3 added 13 characters in body; edited body

There are various forms of the Nakayama lemma. Here is a rather general one; note that it does not involve maximal ideals and is a constructive theorem (Atiyah-MacDonald, Commutative Algebra, Prop. 2.4 ff).

Let $M$ be a finitely generated $A$-module, $\mathfrak{a} \subseteq A$ be an ideal and $\phi \in End_A(M)$ such that $\phi(M) \subseteq \mathfrak{a} M$. Then there is an equation of the form $\phi^n + r_1 \phi^{n-1} + ... + r_n = 0$, where the $r_i$ are in $\mathfrak{a}$.

The proof uses the equality $adj(A) * A adj(X) \cdot X = det(A)$ \text{det}(X)$for quadratic matrices over a ring. I call this an elementary linear algebra fact. Of course, there you only prove it for fields but using function fields implies the result for general rings. If we take$\phi=\text{id}_M$, we get the following form: Let$M$be a finitely generated$A$-module and let$\mathfrak{a} \subseteq A$be an ideal such that$\mathfrak{a} M = M$. Then there exists some$r \in A$such that$rM = 0$and$r \equiv 1$mod$\mathfrak{a}$. In particular, we get: Let$M$be a finitely generated$A$-module and let$\mathfrak{a} \subseteq A$be an ideal such that$\mathfrak{a} M = M$and$\mathfrak{a}$lies in every maximal ideal of$A$. Then$M=0$. Observe that this argument uses Zorn's lemma (namely that every non-unit is contained in a maximal ideal) and is thus nonconstructive. Which is of course not surprising since without Zorn's lemma it is consistent that there are nontrivial rings without any maximal ideals at all. This should convince you that the first form of the Nakayama lemma is the most easy and elementary one. The last form has another short proof, which is standard and given in the question above. Here is another short well-known proof for the last form, which also works if$A$is noncommutative (then we have to replace "maximal ideal" by "maximal left ideal"): Assume$M \neq 0$. Since$M$is finitely generated, an application of Zorn's lemma shows that$M$has a maximal proper submodule$N$. Then$M/N$is simple, thus isomorphic to$R/\mathfrak{m}$A/\mathfrak{m}$ for some maximal left ideal $\mathfrak{m}$. Then $N = \mathfrak{m} M = M$, contradiction.

By the way, I don't know if the first form is true if $A$ is noncommutative. The theory of determinants is not really prosperous over noncommutative rings. Hints?

In many texts about algebraic geometry only the last form of Nakayama's the Nakayama lemma is needed. But the first one is stronger and is used in many results in commutative algebra.

2 added 6 characters in body

There are various forms of the Nakayama lemma. Here is a rather general one; note that it does not involve maximal ideals and is a constructive theorem (Atiyah-MacDonald, Commutative Algebra, Prop. 2.4 ff).

Let $M$ be a finitely generated $A$-module, $\mathfrak{a} \subseteq A$ be an ideal and $\phi \in End_A(M)$ such that $\phi(M) \subseteq \mathfrak{a} M$. Then there is an equation of the form $\phi^n + r_1 \phi^{n-1} + ... + r_n = 0$, where the $r_i$ are in $\mathfrak{a}$.

The proof uses the equality $adj(A) * A = det(A)$ for quadratic matrices over a ring. I call this an elementary linear algebra fact. Of course, there you only prove it for fields but using function fields implies the result for general rings. If we take $\phi=\text{id}_M$, we get the following form:

Let $M$ be a finitely generated $A$-module and let $\mathfrak{a} \subseteq A$ be an ideal such that $\mathfrak{a} M = M$. Then there exists some $r \in A$ such that $rM = 0$ and $r \equiv 1$ mod $\mathfrak{a}$.

In particular, we get:

Let $M$ be a finitely generated $A$-module and let $\mathfrak{a} \subseteq A$ be an ideal such that $\mathfrak{a} M = M$ and $\mathfrak{a}$ lies in every maximal ideal of $A$. Then $M=0$.

Observe that this argument uses Zorn's lemma (namely that every non-unit is contained in a maximal ideal) and is thus nonconstructive. Which is of course not surprising since without Zorn's lemma it is consistent that there are nontrivial rings without any maximal ideals at all. This should convince you that the first form of the Nakayama lemma is the most easy and elementary one. The last form has another short proof, which is standard and given in the question above.

Here is another short well-known proof for the last form, which also works if $A$ is noncommutative (then we have to replace "maximal ideal" by "maximal left ideal"): Assume $M \neq 0$. Since $M$ is finitely generated, an application of Zorn's lemma shows that $M$ has a maximal proper submodule $N$. Then $M/N$ is simple, thus isomorphic to $R/\mathfrak{m}$ for some maximal left ideal $\mathfrak{m}$. Then $N = \mathfrak{m} M = M$, contradiction.

By the way, I don't know if the first form is true if $A$ is noncommutative. The theory of determinants is not really prosperous over noncommutative rings. Hints?

In many texts about algebraic geometry only the last form of Nakayama's lemma is needed. But the first one is stronger and is used in many results in commutative algebra.

1

There are various forms of the Nakayama lemma. Here is a rather general one; note that it does involve maximal ideals and is a constructive theorem (Atiyah-MacDonald, Commutative Algebra, Prop. 2.4 ff).

Let $M$ be a finitely generated $A$-module, $\mathfrak{a} \subseteq A$ be an ideal and $\phi \in End_A(M)$ such that $\phi(M) \subseteq \mathfrak{a} M$. Then there is an equation of the form $\phi^n + r_1 \phi^{n-1} + ... + r_n = 0$, where the $r_i$ are in $\mathfrak{a}$.

The proof uses the equality $adj(A) * A = det(A)$ for quadratic matrices over a ring. I call this an elementary linear algebra fact. Of course, there you only prove it for fields but using function fields implies the result for general rings. If we take $\phi=\text{id}_M$, we get the following form:

Let $M$ be a finitely generated $A$-module and let $\mathfrak{a} \subseteq A$ be an ideal such that $\mathfrak{a} M = M$. Then there exists some $r \in A$ such that $rM = 0$ and $r \equiv 1$ mod $\mathfrak{a}$.

In particular, we get:

Let $M$ be a finitely generated $A$-module and let $\mathfrak{a} \subseteq A$ be an ideal such that $\mathfrak{a} M = M$ and $\mathfrak{a}$ lies in every maximal ideal of $A$. Then $M=0$.

Observe that this argument uses Zorn's lemma (namely that every non-unit is contained in a maximal ideal) and is thus nonconstructive. Which is of course not surprising since without Zorn's lemma it is consistent that there are nontrivial rings without any maximal ideals at all. This should convince you that the first form of the Nakayama lemma is the most easy and elementary one. The last form has another short proof, which is standard and given in the question above.

Here is another short well-known proof for the last form, which also works if $A$ is noncommutative (then we have to replace "maximal ideal" by "maximal left ideal"): Assume $M \neq 0$. Since $M$ is finitely generated, an application of Zorn's lemma shows that $M$ has a maximal proper submodule $N$. Then $M/N$ is simple, thus isomorphic to $R/\mathfrak{m}$ for some maximal left ideal $\mathfrak{m}$. Then $N = \mathfrak{m} M = M$, contradiction.

By the way, I don't know if the first form is true if $A$ is noncommutative. The theory of determinants is not really prosperous over noncommutative rings. Hints?

In many texts about algebraic geometry only the last form of Nakayama's lemma is needed. But the first one is stronger and is used in many results in commutative algebra.