My Math Forum One theorem for rings

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 April 29th, 2017, 05:02 PM #1 Newbie   Joined: Apr 2017 From: Europe Posts: 16 Thanks: 1 One theorem for rings Let $\displaystyle R$ be finite, commutative ring with $\displaystyle 1=/=0$. Statements below are equivalent: a) $\displaystyle a \in U(R)$, where $\displaystyle U(R)$ is a set of invertible elements of $\displaystyle R$ b) $\displaystyle a$ is not a zero divisor c) $\displaystyle a^m=1$, where $\displaystyle m$ is the number of elements that are not zero divisors in $\displaystyle R$ My task is to discuss this theorem in terms of assumptions that are made. To be more precisive - I need to tell which of them are essential for the theorem to work and why + which ones can be omitted and then yet at the same time theorem will work the same way. In this case I also have to give reasons. I came up with one idea. Assumption about $\displaystyle 1=/=0$ is not necessary, because if $\displaystyle 1=0$, then we deal with zero ring, which is finite and commutative, and all statements mentioned in theorem actually happen, because $\displaystyle 0$ is invertible, $\displaystyle 0$ is not a zero divisor, futhermore - $\displaystyle 0^0=1$, but we need to remember that $\displaystyle 1=0$. I also think that assumption about being finite can be loosened up, but I have no really satisfying explanation for that. I have no thoughts on commutative property, but something tells me it's the one that is the most essential in this. I haven't seen using this property in a proof, so I can't really tell. Any ideas?

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