“The fundamental result of commutative algebra, and its power and elegance are second to none.”
―Jürgen Herzog, on NAK
Remark10.44.
We will now show a very simple but extremely useful result known as Nakayama’s Lemma. Nakayama himself claimed that this should be attributed to Krull and Azumaya, but it’s not clear which of the three actually had the commutative ring statement first. So some authors (e.g., Matsumura) prefer to refer to it as NAK. There are actually a range of statements, rather than just one, that go under the banner of Nakayama’s Lemma a.k.a. NAK.
Proposition10.45.NAK 1.
Let \(R\) be a ring, \(I\) an ideal, and \(M\) a finitely generated \(R\)-module. If \(IM = M\text{,}\) then:
there is an element \(r\in 1 + I\) such that \(rM=0\text{,}\) and
there is an element \(a\in I\) such that \(am=m\) for all \(m\in M\text{.}\)
Proof.
Let \(M = R m_1 + \cdots + R m_s\text{.}\) By assumption, we have equations
with \(a_{ij}\in I\text{.}\) Setting \(A=[a_{ij}]\) and \(v=[m_i]\text{,}\) we have a matrix equation \(Av=v\text{.}\) By the [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Determinantal Technique|Determinantal Technique]], the element \(\det(I_{s \times s} - A) \in R\) kills each \(m_i\text{,}\) and hence it kills \(M\text{.}\) Since \(\det(I_{s \times s}-A)\equiv \det(I_{s \times s}) \equiv 1 (\!\pmod I)\text{,}\) this determinant is the element \(r\) we seek for the first statement. For the latter statement, set \(a=1-r\text{,}\) which is in \(I\) and satisfies \(am=m-rm=m\) for all \(m\in M\text{.}\)
Theorem10.46.NAK 2.
Let \((R,\fm,k)\) be a local ring, and \(M\) be a finitely generated module. If \(M=\fm M\text{,}\) then \(M=0\text{.}\)
Proof.
By Proposition 10.45, there exists an element \(r\in 1+\fm\) that annihilates \(M\text{.}\) Notice that \(1 \notin \fm\text{,}\) so any such \(r\) must be outside of \(\fm\text{,}\) and thus a unit. Multiplying by its inverse, we conclude that \(1\) annihilates \(M\text{,}\) or equivalently, that \(M=0\text{.}\)
Theorem10.47.NAK 3.
Let \((R,\fm,k)\) be a local ring, and \(M\) be a finitely generated module, and \(N\) a submodule of \(M\text{.}\) If \(M = N + \fm M\text{,}\) then \(M = N\text{.}\)
Proof.
By taking the quotient by \(N\text{,}\) we see that
Let \((R,\fm,k)\) be a local ring, and \(M\) be a finitely generated module. For \(m_1,\dots,m_s\in M\text{,}\)
\begin{equation*}
m_1,\dots, m_s \text{ generate } M \Longleftrightarrow \overline{m_1},\dots,\overline{m_s} \text{ generate } M/\fm M.
\end{equation*}
Thus, any generating set for \(M\) consists of at least \(\dim_k (M/\fm M)\) elements.
Proof.
The implication \((\implies)\) is clear. If \(m_1,\dots, m_s \in M\) are such that \(\overline{m_1},\dots,\overline{m_s}\) generate \(M/\fm M\text{,}\) consider \(N= R m_1 + \dots + R m_s \subseteq M\text{.}\) Since \(M/\fm M\) is generated by the image of \(N,\) we have \(M= N + \fm M\text{.}\) By Theorem 10.46, \(M=N\text{.}\)
Remark10.49.
Since \(R/\fm\) is a field, \(M/\fm M\) is a vector space over the field \(R/\fm\text{.}\)
Definition10.50.Minimal Generators.
Let \((R,\fm)\) be a local ring, and \(M\) a finitely generated module. A set of elements \(\{m_1,\dots,m_t\}\) is a minimal generating set of \(M\) if the images of \(m_1,\dots,m_t\) form a basis for the \(R/\fm\) vector space \(M/\fm M\text{.}\)
Lemma10.51.Min Gen Sets have Same Cardinality.
Let \((R,\fm,k)\) be a local ring and \(M\) be a finitely generated module. Any generating set for \(M\) contains a minimal generating set, and every minimal generating set has the same cardinality.
Proof.
As a consequence of basic facts about basis for vector spaces
Definition10.52.Minimal Number of Generators.
Let \((R,\fm)\) be a local ring and \(N\) an \(R\)-module. The minimal number of generators of \(M\) is
\begin{equation*}
\mu(M) := \dim_{R/\fm} \left( M / \fm M \right).
\end{equation*}
Equivalently, this is the number of elements in a minimal generating set for \(M\text{.}\)
Definition10.53.Minimal Number of Generators (Graded).
Let \(R\) be an \(\N\)-graded ring with \(R_0\) a field, and \(M\) a finitely generated \(\Z\)-graded \(R\)-module. The minimal number of generators of \(M\) is
\begin{equation*}
\mu(M) := \dim_{R/R_+} \left( M / R_+ M \right).
\end{equation*}
Proposition10.54.GNAK1.
Let \(R\) be an \(\N\)-graded ring, and \(M\) a \(\Z\)-graded module such that \(M_{<a}=0\) for some \(a\text{.}\) If \(M= (R_+) M\text{,}\) then \(M=0\text{.}\)
Proof.
If \(M \neq 0\text{,}\) then \(M\) has a nonzero homogeneous element. Suppose \(M_a \neq 0\text{.}\) On the one hand, the homogeneous elements in \(M\) live in degrees at least \(a\text{,}\) but \((R_+)M\) lives in degrees strictly bigger than \(a\text{.}\) Thus \((R_+)M \neq M\text{.}\)
Remark10.55.
If \(M\) is finitely generated, then it can be generated by finitely many homogeneous elements, the homogeneous components of some finite generating set. If \(a\) is the smallest degree of a homogeneous element in a homogeneous generating set, since \(R\) lives only in positive degrees we must have \(M \subseteq R M_{\geqslant a} \subseteq M_{\geqslant a}\text{,}\) so \(M_{<a} = 0\text{.}\)
Proposition10.56.Graded Rings and Generators.
Let \(R\) be an \(\N\)-graded ring, with \(R_0\) a field, and \(M\) a \(\Z\)-graded module such that \(M_{<a}=0\) for some degree \(a\text{.}\) A set of elements of \(M\) generates \(M\) if and only if their images in \(M/(R_+)M\) span \(M/(R_+)M\) as a vector space over \(R_0\text{.}\) Since \(M\) and \((R_+)M\) are graded, \(M/(R_+)M\) admits a basis of homogeneous elements.
Proof.
Just as above, we obtain the following:
Macaulay2.
In Macaulay2, the command mingens returns the a minimal generating set of the given module (as a list), while numgens returns the minimal number of generators. Notice that this computation is only reliable if the ring and module you are considering are defined to be graded.
Remark10.57.
Note that we can use [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Proposition - GNAK1|NAK]] to prove that certain modules are finitely generated in the graded case; in the local case, we cannot.
