Timeline for when tensor complex resolves S/I+J?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| S Feb 21 at 17:47 | history | suggested | user5826 | CC BY-SA 4.0 |
corrected typeset errors
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| Feb 21 at 17:02 | review | Suggested edits | |||
| S Feb 21 at 17:47 | |||||
| Jan 12, 2012 at 1:27 | answer | added | Mariano Suárez-Álvarez | timeline score: 2 | |
| Jan 11, 2012 at 20:25 | vote | accept | today user | ||
| Jan 11, 2012 at 11:56 | answer | added | Daniel Erman | timeline score: 7 | |
| Jan 11, 2012 at 7:08 | comment | added | today user | @Mariano, I didn't get your argument for minimality and exactness. would you please explain the idea of your proof? | |
| Jan 11, 2012 at 6:25 | comment | added | Mariano Suárez-Álvarez | Ah. The thing is, when I say linear I mean the module in the $i$th position is generated by elements of degree $i$, which is the definition I am used to. If you only mean the degrees of the generators of the modules in the resolution grow by one, then of course it is not true that linear otimes linear is linear. But in any case, the argument I sketcked above (that is, the Künneth formula) shows that the tensor product of your resolutions is exact, and minimality follows from the definition of its differential and the minimality of the two resolutions you started with. | |
| Jan 11, 2012 at 6:04 | comment | added | today user | @Yemon No, I just gave a counterexample for the statement mentioned by Mariano.. But the tensor product still gives the minimal resolution for that example... | |
| Jan 11, 2012 at 5:14 | comment | added | Yemon Choi | It sounds to me as if you are saying that you have answered your own question. Is that right? | |
| Jan 11, 2012 at 4:42 | comment | added | today user | @Yeman, $S=k[x_1,\ldots,x_n,y_1,\ldots,y_m]$. I mean is the $F_\cdot\otimes G_\cdot$ is the minimal free resolutiuon of $S/I+J$. The above example shows two ideals with linear resolution such that $I+J$ does not have linear resolution! | |
| Jan 11, 2012 at 4:38 | comment | added | Yemon Choi | If $I$ and $J$ are ideals in DIFFERENT rings then, as Mariano says, what is the ambient ring $S$ into which you are embedding $I$ and $J$? Also, what do you mean by "resolving the resolution" of $I+J$? | |
| Jan 11, 2012 at 4:25 | comment | added | today user | of course it's not true Mariano, here is an example: \\ $I1:=Ideal(x[1]x[2],x[2]x[3],x[1]x[3])$ \\ $I2:=Ideal(y[1]y[2],y[2]y[3],y[1]y[3])$ Then we have:\\ $Res(S/I1): 0\rightarrow S(−3)^2\rightarrow S(−2)^3\rightarrow S$ \\ $Res(S/I2):\ 0\rightarrow S(−3)^2\rightarrow S(−2)^3\rightarrow S $\\ $Res(S/I1+I2): 0\rightarrow S(−6)^4\rightarrow S(−5)^12\rightarrow S(−3)^4+S(−4)^9\rightarrow S(−2)^6\rightarrow S$ | |
| Jan 11, 2012 at 4:03 | comment | added | Mariano Suárez-Álvarez | In any case, the tensor product of linear complexes is linear; if they are both exact and composed of free modules, their tensor product is exact. Since it is both exact and linear, it is minimal. But you should look for at few examples to see what exactly it resolves. | |
| Jan 11, 2012 at 3:57 | comment | added | Mariano Suárez-Álvarez | What is $S{}{}$? | |
| Jan 11, 2012 at 3:50 | history | asked | today user | CC BY-SA 3.0 |