviously false: for example, there are non-projective abelian groups, like $\Z/n$ for $n\not=0$, and yet all ideals of $\Z$ are projective.

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