**Claim:**

ℚ(i) and ℚ(√2) are isomorphic vector spaces over ℚ.

**Proof:**

The minimal polynomial of i is X^{2} + 1 and the minimal polynomial of √2 is X^{2} – 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.

**Claim:**

ℚ(i) and ℚ(√2) are not isomorphic fields.

**Proof:**

Assume ℚ(i) and ℚ(√2) are isomorphic fields.

Then there is a field isomorphism

φ: ℚ(i) → ℚ(√2)

We have:

φ(1) = 1 ⇒ -φ(1) = φ(-1) = -1

Then holds:

(φ(i))^{2} = φ(i^{2}) = φ(-1) = -1

There must be an element x = φ(i) in ℚ(√2), with x^{2} = -1.

That can’t be true, since x^{2} ≥ 0 for all x ∈ ℚ(√2).

We have a contradiction, the two fields can’t be isomorphic.

*q.e.d.*