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In Hulek's Elementary Algebraic Geometry, Nakayama's lemma is stated as follows: Let $A \neq 0$ be a finite $B$-algebra. Then for all proper ideals $m$ of $B$, we have $mA \neq A$. (Here, $A$ and $B$ are both commutative rings with identity.) My question is: how is that related to the usual statements, e.g. Statement 1 in wikipedia: http://en.wikipedia.org/wiki/Nakayama_lemma#Statement? Specifically, does the wikipedia statement imply Hulek's, and if so, how?

Edit: While the statement is cited correctly, it is misleading, because Hulek's definition of what it means for $A$ to be a finite $B$-algebra is that (1) $B\subset A$ is a subring, and (2) there are $a_i \in A$ such that $A=Ba_1+\cdots+Ba_n$.

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1 Answer 1

up vote 7 down vote accepted

There is something wrong with the result that you cite. E.g. $A = \mathbb Z/3\mathbb Z$ is a finite algebra over $B = \mathbb Z/6\mathbb Z$ (via the natural surjection), and $m = 2B$ is a proper ideal of $B$. However, $m A = 2 A = A$.

What is true is the following: if A is a non-zero finite $B$-algebra, then $A_{\mathfrak m}$ is non-zero for at least one maximal ideal of $B$ (a general property of non-zero modules over a ring), therefore $A/\mathfrak m A$ is non-zero for at least one maximal ideal of $B$ (Nakayama's lemma, in its usual formulation, applied to the finitely generated $B_{\mathfrak m}$-module $A_{\mathfrak m}$), and so $A/m A \neq 0$ for all ideals $m \subset \mathfrak m$.

Perhaps you meant to ask that $B$ be local, so that all proper ideals $m$ of $B$ lie in just a single maximal ideal $\mathfrak m$?

[Added in light of OP's comment below:] Suppose that $B \to A$ is an embedding (as the author cited by the OP apparently does). Then for any maximal ideal $B_{\mathfrak m} \to A_{\mathfrak m}$ is again an embedding, and thus $A_{\mathfrak m}$ is non-zero. Thus by (usual) Nakayama the quotient $A/\mathfrak m A = A_{\mathfrak m }/\mathfrak m A_{\mathfrak m}$ is non-zero, and hence for any proper ideal $m$ of $B$, we have $A/m A$ is non-zero (because $m$ is contained in some maximal ideal $\mathfrak m$).

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Oh, I'm sorry! Earlier on, his definition of being a finite $B$-algebra of $A$ includes the stipulation that $B$ is a subring of $A$. On the other hand, everything, including the statement of Nakayama's lemma, comes before any mention of local rings, etc. –  Cantor's Paradise May 13 '11 at 3:54
    
@Cantor: Thanks for the clarification. I've added a proof of the claim under this additional assumption. Regards, –  Matt E May 13 '11 at 3:58

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