Rings, Ideals, and Modules - Key Proof

We present the detailed proof of Nakayama's Lemma, one of the most important technical tools in commutative algebra.

TheoremNakayama's Lemma (Full Statement)

Let $R$ be a commutative ring, $I$ an ideal contained in the Jacobson radical $J(R)$ , and $M$ a finitely generated $R$ -module. If $IM = M$ , then $M = 0$ .

ProofProof of Nakayama's Lemma

Suppose $M \neq 0$ and let $\{m_1, \ldots, m_n\}$ be a minimal generating set for $M$ . Since $IM = M$ , we have: $m_n = \sum_{i=1}^n a_i m_i$ for some $a_i \in I \subseteq J(R)$ .

Rearranging: $(1 - a_n)m_n = \sum_{i=1}^{n-1} a_i m_i$

Since $a_n \in J(R)$ , the element $1 - a_n$ is a unit in $R$ . Multiplying both sides by $(1 - a_n)^{-1}$ : $m_n = \sum_{i=1}^{n-1} (1-a_n)^{-1}a_i m_i$

This expresses $m_n$ as a combination of $m_1, \ldots, m_{n-1}$ , contradicting minimality of the generating set. Therefore $M = 0$ .

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Remark

The key insight is that elements of the Jacobson radical have the property that $1 - a$ is always a unit. This allows us to "divide" and reduce the size of minimal generating sets, ultimately showing no non-trivial module can be generated by elements in $IM$ if $I \subseteq J(R)$ .

ExampleCommon Application

If $M$ is finitely generated and $\mathfrak{m}$ is maximal, consider the quotient $M/\mathfrak{m}M$ , which is a vector space over the field $k = R/\mathfrak{m}$ . If $\{m_1, \ldots, m_n\}$ generates $M/\mathfrak{m}M$ as a $k$ -vector space, then by Nakayama's Lemma, $\{m_1, \ldots, m_n\}$ generates $M$ as an $R$ -module.

This "lifting" principle is ubiquitous in local ring theory and deformation theory.

ProofProof of Existence of Maximal Ideals

Let $R$ be a non-zero ring and $I \subsetneq R$ a proper ideal. Consider the set: $\mathcal{S} = \{J : J \text{ is an ideal, } I \subseteq J \subsetneq R\}$

This set is non-empty since $I \in \mathcal{S}$ . We partially order $\mathcal{S}$ by inclusion.

For any chain $\{J_\alpha\}$ in $\mathcal{S}$ , let $J = \bigcup_\alpha J_\alpha$ . We verify $J$ is an ideal:

If $a, b \in J$ , then $a \in J_\alpha$ and $b \in J_\beta$ for some $\alpha, \beta$ . Since the chain is totally ordered, one contains the other, so $a + b \in J$ .
If $a \in J$ and $r \in R$ , then $a \in J_\alpha$ for some $\alpha$ , so $ra \in J_\alpha \subseteq J$ .

Moreover, $J \neq R$ since $1 \notin J_\alpha$ for any $\alpha$ implies $1 \notin J$ .

By Zorn's Lemma, $\mathcal{S}$ has a maximal element $\mathfrak{m}$ , which is a maximal ideal containing $I$ .

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Remark

The use of Zorn's Lemma makes this proof non-constructive. In constructive mathematics, the existence of maximal ideals cannot be proven in general. However, for specific rings like $\mathbb{Z}$ or polynomial rings over fields, maximal ideals can be constructed explicitly.

These proofs illustrate fundamental techniques in commutative algebra: the Jacobson radical property for Nakayama, and Zorn's Lemma for existence results involving maximality conditions.