Timeline for A sufficient condition for a set of primes to be the set of reducibility of an integer polynomial
Current License: CC BY-SA 4.0
18 events
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| Feb 24, 2020 at 17:51 | history | edited | user6976 | CC BY-SA 4.0 |
added 82 characters in body
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| Feb 17, 2020 at 3:21 | history | edited | user6976 | CC BY-SA 4.0 |
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| Feb 17, 2020 at 2:49 | comment | added | user6976 | The terminology in "serious" textbooks on ring theory is beyond good and evil. For example "noncommutative rings" in these books means rings that are not necessarily commutative. So commutative rings are non-commutative. The same happens to associative/non-associative. I understand why 1 is not a prime number but calling 1 composite and polynomial 1 reducible is extremely bad. | |
| Feb 17, 2020 at 2:41 | history | edited | user6976 | CC BY-SA 4.0 |
added a link to a concrete answer
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| Feb 16, 2020 at 3:58 | comment | added | YCor | I haven't yet seen any serious textbook choosing the convention that a nonzero constant polynomial is irreducible (or that $1$ is prime). | |
| Feb 9, 2020 at 4:25 | comment | added | user6976 | The OP has to decide what is an irreducible polynomial. But since he disappeared, I prefer the more convenient definition. | |
| Feb 9, 2020 at 3:09 | comment | added | Robert Israel | Depends. Many authors disagree, and require an irreducible polynomial to be non-constant (or, over a ring, not a unit). See Wikipedia for example. | |
| Feb 7, 2020 at 22:38 | comment | added | user6976 | 1 is an irreducible polynomial by definition, as is any constant. | |
| Feb 7, 2020 at 22:29 | comment | added | Robert Israel | $4 x^2 + 1$ is a unit mod $2$, so I don't know if that counts as irreducible. | |
| Feb 7, 2020 at 22:21 | comment | added | user6976 | @RobertIsrael: $4x^2+1$ then? | |
| Feb 7, 2020 at 22:01 | comment | added | Robert Israel | $x^2+1$ is reducible mod $2$ as well as the primes $4k+1$. | |
| Feb 7, 2020 at 21:33 | comment | added | user6976 | In fact if $S$ is just the set of all primes of the form $4k+3$ then what is the polynomial? For $4k+1$ it is $x^2+1$. | |
| Feb 7, 2020 at 20:57 | comment | added | user6976 | @RobertIsrael: The OP does not know much about the question. It could be that there is no polynomial reducble mod $p$ iff $p$ is 1 mod 4 except 5 and 13. | |
| Feb 7, 2020 at 20:40 | comment | added | Robert Israel | Answer: no. There must be an algorithm for deciding membership in $S$ of time complexity no worse than that for determining whether the given polynomial factors mod $p$. | |
| Feb 7, 2020 at 20:08 | comment | added | Robert Israel | To make things somewhat less trivial, what if we restrict to $S$ that is decidable, i.e. such that there exists an algorithm for determining whether or not a given integer is in $S$? | |
| Feb 7, 2020 at 18:46 | comment | added | YCor | "Not always". The OP's question is ambiguous on whether it means "does it exist for every $S$" or "does it exist for some $S$" (without thinking I actually understood the second meaning and the first meaning doesn't make a serious question according to your remark). | |
| Feb 7, 2020 at 18:41 | history | edited | user6976 | CC BY-SA 4.0 |
misprints fixed
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| Feb 7, 2020 at 6:16 | history | answered | user6976 | CC BY-SA 4.0 |