# isomorphic

• May 13th 2012, 05:06 AM
saravananbs
isomorphic
$(i)Q[x]/$ $is$ $isomorphic$ $to$ $Q[x]/$
$(ii)R[x]/$ $is$ $isomorphic$ $to$ $R[x]/$

i think (i) is correct, but i don't know the reason, can you explain please.
• May 13th 2012, 10:46 AM
Deveno
Re: isomorphic
let's look at the second one, first.

the coset of x in R[x]/<x2+1>, x+<x2+1> has the following property:

writing [x] for x+<x2+1>, we get:

[x]2 + [1] = [x2+1] = [0] (since x2+1 is clearly in <x2+1>).

this means that [x]2 = [-1].

identifying R with the isomorphic copy of it inside R[x]/<x2+1> (via the isomorphism a→[a] = a + <x2+1>), we have:

[x]2 = -1.

now all cosets of R[x]/x2+1> are of the form:

[a+bx] (= a + bx + <x2+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]<x2+1> is isomorphic to C = R(i).

in the same manner, R[x]/<x2+x+1> is isomorphic to R(ω), where ω = -1/2 + i(√3/2)

(note that ω2+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
(a2 - b2) + i(2ab) = -3 = -3 + i0.

comparing sides, we must have that:

2ab = 0
a2 - b2 = -3

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

then a2 = -3, which is not possible, since Q is an ordered field, and a2 ≥ 0 > -3.

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

b2 = 3, implying that Q contains a square root of 3, which it does not.

since Q(ω) has a root to x2 + 3, and Q(i) does not, they cannot be isomorphic fields

(if x2 + 3 factors in Q(ω), then it must have a corresponding factorization in Q(i), if the 2 are isomorphic).