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r_{1} - 9\cdot r_{2}$ &$\boxed{1 = \mathbf{10}\cdot a + \mathbf{-49}\cd Für jedes $m\in\intgr$ mit $p\ndivides m$ ist die Abbildung ${M_{m}:\intgr/p\intgr\to\intgr/p\intgr}$, - die vermöge $M_{m}([x])=[m]\cdot[x](=[mx])$ + die vermöge $M_{m}([x])=[m]\cdot[x](=[mx])$ definiert wird, wohldefiniert und injektiv. Insbesondere existert ein $[x]\in\intgr/p\intgr$, so dass $[m]\cdot [x]=[1]$. @@ -5819,10 +5819,10 @@ $r_{3} = r_{1} - 9\cdot r_{2}$ &$\boxed{1 = \mathbf{10}\cdot a + \mathbf{-49}\cd \uwave{{\bfseries Injektivität:}}\\ Seien $[x],[x']\in\intgr/p\intgr$. - Angenommen $M_{m}([x])=M_{m}([x'])$.\\ + Angenommen $M_{m}([x])=M_{m}([x'])$. \textbf{Zu zeigen:} $[x]=[x']$.\\ Aus $M_{m}([x])=M_{m}([x'])$ - folgt $[mx]=[mx']$ per Konstruktion von der Abbildung $M_{m}$.\\ + folgt $[mx]=[mx']$ per Konstruktion der Abbildung $M_{m}$.\\ Per Definition der Äquivalenzklassen gilt somit $mx\equiv mx'$ modulo $p$.\\ Daraus folgt ${p\divides (mx-mx')}$, @@ -5845,11 +5845,9 @@ $r_{3} = r_{1} - 9\cdot r_{2}$ &$\boxed{1 = \mathbf{10}\cdot a + \mathbf{-49}\cd \textbf{Bemerkung.} Die letzte Aussage in diesem Satz gilt auch allgemeiner: -Sind $n,m\in\intgr$ teilerfremd, dann ist $[m]$ innerhalb $\intgr/n\intgr$ -invertierbar. -Aber wenn $n$ nicht prim ist, können wir das o.\,s. Argument nicht verwenden, -da das Zwischenresultat der Injektivität nicht mehr gilt. -Stattdessen müssen wir schon das Lemma von B\'ezout anwenden. +Sind $n,m\in\intgr$ teilerfremd, dann ist $[m]$ innerhalb $\intgr/n\intgr$ invertierbar. +Falls $n$ nicht prim ist, muss man sich allerdings bei der Injektivitätsargumentation mehr bemühen. +Einfacher ist also natürlich die Anwendung von dem Lemma von B\'ezout. \setcounternach{part}{3} \part{Quizzes}