Timeline for Ring where nonlinear polynomials have at least one root
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 4, 2018 at 20:41 | vote | accept | Alex Meiburg | ||
| Feb 4, 2018 at 20:26 | comment | added | Emil Jeřábek | I note that the condition that all monic polynomials have roots is preserved under quotients. Thus, e.g., if $R$ is the non-field ring from @tj_ 's comment, and $I$ is a non-prime ideal of $R$, then the ring $R/I$ also satisfies the condition, and it is not even an integral domain. | |
| Feb 4, 2018 at 19:40 | answer | added | tj_ | timeline score: 9 | |
| Feb 4, 2018 at 18:22 | comment | added | tj_ | Concerning the question: "Is there a ring in which all monic polynomials have a root, that is not a field ?" Yes, for example the subring of $\overline{\mathbb{Q}}$ of elements integral over $\mathbb{Z}$. (for a proof note that if an integral extension is a field, the base ring is also a field). | |
| Feb 4, 2018 at 13:46 | comment | added | YCor | @AlexMeiburg please edit the question (including title). abx made a comment of his remark because it's not worth an answer, so it can just ask the new one (with monic) and add a remark reflecting the previous comments. | |
| Feb 4, 2018 at 10:49 | answer | added | Aaron Meyerowitz | timeline score: 2 | |
| Feb 4, 2018 at 7:02 | comment | added | Alex Meiburg | @abx, you're totally right! I'm sorry, this question was silly then. The monic question is still interesting though. | |
| Feb 4, 2018 at 6:50 | comment | added | abx | Doesn't $\ ax^2-1=0\ $ give an inverse to each $a\neq 0$? | |
| Feb 4, 2018 at 6:08 | comment | added | R. van Dobben de Bruyn | Another interesting variant would be to restrict to monic polynomials, so that the inverse trick no longer works. | |
| Feb 4, 2018 at 5:12 | history | asked | Alex Meiburg | CC BY-SA 3.0 |