Timeline for Polynomial roots in the ring extension
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Aug 25, 2022 at 8:59 | history | edited | darij grinberg | CC BY-SA 4.0 |
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| Jul 29, 2017 at 20:05 | vote | accept | Mikhail Goltvanitsa | ||
| Jul 29, 2017 at 20:05 | vote | accept | Mikhail Goltvanitsa | ||
| Jul 29, 2017 at 20:05 | |||||
| Jul 29, 2017 at 20:03 | vote | accept | Mikhail Goltvanitsa | ||
| Jul 29, 2017 at 20:05 | |||||
| Jul 29, 2017 at 20:03 | vote | accept | Mikhail Goltvanitsa | ||
| Jul 29, 2017 at 20:03 | |||||
| Jul 29, 2017 at 19:35 | answer | added | Mikhail Goltvanitsa | timeline score: 4 | |
| May 4, 2016 at 19:22 | vote | accept | Mikhail Goltvanitsa | ||
| Jul 29, 2017 at 20:03 | |||||
| Apr 30, 2016 at 16:42 | history | edited | Mikhail Goltvanitsa | CC BY-SA 3.0 |
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| Apr 30, 2016 at 16:37 | review | Close votes | |||
| May 1, 2016 at 7:59 | |||||
| Apr 30, 2016 at 16:34 | comment | added | YCor | I know, but this was addressed to other readers. | |
| Apr 30, 2016 at 16:31 | comment | added | Mikhail Goltvanitsa | @YCor thank you. But I think that we understand each other ) | |
| Apr 30, 2016 at 16:29 | comment | added | YCor | Note: "unitary polynomial" means "monic polynomial". | |
| Apr 30, 2016 at 16:25 | history | edited | Mikhail Goltvanitsa | CC BY-SA 3.0 |
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| Apr 30, 2016 at 16:23 | comment | added | Mikhail Goltvanitsa | I am interesting only in the unitary polynomials. That is $f(x) = x^m+\sum\limits_{j= 0}^{m-1}f_jx^j$ | |
| Apr 30, 2016 at 16:22 | comment | added | Uri Bader | My comment was very stupid. I came back to erase it, but figure its too late... | |
| Apr 30, 2016 at 16:12 | comment | added | Todd Leason | In general, even in the commutative case, polynomials don't have roots in an extension. Example: $f=1 + 2x\in (\mathbb{Z}/4)[x]$. The problem is that $R[x]/(f)$ is in general no extension of $R$. | |
| Apr 30, 2016 at 16:11 | comment | added | Pace Nielsen | @user89334 The example in my answer below does just that. The ideal generated by $1+ax$ contains $1=(1-ax)(1+ax)$. | |
| Apr 30, 2016 at 16:10 | answer | added | Pace Nielsen | timeline score: 7 | |
| Apr 30, 2016 at 15:56 | comment | added | Uri Bader | How could $R[x]f(x)R[x]$ be equal to $R[x]$ when the degree of $f$ is not 0? | |
| Apr 30, 2016 at 15:45 | history | edited | YCor |
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| Apr 30, 2016 at 15:43 | history | asked | Mikhail Goltvanitsa | CC BY-SA 3.0 |