Timeline for The existence of non-trivial homomorphisms $\prod_{n=1}^{\infty}\mathbb{Z}/\bigoplus_{n=1}^{\infty}\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2014 at 7:17 | history | edited | Adam Przeździecki | CC BY-SA 3.0 |
a related open problem is added
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| Apr 28, 2014 at 0:08 | vote | accept | Jeremy Brazas | ||
| Apr 27, 2014 at 23:56 | comment | added | Mark Wildon | @JeremyBrazas Write $N_k(a)$, $N_\ell(b)$ and $N_{k+\ell}(a+b)$ for the sets. Clearly $N_{k+\ell}(a+b)$ contains $N_k(a) \cap N_\ell(b)$ and this intersection is in $F$. Since $F$ is a filter $N_{k+\ell}(a+b) \in F$. | |
| Apr 27, 2014 at 23:17 | comment | added | Jeremy Brazas | Perhaps a naive question: if you also send $(b_n)$ to $\ell$ where $N_{\ell}\in F$, how to you know that $N_{(k+\ell) \text{mod}p }=\{n|a_n+b_n=(k+\ell) \text{mod}p\}\in F$? | |
| Apr 27, 2014 at 21:25 | comment | added | YCor | OK. In a sense, you retrieve this way the ultraproduct, since the homomorphism factors through the ultrapower of $\mathbf{Z}$ with respect to $F$. So basically it's a restatement of the ultrapower argument. (The ultrapower has a unique nonzero homomorphism onto $\mathbf{Z}/p\mathbf{Z}$ up to scalar multiplication in $\mathbf{Z}/p\mathbf{Z}^*$, [and unique if prescribed to have the value 1 on the constant sequence 1], so SJR's answer provides the same homomorphism.) | |
| Apr 27, 2014 at 20:05 | comment | added | Adam Przeździecki | @Yves - thanks, I missed the "ultra" part. | |
| Apr 27, 2014 at 20:04 | history | edited | Adam Przeździecki | CC BY-SA 3.0 |
added 5 characters in body
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| Apr 27, 2014 at 18:30 | comment | added | YCor | I'm not sure what you mean: if $F$ is not an ultrafilter there is some $(a_n)$ such that no $N_k$ belongs to $F$. | |
| Apr 27, 2014 at 17:16 | history | answered | Adam Przeździecki | CC BY-SA 3.0 |