>>10800013

A ring homomorphism is just a function f(x) that takes elements of one ring and gives out an element of another ring, while also satisfying:

>f(x + y) = f(x) + f(y)

>f(x * y) = f(x) * f(y)

>f(1) = 1 (where 1 is just the multiplicative identity)

For (a), another anon gave a good explanation - it's incredibly simple, but it's designed to test if you actually understand what you're supposed to be doing. It's like asking what 1+2 is. It's a simple question, but doing it shows that you understand that it's addition and not multiplication.

A ring isomorphism g(x) is a ring homomorphism that is ALSO bijective, i.e.:

>injective or one-to-one: g(x) = g(y) ==> x=y

>surjective or onto: for any element r of the domain R (the second ring that the function gives out values for), you have some element x in S such that g(x) = r. Interestingly, this means that whether or not a function is surjective is only determined by how you define the domain of the function. For instance, sin : R -> [-1, 1] is surjective, but sin : R -> R is not.

Thus, to show that phi (I'll just use p(x)) is not an isomorphism, you need it to simply not satisfy ONE of those conditions. Specifically for (b), we will be using a proof by contradiction:

>suppose that p(x) : S -> R with p(x) = x for all s in S is an isomorphism

>since S is a proper subring of R, there exists some element u in R such that u is not in S

>since u is in R and p is surjective, there exists some element s in S such that p(s) = u

>however, since p(s) = s for any s in S, we must have s=u, which contradicts our statement that u is not in S

>thus, p cannot be an isomorphism