ℚ(i) and ℚ(√2) are isomorphic ℚ vector spaces.
i has the minimal polynomial X2 + 1 and √2 has the minimal polynomial X2 – 2. Both have the degree 2, so the elements are algebraic via ℚ.
Thus ℚ(i) and ℚ(√2) are ℚ vector spaces of the same dimension 2 and thus isomorphic.
ℚ(i) and ℚ(√2) are not isomorphic fields.
Assuming ℚ(i) and ℚ(√2) are isomorphic fields.
Then there is a field isomorphism
φ: ℚ(i) → ℚ(√2)
The following applies:
φ(1) = 1 ⇒ -φ(1) = φ(-1) = -1
This still applies:
(φ(i))2 = φ(i2) = φ(-1) = -1
That means there must be an element x = φ(i) in ℚ(√2), with x2 = -1.
But this cannot be true, because it is x2 ≥ 0 for all x ∈ ℚ(√2).
We have a contradiction and the two fields cannot be isomorphic.