Show that in \(\mathbb{C}\), the fields \(\mathbb{Q}(i) \cong \mathbb{Q}(\sqrt 2)\) as vector spaces, but not as fields.
Solution
Both fields are regarded as vector spaces over \(\QQ\).
1. The sets\[ \{1,i\} \qquad\text{and}\qquad \{1,\sqrt2\} \]are \(\QQ\)-bases of \(\QQ(i)\) and \(\QQ(\sqrt2)\), respectively.
The element \(i\) has minimal polynomial \(X^2+1\) over \(\QQ\), while \(\sqrt2\) has minimal polynomial \(X^2-2\). Both polynomials are irreducible over \(\QQ\), so both simple extensions have degree \(2\). The standard power bases are therefore the displayed pairs.
2. The map\[ T:\QQ(i)\longrightarrow\QQ(\sqrt2), \qquad T(a+bi)=a+b\sqrt2 \]is a \(\QQ\)-vector-space isomorphism.
The map sends the basis \(\{1,i\}\) from 1 bijectively to the basis \(\{1,\sqrt2\}\). Hence it is a \(\QQ\)-linear isomorphism.
3. There is no field isomorphism\[ \QQ(i)\longrightarrow\QQ(\sqrt2). \]
Suppose \(\varphi\) were such an isomorphism. Every field homomorphism fixes the prime field, so \(\varphi(-1)=-1\). Since \(i^2=-1\), \[ \varphi(i)^2=\varphi(i^2)=\varphi(-1)=-1. \] But \(\QQ(\sqrt2)\subseteq\RR\), and no real number has square \(-1\). This is a contradiction.
4. Thus \(\QQ(i)\) and \(\QQ(\sqrt2)\) are isomorphic as \(\QQ\)-vector spacesbut not as fields.
Combine 2 and 3.