let's look at the second one, first.

the coset of x in R[x]/<x^{2}+1>, x+<x^{2}+1> has the following property:

writing [x] for x+<x^{2}+1>, we get:

[x]^{2}+ [1] = [x^{2}+1] = [0] (since x^{2}+1 is clearly in <x^{2}+1>).

this means that [x]^{2}= [-1].

identifying R with the isomorphic copy of it inside R[x]/<x^{2}+1> (via the isomorphism a→[a] = a + <x^{2}+1>), we have:

[x]^{2}= -1.

now all cosets of R[x]/x^{2}+1> are of the form:

[a+bx] (= a + bx + <x^{2}+1>), for a,b in R. if we write [x] = i, then:

[a+bx] = [a] + [b][x] = a + [x]b = a + ib, and these cosets obey the same algebraic rules as the complex numbers. so R[x]<x^{2}+1> is isomorphic to C = R(i).

in the same manner, R[x]/<x^{2}+x+1> is isomorphic to R(ω), where ω = -1/2 + i(√3/2)

(note that ω^{2}+ω+1 = (-1/2 + i(√3/2))^{2}+ (-1/2 + i(√3/2)) + 1

= (1/4 - 3/4) + i[2(-1/2)(√3/2)] -1/2 + i(√3/2) + 1 = -1/2 - i(√3/2) - 1/2 + i(√3/2) + 1 = 1 - 1 + i0 = 0 + i0 = 0).

now ω is a complex number, so R(ω) is clearly contained in C = R(i).

but i = (2/√3)ω + √3, and since 2/√3 and √3 are real numbers, we see that R(i) is contained in R(ω), hence R(i) = R(ω).

since both quotient rings are isomorphic to the same field (the complex numbers), they must be isomorphic to each other.

now let's consider Q(i) and Q(ω). note that Q(ω) contains i√3 (= √(-3)), which is 2ω + 1.

now Q(i) = {a + ib : a,b in Q}. can we find a,b in Q such that (a + ib)^{2}= -3? suppose we could:

(a + ib)^{2}= -3

(a^{2}- b^{2}) + i(2ab) = -3 = -3 + i0.

comparing sides, we must have that:

2ab = 0

a^{2}- b^{2}= -3

so either a or b must be 0 (since Q is a field, and thus an integral domain). suppose b = 0.

then a^{2}= -3, which is not possible, since Q is an ordered field, and a^{2}≥ 0 > -3.

thus it could only possibly be that a = 0, in which case:

b^{2}= 3, implying that Q contains a square root of 3, which it does not.

since Q(ω) has a root to x^{2}+ 3, and Q(i) does not, they cannot be isomorphic fields

(if x^{2}+ 3 factors in Q(ω), then it must have a corresponding factorization in Q(i), if the 2 are isomorphic).