One theorem that I have learned in algebra and always forget the statement(s) of is the famous Nakayama’s Lemma. A quick Google search shows that I am probably not the only the person who has this issue. In any case, there is a particularly nice case of Nakayama’s Lemma that I use to reconstruct the more general version via quiver representation theory that I thought would be cool to write about.

On this page:

• Statement of Nakayama’s Lemma
• Quivers Representations and Path Algebras
• Nakayama’s Lemma for Path Algebras

Statement of Nakayama’s Lemma

When someone says “Nakayama’s Lemma,” there are at least ten different statements that I am somewhat aware of (I can never remember them anyway!) that come to mind. I will give the one that I have seen stated the most (not just in the land of representation theory). For this entire section, we will let $R$ be a (not necessarily commutative) unital ring.

[definition] [deftitle]Definition %counter% (Jacobson radical)[/deftitle]

The Jacobson radical $\rad R$ of $R$ is the intersection of all the maximal right ideals of $R$. In the special case where $R = 0$, we define $\rad R = 0$. [/definition]

As it turns out, $\rad R$ is a two-sided ideal. Not only that, but the left and right-Jacobson radicals coincide (hence, why we just say the Jacobson radical).

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The left and right Jacobson radicals coincide. In particular, the Jacobson radical is a two-sided ideal of $R$. [/theorem]

Now we state Nakayama’s Lemma:

[theorem] [thmtitle]Theorem %counter% (Nakayama’s Lemma)[/thmtitle]

Let $J \subseteq R$ be a right ideal contained in $\rad R$ and $M_R$ be a finitely-generated right $R$-module. If $MJ = M$, then $M = 0$. [/theorem]

Quivers Representations and Path Algebras

While I have no idea what the geometric intuition for Nakayama’s Lemma is, I am much more familiar with how the statement can be understood from the representation theory of finite-dimensional algebras via quivers (direct graphs with arcs of multiplicity allowed).

[definition] [deftitle]Definition %counter% (Quivers)[/deftitle]

A quiver $Q$ is an ordered quadruple $(Q_0, Q_1, s, t)$ where

• $Q_0$ is the set of vertices.
• $Q_1$ is the set of arrows (arcs).
• $s:Q_1 \to Q_0$ is the “source” map that sends any arrow to its tail.
• $t:Q_1 \to Q_0$ is the “target” map that sends any arrow to its head. [/definition]

[definition] [deftitle]Definition %counter% (Path algebra)[/deftitle]

Let $k$ be an algebraically closed field. Then $kQ$ is a $k$-algebra, called the path algebra of $Q$, with basis labeled by the (oriented) paths in $Q$ and multiplication given by path concatenation. [/definition]

In the case of acyclic quivers, it turns out that it is extremely easy to describe the Jacobson radical.

[theorem] [thmtitle]Theorem %counter% (Jacobson radical is spanned by arrows)[/thmtitle]

Let $Q$ be an acyclic quiver. Then $\rad kQ$ is the two-sided ideal generated by all the arrows. [/theorem]

In the more general case where we have a quiver with relations (i.e. a path algebra modded by an admissible ideal), this statement is also true.

Now we turn to the other side of the picture: quiver representations.

[definition] [deftitle]Definition %counter% (Representations of quivers)[/deftitle]

A representation $V$ of the quiver $Q$ is an assignment of a vector space $V(i)$ to each vertex $i \in Q_0$ and a $k$-linear map $V(\alpha):V(s\alpha) \to V(t\alpha)$ to every arrow $\alpha:i \to j$ in the quiver $Q$.

We say that $f:V \to W$ is a morphism of the representations $V$ and $W$ of $Q$ provided that $f$ assigns a $k$-linear map to each vertex of $Q$ so that the diagram

commutes for every arrow $\alpha$ of $Q$.

Under this construction, we obtain the abelian $k$-category $\rep Q$ where the objects are the representations above and the morphisms are the morphisms defined above. [/definition]

A fundamental theorem of the representations of quivers is that representations of quivers naturally correspond to f.g. modules of the path algebra. Accordingly, we can construct all of the f.g. modules of various algebras by literally drawing pictures and contravariantly associating vector spaces and linear transformations between them to the associated quiver.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The category $\rep Q$ is equivalent to $\rmod{kQ}$ where $\rmod{kQ}$ is the category of finitely generated right $kQ$-modules. [/theorem]

The functor $\mathcal{G}:\rep Q \to \rmod kQ$ that takes representations to modules over the path algebra is the interesting one (the converse functor tells us these are the only ones we really care about). In particular, we are interested in how it maps objects. Here’s the construction.

Given a representation $V$ of $Q$, we define $\mathcal{G}(V) = \bigoplus_{i\in Q_0} V(i)$. Then we define a $kQ$-action on $\mathcal{G}(V)$ as follows: Let $p \in kQ$ be a path and $v \in V(i)$. Then

where $V(p)$ is the composition of the $V(\alpha_j)$’s where $p = \alpha_1 \cdots \alpha_r$ is the arrows making up the path $p$. Furthermore, $vp$ sits in the $ti$-component of $\mathcal{G}(V)$.

Nakayama’s Lemma for Path Algebras

Now we bring back Nakayama’s Lemma and apply it to a path algebra. Under everything we mentioned above, we can construct a simple example from which Nakayama’s Lemma is actually completely trivial.

Consider the quiver

with representation

Then the module $M_{\CC Q}$ associated with $V$ is the $\CC$-vector space

with the $\CC Q$-action given above. For example,

gives

That is, the $\CC Q$-action on $M$ just applies the corresponding maps in $V$ and then pushes the result along the arrows (which is expected due to the construction of the naturality of the equivalence). The fact that the action keeps “pushing” things along the arrows means that any ideal $J$ sitting inside $\rad \CC Q = (\alpha, \beta, \alpha\beta)$ necessarily cannot have the property that $MJ = M$ as $Q$ is acyclic. So Nakayama’s Lemma is actually a completely trivial statement here. This idea generalizes as well — the lack of cycles prevents us from “landing back where we started” (i.e. $MJ = M$).

So whenever I need to reconstruct the statement of Nakayama’s Lemma, I just think about examples like this one and note that $M \neq 0$, of course, implies that any ideal $J$ sitting inside $\rad \CC Q$ is not gonna result in $MJ = M$ for acyclic $Q$. So the general statement then is fairly immediate.