Timeline for Algebra with a certain abelian group as the multiplicative group
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2020 at 13:13 | history | edited | Alex B. | CC BY-SA 4.0 |
Simplified the argument a bit following LSpice's comment.
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| Oct 8, 2020 at 13:10 | comment | added | Alex B. | @LSpice: we are assuming that $R^\times=\{1,u,\ldots,u^4\}$, so we must have $-u=u^i$ for some $i$. Since $u$ is a unit, this means that $-1=u^{i-1}$ for some $i$. But $-1$ has order dividing $2$, and we are in a cyclic group of order $5$, so $-1$ must have order $1$, i.e. be equal to $1$. I guess, I could have bypassed the whole $-u$ business. Let me edit. | |
| Oct 8, 2020 at 12:48 | comment | added | LSpice | Why does $-u \in R^\times$ force $1 = -1$? | |
| Oct 8, 2020 at 11:49 | history | answered | Alex B. | CC BY-SA 4.0 |