Timeline for Uncountable cardinals and Prufer $p$-groups
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 15, 2016 at 3:36 | history | edited | YCor | CC BY-SA 3.0 |
clarified introduction (answer is no, not yes).
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| Apr 21, 2015 at 10:18 | comment | added | YCor | PS I mean $\mathrm{Aut}_{\mathbf{F}_p}(A)$ instead of $\mathrm{Aut}_{\mathbf{Z}}(A)$ in my third comment, 1st line, and $\mathbf{F}_p$ is the field with $p$ elements. | |
| Apr 21, 2015 at 9:56 | comment | added | YCor | The decomposition lemma I used is the following one (of undergraduate level): let $K$ be a field, $M$ a vector space over $K$ (no dimension restriction). Let $u$ be a self-operator on $M$. Let $P\in K[t]$ be a polynomial such that $P(u)=0$. Assume that $P=\prod P_i$ with the $P_i$ pairwise prime to each other. Then $M=\bigoplus_i\mathrm{Ker}(P_i(u))$. This can usually be used when $P_i$ is a power of an irreducible polynomial; in the precise case here, the $P_i$ can themselves be chosen irreducible because $t^{q^n}-1$ has simple roots in any characteristic $\neq q$. | |
| Apr 21, 2015 at 9:52 | comment | added | YCor | The action of $C_{q^\infty}$ on $A$ is given by a group homomorphism $\phi:A\to\mathrm{Aut}_{\mathbf{Z}}(A)$, where $\mathrm{Aut}_{\mathbf{F}_p}$ means automorphisms as $\mathbf{F}_p$-modules and is contained in the group of units of the ring $\mathrm{End}_{\mathbf{F}_p}$. For $u\in A$ and $P\in\mathbf{F}_p[t]$, by $\mathrm{Ker}(P(u))$ I mean $\mathrm{Ker}(P(\phi(u)))$, where $P(\phi(u))$ is computed in the $\mathbf{F}_p$-algebra $\mathrm{End}_{\mathbf{F}_p}$. | |
| Apr 21, 2015 at 8:43 | comment | added | W4cc0 | I was re-reading the answer and I find that it is unclear to me how to pass from the fact that $x_n$ is an operator on $A$ vanished by... to the decomposition. What you meant by $Ker(P(x_n))$? | |
| Apr 21, 2015 at 6:44 | comment | added | YCor | At the time I first posted, I indeed only found a proof for $p=q$... | |
| Apr 21, 2015 at 4:58 | comment | added | W4cc0 | I thought that you don't know about the case $p\neq q$. Anyway still Thanks! | |
| Apr 20, 2015 at 21:36 | history | edited | YCor | CC BY-SA 3.0 |
added case $p\neq q$
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| Apr 20, 2015 at 17:14 | vote | accept | W4cc0 | ||
| Apr 20, 2015 at 17:14 | comment | added | W4cc0 | First of all thanks. When $p\neq q$ nothing is known? | |
| Apr 20, 2015 at 14:59 | history | answered | YCor | CC BY-SA 3.0 |