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Dec
24
comment Is Kaplansky's theorem for hereditary rings a characterization?
Actually I think this follows from the fact that $A = k[X]/(X^2)$ is a quotient of the PID $k[X]$. Any free module over $A$ is a module over $k[X]$, and the structure theorem for PIDs plus some basic algebra is enough to prove the result. The same should work for any quotient of $k[X]$.
Dec
21
awarded  Student
Dec
21
asked Is Kaplansky's theorem for hereditary rings a characterization?
May
15
awarded  Supporter
Dec
6
comment Which group is meant by $\mathbb Z/2 \times \mathbb Z/2.$
Yes, it is the same group. The $\times$ means that you take the product set with componentwise operations, and the product and direct sum of two abelian groups are the same.
Nov
5
awarded  Teacher
Nov
5
answered Can $\Bbb N$ be topologized to be a compact Hausdorff space?