Take A as an R-module, and B is its submodule. We can get a submodule $B_1$ of A generated by B and an element $b_1 \in A\backslash$B. By the similar way we can get a submodule $B_2$ of A generated by $B_1$ and an element $b_2 \in A\backslash$$B_1$. Keep doing like this, we can get a sequence B$\subset B_1\subset B_2\subset B_3\subset B_4...$. These sequence is ordered by inclusion and up bounded by A. By Zorn's Lemma, we can get a $B_n$ in the sequence which is the maximal element in this sequence,thus is also equal to A. In this sense, any module A can be generated by a submodule B and finite elements in A. But if A is not finite generated and B is not finite generated, it seems to be a contradiction. What I missed in my proof? Thank you!
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$\begingroup$ $A$ is not necessarily in your ordered sequence, so Zorn's lemma doesn't apply. $\endgroup$ – user64687 Apr 1 '15 at 20:31
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Zorn's lemma says that given any partially ordered set $S$ such that every chain in $S$ is bounded above in $S$, $S$ has a maximal element. Your $S$ is $\{B_i\}$, and there's no reason the chain $S$ should have an upper bound in $S$: claiming $A$ as such a bound is claiming $A\in S$, which is the conclusion you're trying to prove.
answered Apr 1 '15 at 20:31
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$\begingroup$ Thank you, could you give me an example or a quick answer about why A is not necessary in the sequence? thank you! $\endgroup$ – 6666 Apr 1 '15 at 20:47
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$\begingroup$ Take any module that's not finitely generated and $B=0$. For instance, $A=R$ for $R$ a ring of polynomials in infinitely many variables. $\endgroup$ – Kevin Arlin Apr 2 '15 at 0:29
