Timeline for Irreducibility of the Sylvester resultant
Current License: CC BY-SA 3.0
7 events
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| Mar 17, 2018 at 5:22 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Mar 15, 2018 at 8:28 | comment | added | François Brunault | If $R=R_1 \times R_2$ for some integral domains $R_1,R_2$ then the resultant $R_{r,s}$ over $R$ is reducible for a rather silly reason: using the isomorphism $R[a_i,b_j] = R_1[a_i,b_j] \times R_2[a_i,b_j]$ we have $R_{r,s} = (R_{r,s},1) \cdot (1,R_{r,s})$ and none of the factors is a unit. Irreducibility is not well-behaved for such rings, one may try to replace it by some kind of local irreducibiliy (note $\operatorname{Spec} R = \operatorname{Spec} R_1 \sqcup \operatorname{Spec} R_2$) as in the answers of this question mathoverflow.net/questions/63899 | |
| Mar 14, 2018 at 9:40 | comment | added | Jason Starr | In many such question, e.g., the one linked below, the question is not about irreducibility of resultants of polynomials whose coefficients are each variables, but results of polynomials "derived" from some other "universal" object: mathoverflow.net/questions/156586/… | |
| Mar 14, 2018 at 8:27 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Mar 14, 2018 at 8:24 | comment | added | François Brunault | One has to exclude the cases $r=0$ and $s=0$ because $\operatorname{res}_{r,0}(f,b_0) = b_0^r$ and $\operatorname{res}_{0,s}(a_0,g) = a_0^s$. | |
| Mar 14, 2018 at 8:10 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Mar 14, 2018 at 7:53 | history | asked | Mariano Suárez-Álvarez | CC BY-SA 3.0 |