
isomorphic
$\displaystyle (i)Q[x]/<x^2+1>$ $\displaystyle is$ $\displaystyle isomorphic$ $\displaystyle to$ $\displaystyle Q[x]/<x^2+x+1>$
$\displaystyle (ii)R[x]/<x^2+1>$ $\displaystyle is$ $\displaystyle isomorphic$ $\displaystyle to$ $\displaystyle R[x]/<x^2+x+1>$
either of the statement is correct. justify your answer.
i think (i) is correct, but i don't know the reason, can you explain please.

Re: isomorphic
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).