Timeline for Why are polynomials easier to handle with than integers?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2010 at 16:10 | answer | added | Laurent Moret-Bailly | timeline score: 5 | |
| Oct 29, 2010 at 14:06 | vote | accept | zhaoliang | ||
| Oct 29, 2010 at 11:19 | answer | added | Bugs Bunny | timeline score: 4 | |
| Oct 29, 2010 at 9:04 | answer | added | Denis Serre | timeline score: 4 | |
| Oct 29, 2010 at 8:36 | comment | added | QuantumBrian | Long division with polynomials is much easier than long division of integers (do students even learn that any more?). That's also probably due to the "no carry-over" mentioned in Amri's comment. | |
| Oct 29, 2010 at 8:34 | history | edited | Jose Brox |
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| Oct 29, 2010 at 8:33 | answer | added | Jose Brox | timeline score: 0 | |
| Oct 29, 2010 at 6:20 | comment | added | KConrad | There is differentiation on polynomials. This is no simple operation like this on integers, although people do think about getting good analogues of it (e.g., Buium). For example, a polynomial in Q[x] is squarefree iff it is relatively prime to its derivative, and gcd(f,f') can be computed rather efficiently. There is no simple way to determine if an integer is squarefree without in some way factoring it. | |
| Oct 29, 2010 at 4:17 | comment | added | Amritanshu Prasad | The most elementary thing that always simplifies "function field number theory" in my experience the following: In a polynomial $a_0+a_1x+\cdots+a_nx^n$, we may think of $a_i$ as the "$i$th digit" in the expansion of the polynomial. When you add polynomials, there is no carry-over. This makes life much simpler. For example, $Z/p^2\to \Z/p$ is not a split epimorphism of abelian groups, but $Z/p[t]/(t^2)\to Z/p$ is split. | |
| Oct 29, 2010 at 3:59 | history | asked | zhaoliang | CC BY-SA 2.5 |