If a and e are in ring R (e not 1), then is it true that ae=a implies e^2=e?

Question

Prove if true.

-R is an arbitrary ring (not the reals)

-a is an arbitrary element of R and hence 1*e=1 must be true so the case where a=0 doesn't need to be considered because it alows for any e.

-I don't want to assume a has an inverse so ae=a does not mean e=1.

Answer 1

ok

is this for every a? i understand that this is the case.
so in particular take a=e

then ee=e, e^2=e

Answer 2

If I'm reading the question correctly, then I don't believe that ae=a implies e^2=e. For example, a could be zero. Then e could be anything, and e^2 would not necessarily have to equal e. Also, if a is nonzero, then you can divide both halves of the equation by a, leaving you with e=1, which contradicts a presupposition in this problem. Therefore if e is not 1, then a must be zero, and e can be anything (besides 1), which means that it is impossible to tell if e^2=e. (Also, just to ramble a bit more, the only other possibility for e^2=e is e=0, but that isn't necessary for the problem.)

I hope I read the problem right, and that I've helped at least somewhat. B'bye now!

Answer 3

This is not true. Let R be the integers modulo 10 and let a=5 and e=3. then ae=5*3=5 (remember 15 is congruent to 5 (modulo 10)),
but e^2=3^2=9, which is not congruent to 3 (modulo 10).

Answer 4

the identity in Real numbers is 1
so yes its true that if ae = e => a^2 = e => e=1