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user91132
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I usually find the statement of Nakayama's Lemma easy to remember because of its proof, which is really nothing more than the definition of the Jacobson radical plus the existence of maximal left ideals in a ring.

Every non-zero finitely generated module $M$ admits a non-zero cyclic quotient module, which in turn (by a Zornication) admits a non-zero simple quotient module. So we can find a submodule $N$ of $M$ with $M/N$ simple. But now $J . (M/N) = 0$ since the Jacobson radical $J$ of the ring kills every simple module, so $JM \leq N < M$ which says that $JM$ is a proper submodule of $M$.

Note that this general form of the Lemma doesn't need any complicated determinant-type arguments. In the commutative case, other forms of the Lemma can easily be obtained from this general form "$JM < M$" by considering localisation.

I usually find the statement of Nakayama's Lemma easy to remember because of its proof, which is really nothing more than the definition of the Jacobson radical plus the existence of maximal left ideals in a ring.

Every non-zero finitely generated module $M$ admits a non-zero cyclic quotient module, which in turn (by a Zornication) admits a non-zero simple module. So we can find a submodule $N$ of $M$ with $M/N$ simple. But now $J . (M/N) = 0$ since the Jacobson radical $J$ of the ring kills every simple module, so $JM \leq N < M$ which says that $JM$ is a proper submodule of $M$.

Note that this general form of the Lemma doesn't need any complicated determinant-type arguments. In the commutative case, other forms of the Lemma can easily be obtained from this general form "$JM < M$" by considering localisation.

I usually find the statement of Nakayama's Lemma easy to remember because of its proof, which is really nothing more than the definition of the Jacobson radical plus the existence of maximal left ideals in a ring.

Every non-zero finitely generated module $M$ admits a non-zero cyclic quotient module, which in turn (by a Zornication) admits a non-zero simple quotient module. So we can find a submodule $N$ of $M$ with $M/N$ simple. But now $J . (M/N) = 0$ since the Jacobson radical $J$ of the ring kills every simple module, so $JM \leq N < M$ which says that $JM$ is a proper submodule of $M$.

Note that this general form of the Lemma doesn't need any complicated determinant-type arguments. In the commutative case, other forms of the Lemma can easily be obtained from this general form "$JM < M$" by considering localisation.

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user91132
  • 3.7k
  • 1
  • 24
  • 26

I usually find the statement of Nakayama's Lemma easy to remember because of its proof, which is really nothing more than the definition of the Jacobson radical plus the existence of maximal left ideals in a ring.

Every non-zero finitely generated module $M$ admits a non-zero cyclic quotient module, which in turn (by a Zornication) admits a non-zero simple module. So we can find a submodule $N$ of $M$ with $M/N$ simple. But now $J . (M/N) = 0$ since the Jacobson radical $J$ of the ring kills every simple module, so $JM \leq N < M$ which says that $JM$ is a proper submodule of $M$.

Note that this general form of the Lemma doesn't need any complicated determinant-type arguments. In the commutative case, other forms of the Lemma can easily be obtained from this general form "$JM < M$" by considering localisation.