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Jul 24 at 15:35 history edited Ira Gessel CC BY-SA 4.0
Fixed a typo (spelling of Cayley).
Jun 8, 2015 at 17:26 comment added Ben Wieland If you restrict to rational coefficients, you get a $\mathbb Q$-monoid. If you work formally subject to $x^n=0$, this example still works, but now it is a finite-dimensional nilpotent Lie group, quite like André's example.
Jun 4, 2015 at 14:55 history answered Gerald Edgar CC BY-SA 3.0