Let R be a ring with a unit element. Using its elements, we define a ring T by defining operations ! and @ on T such that a ! b = a+b+1 and a@b = ab+a+b where the addition and multiplication on right hand side of these relations are those of R. (Note that the ring T forms a group with the operation !). Prove that the rings R and T are isomorphic.
I am thinking that a good place to start is to figure out what are the neutral elements with respect to the operations ! and @ in T: For what element x in the ring R will x ! a = a for all a? For what element y in R will y@a = a for all a in R? (Notice that both ! and @ are commutative, so it suffices to check that x and y work from the left).
Then you could find inverses with respect to !, i.e. for a given a, what element b satisfied that a ! b = x, where x is the neutral element from before?
Hi,
Thank you for your post. I have found out the identity elements for ! and @ are -1 and 0 respectively. But I do not know how finding the identity elements or inverses would help me establish an isomorphism from R to T. The method I am trying to use is to establish the result is to establish a homomorphism from R onto T and then prove the kernel of this homomorphism is -1 (the identity element of !). Please correct my approach if it is wrong.
I agree with the identity elements you found. Though you probably meant the kernel of the homomorphism to be {0} (remember the kernel is a subset of R), and not {-1}. Let me tell you my approach.
So given those identity elements, you know that 0 in R must map to -1 in T, and that 1 in R must map to 0 in T. In fact, this suggests that we try mapping any x in R to x-1 in T.
Does this give a ring homomorphism? Call the map f. Then
f(x+y) = x+y-1,
while
f(x) ! f(y) = (x-1)!(y-1) = x-1+y-1+1 = x+y-1,
the same thing as above.
What about f(x*y) = xy-1. Is this equal to f(x)&f(y)?
Finally you just need to check that f is a bijection.
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