## Assertion

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

## Proof

i has the minimal polynomial X^{2} + 1 and √2 has the minimal polynomial X^{2} – 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.

## Assertion

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

## Proof

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(i^{2}) = φ(-1) = -1

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

But this cannot be true, because it is x^{2} ≥ 0 for all x ∈ ℚ(√2).

We have a contradiction and the two fields cannot be isomorphic.