proof of Nakayama's lemma

Question

Let $R$ a local ring with maximal ideal $\mathfrak m$. Suppose $M$ is finitely generated $R-$module s.t. $\mathfrak m M=M$. Prove that $M=0$.

Proof : Let $m_1,...,m_n$ be generator of $M$. As $$\mathfrak m M=M,$$ there is a matrix $A$ with entries in $\mathfrak m$ s.t. $$A\bar m=\bar m$$ where $\bar m=(m_1,...,m_n)^T$.

Q1) Why is there such a matrix ? Where does it come from ?

Multiplying by the adjugate of the matrix $A-I$ implies that if $D=\det(A-I)$, then $Dm_i=0$ for all $i$.

Q2) I don't understand what do they mean by "multipliying by adjugate of the matrix $A-I$". I saw on wikipedia what is the adjugate, but it doesn't help.

By expanding the determinant, we can see that $D=\det (A-I)$ cannot be in $\mathfrak m$, for all the component of $A$ are in $\mathfrak m$, and so, there is only one term of the expansion of $\det(A-I)$ which is not, and this is the 1 which come from the $\prod(a_{ii}-1)$ terms.

Q3) I still don't understand what they mean, it looks really confuse...

It's not necessary to put the rest, since here I understand nothing of what they do. If it's clear for someone, I would appriciate some explanations...

Answer 1

The matrix exists because of the condition $\mathfrak{m}M=M$. That means that there are $a_{i1},a_{i2},\ldots,a_{in}\in\mathfrak{m}$ such that $$\sum_{j=1}^n{a_{ij}m_j}=m_i$$ for all $i$. This is the same thing as the existence of such a matrix.

The adjugate $B'$ matrix of a matrix $B$ has the property that $$BB'=\det(B)I$$ where $I$ is the identity matrix. This is what is important here. Since $$(A-I)\overline{m}=0$$ we must have that $Dm_i=0$ for all $i$.

What they mean for Q3 is that if you take the formula for the determinant as a sum $$\sum_{\sigma\in S_n}{(-1)^{\mathrm{sgn}(\sigma)}b_{1\sigma(1)}\cdots b_{n\sigma(n)}}$$ then since the diagonal entries are not in $\mathfrak{m}$ but every other entry is we must have that every term in is $\mathfrak{m}$ (because $\mathfrak{m}$ is an ideal) except for the term where $\sigma$ is the identity, when none of the elements of the product are in $\mathfrak{m}$; this means that $D+\mathfrak{m}\neq\mathfrak{m}$, so $D\notin\mathfrak{m}$.

Answer 2

Matt Samuel's answers explains pretty well the situation.

However, since Nakayama's lemma holds in a much more general setting, it could help your intuition if you see a different proof.

Let $R$ be a ring (not necessarily commutative, but with identity). Let $J(R)$ be its Jacobson radical and suppose $M$ is a finitely generated left $R$-module with $M=J(R)M$. Then $M=0$.

The Jacobson radical is the intersection of all maximal left ideals (it is the same as the intersection of all maximal right ideals). The property we need is that

if $r\in J(R)$, then $1-r$ is left invertible

which is rather easy to show: if $R(1-r)\ne R$, then $R(1-r)$ is contained in a maximal left ideal $I$; on the other hand $r\in J(R)$ implies $r\in I$, so $1=(1-r)+r\in I$: contradiction.

Let $\{x_1,\dots,x_k\}$ be a minimal generating set for $M$, in the sense that no proper subset of it generates $M$; assume $k\ge1$. Since $M=J(R)M$, we can write $$ x_k=\sum_{i=1}^k r_ix_i $$ with $r_i\in J(R)$ and, in particular $$ (1-r_k)x_k=\sum_{i=1}^{k-1}r_ix_i $$ Take a left inverse $s$ of $1-r_k$; then $$ x_k=\sum_{i=1}^{k-1}sr_ix_i $$ contradicting the mimimality of $\{x_1,\dots,x_k\}$. QED

Such a proof is not “constructive”, as it relies on existence of maximal ideals. However, it can be very easily adapted to your situation and it doesn't require any determinant based argument.

---

The last part can be explained more easily, too. If you look at $A-I$ modulo $\mathfrak{m}$, then it is $-I$, so its determinant modulo $\mathfrak{m}$ is $(-1)^n$. Therefore the determinant of $A-I$ does not belong to $\mathfrak{m}$, so it is invertible in $R$