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April 29th, 2017, 06:35 PM  #1 
Newbie Joined: Apr 2017 From: Europe Posts: 16 Thanks: 1  Ring theory theorem
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? 
April 30th, 2017, 09:09 AM  #2 
Senior Member Joined: Aug 2012 Posts: 1,681 Thanks: 437 
I'd certainly start by showing the equivalence of a, b, and c and seeing which assumptions you use. ps  For example in the integers 2 is neither invertible nor a zero divisor, so finiteness is essential. pps  Is U(R) "a" set of invertible elements? Or THE set of invertible elements? Step one, write everything down with precision. Last edited by Maschke; April 30th, 2017 at 09:58 AM. 
April 30th, 2017, 11:55 AM  #3  
Newbie Joined: Apr 2017 From: Europe Posts: 16 Thanks: 1 
Well, I also thought about this: Quote:
Well, maybe none of them, but then my teacher wouldn't form such a problem... or would he? Is there really any difference between "a set" and "the set"? Sorry, I didn't know. I don't study mathematics in english, so it obviously can cause a little misunderstanding. The example with integers is a nice observation. Thank you. I kinda want to boast about my own example on commutative property, in which turns out that it's necessary. Let's take such a ring: $\displaystyle M(2, \mathbb{Z}_{2})$. It is finite (16 elements exactly), not commutative and has $\displaystyle 1$ and $\displaystyle 0$. I found the element that isn't a zero divisor or invertible. Only for 2 of them statements from the theorem happen. So... it turns out that finiteness and commutativity are essential. What do you think about $\displaystyle 1=0$ case? Can you say what happens if $\displaystyle R$ is a field?  

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