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Universität/Hochschule Isomorphismen heben
xiao_shi_tou_
Senior Letzter Besuch: in der letzten Woche
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Hallo.

Kann man Isomorphismen mit Hensel's Lemma heben?
Etwa so:

Sei \(R\) ein (kommutativer) Ring und sei \(\varphi\colon A\to B\) ein \(R\)-Algebren Homomorphismus.

Seien \(A\) und \(B\) vollständig bezüglich der \(I\)-adischen Vervollständigung.

Angenommen der von \(\varphi\) induzierte Morphismus \(A/IA\to B/IB \) ist ein Isomorphismus. Ist dann schon \(\varphi\) ein Isomorphismus?

Vielleicht braucht man noch zusätzliche Bedingungen an \(A\).

Grüße.




-----------------
”己所不欲,勿施于人“(Konfuzius)
PS: Falls ich plötzlich aufhöre in einem Thread zu antworten, dann kann es sein, dass ich es vergessen habe. Ihr könnt mir in diesem Fall eine Private Nachricht schicken.
\(\endgroup\)


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Dune
Senior Letzter Besuch: im letzten Monat
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Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag  Beitrag No.1, eingetragen 2018-11-24


Moin,

betrachte die triviale $K[X]$-Algebra K (d.h. $X \cdot K = 0$) sowie die $K[X]$-Algebra $K[[X]]$ der formalen Potenzreihen über K. Beide sind X-adisch vollständig. Der stetige Homomorphismus $K[[X]] \to K$ mit $X \mapsto 0$ sollte nun ein Gegenbeispiel sein.

VG Dune



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xiao_shi_tou_
Senior Letzter Besuch: in der letzten Woche
Dabei seit: 12.08.2014
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Aus: Bonn
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag  Beitrag No.2, vom Themenstarter, eingetragen 2018-11-24

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Hallo Dune.
Danke für die Antwort.

Die Algebrastruktur ist gegeben durch die natürliche Projektion:
\(K[X]\twoheadrightarrow K[X]/(X)\cong K\), das man gibt identifiziert \(K\) mit der \(K[X]\)-Algebra \(K[X]/(X)\) über den Isomorphismus \(K[X]/(X)\cong K\). Demnach entspricht \((X)K[X]=(X)/(X)\) dem Nullideal  \((0)\) in \(K\).

Nun haben wir \(\underset{n\geq 1}{\varprojlim K/{0^n}}\cong \underset{n\geq 1}{\prod}K=K^\mathbb{N}\colon (x_i)=(x_j)\forall i,j\cong K\), daher ist \(K\) in der Tat \((X)\) vollständig.

Demnach hast du völlig Recht mit deinem Gegenbeispiel.

Dennoch ist dieses Gegenbeispiel etwas "unbefriedigend", da die Filtration recht trivial ist \(K\supset 0=0=0=0=0\cdots\). Interessant wäre ein Gegenbeispiel mit einer "eigentlichen" Filtration \(A\supset F^1A\supset F^2A\supset \cdots\) die man sich wirlich als ineinander verschachtelte Umgebungen vorstellen kann die strikt ineinander enthalten sind.

Desweiteren würde mich auch interessieren, ob man bestimmte (sinnvolle, also nicht etwas wie \(R=0\)) Bedingungen an \(R\) angeben kann, sodass die Aussage wahr ist.

Viele Grüße
\(\endgroup\)


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Du musst schon fordern, dass für hinreichend große $n$ der Pfeil $A/I^n A \to B/I^n B$ ein Isomorphismus ist; weniger zu fordern wird in der Allgemeinheit nicht ausreichen.



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