Timeline for Polynomial roots in the ring extension
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 29, 2017 at 20:05 | vote | accept | Mikhail Goltvanitsa | ||
| Jul 29, 2017 at 20:05 | |||||
| May 4, 2016 at 19:22 | vote | accept | Mikhail Goltvanitsa | ||
| Jul 29, 2017 at 20:03 | |||||
| May 4, 2016 at 14:35 | comment | added | Pace Nielsen | @MikhailGoltvanitsa The elements of $R$ don't commute with $t$, because this is the coproduct of rings. For instance the ring $\mathbb{Z}[s]\coprod_{\mathbb{Z}}\mathbb{Z}[t] = \mathbb{Z}\langle s,t\rangle$ has $s,t$ non-commuting. I can't explain the intricacies of the coproduct of rings in this comment, so I recommend finding a good book on the subject. | |
| May 4, 2016 at 14:12 | comment | added | Mikhail Goltvanitsa | @Pace Nielsen, Thank you. Can you explain more circumstantially what is the construction of $R\coprod_{\mathbb{Z}}\mathbb{Z}[t]/(f(t))$. Also I don't understand, why $t$ don't commute with elements of $R$. | |
| May 2, 2016 at 13:02 | history | edited | Pace Nielsen | CC BY-SA 3.0 |
added 1446 characters in body
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| Apr 30, 2016 at 16:24 | comment | added | Mikhail Goltvanitsa | @Pace Nielsen, Thank you very much, but I'm interesting only in the unitary polynomials. I will clarify my question. | |
| Apr 30, 2016 at 16:15 | comment | added | Todd Leason | I found the simpler example in the commutative case and beat you for 2 minutes. -:) | |
| Apr 30, 2016 at 16:10 | history | answered | Pace Nielsen | CC BY-SA 3.0 |