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As the title says, I'm looking for an induction example, where the inductive step is rather simple to prove, but it's hard to find a base case for which the theorem holds.

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marked as duplicate by Micah, Parcly Taxel, Alexander Konovalov, suomynonA, Alexis Olson Oct 16 '16 at 17:12

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  • $\begingroup$ @Micah Thank you, that's exactly what I was looking for (: $\endgroup$ – user326377 Oct 16 '16 at 14:13
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This is an example not given in the link and it may count: every finitely generated ideal $I$ in a Von-Neumann regular ring $R$ is generated by an idempotent. The usual proof for induction uses the number of generators of $I$, let's say $n$. The base case $n=1$ is easy, and most of the work is done in the case $n=2$. It isn't hard but is kind of operative and in contrast, the inductive step is almost trivial. That's why in most of the books where this result is stated the authors just prove the case $n=2$ and leave the easy step induction for the readers.

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    $\begingroup$ Tank you, nice example! $\endgroup$ – user326377 Oct 16 '16 at 15:17