Ideal and Factor Ring Problem: Proving A=R When 1 is an Element of A

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Homework Statement



If A is an ideal of a ring R and 1 belongs to A, prove that A=R.

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The Attempt at a Solution



I said that r should an element of R. and since A is ideal to ring R and 1 is an element of A , then ar should be an element of A . 1 must be an element of a which is an element of ar which is an element of A. Therefore 1*ra=ar*1=> 1 is an element of R. Therefore,R=A
 
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1 must be an element of a doesn't mean anything... what the heck is a supposed to be anyway? I'm assuming it's an element of A maybe... at any rate, nothing can be an element of ar as ar is simply a member of the ring, and you have no reason to believe it's a set.

You realize an ideal is defined such that if a is in A, then for all x in R, x*a is in A?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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