My Math Forum Ring theory theorem

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 April 29th, 2017, 05:35 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, 08:09 AM #2 Senior Member   Joined: Aug 2012 Posts: 2,355 Thanks: 737 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 08:58 AM.
April 30th, 2017, 10:55 AM   #3
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Quote:
 I'd certainly start by showing the equivalence of a, b, and c and seeing which assumptions you use.
, but proof presented on a lecture actually uses every assumption. And yet at the same time I need to say which ones can be omitted. That is the problem. :P
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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