ℚ(i) and ℚ(√2) are isomorphic vector spaces over ℚ.
The minimal polynomial of i is X2 + 1 and the minimal polynomial of √2 is X2 – 2. Both have degree 2, the elements are therefore algebraic over ℚ.
ℚ(i) and ℚ(√2) are vector spaces over ℚ and they have the same dimension 2, therefore they are isomorphic.
ℚ(i) and ℚ(√2) are not isomorphic fields.
Assume ℚ(i) and ℚ(√2) are isomorphic fields.
Then there is a field isomorphism
φ: ℚ(i) → ℚ(√2)
φ(1) = 1 ⇒ -φ(1) = φ(-1) = -1
(φ(i))2 = φ(i2) = φ(-1) = -1
There must be an element x = φ(i) in ℚ(√2), with x2 = -1.
That can’t be true, since x2 ≥ 0 for all x ∈ ℚ(√2).
We have a contradiction, the two fields can’t be isomorphic.