Show that \(\mathbb{Q}(\sqrt{2})\) and \(\mathbb{Q}(\sqrt{3})\) are not isomorphic.
Solution
1. Every field isomorphism between these fields fixes \(\mathbb Q\).
A field isomorphism sends \(1\) to \(1\), hence fixes every integer. It preserves inverses, so it also fixes every quotient of integers with nonzero denominator. Thus it fixes the prime subfield \(\mathbb Q\) pointwise.
2. The field \(\mathbb Q(\sqrt3)\) contains no element whose square is \(2\).
For a prime \(\ell\) and a nonzero rational number \(r\), let \(v_\ell(r)\) be the exponent of \(\ell\) in its numerator minus the exponent in its denominator. Unique prime factorization gives \(v_\ell(r^2)=2v_\ell(r)\), an even integer. Consequently \(3\) is not a rational square, since \(v_3(3)=1\). Hence \(1,\sqrt3\) are linearly independent over \(\mathbb Q\) and form a basis of the quadratic field \(\mathbb Q(\sqrt3)\).
Suppose \((a+b\sqrt3)^2=2\) with \(a,b\in\mathbb Q\). Expanding and comparing the two basis coefficients gives \[ a^2+3b^2=2,\qquad 2ab=0. \] If \(b=0\), then \(a^2=2\), contradicting \(v_2(2)=1\). If \(a=0\), then \(b^2=2/3\), contradicting \(v_2(2/3)=1\). The equation \(2ab=0\) forces at least one of these cases, so there is no such element.
3. The fields are not isomorphic.
If an isomorphism \(\sigma:\mathbb Q(\sqrt2)\to\mathbb Q(\sqrt3)\) existed, step 1 would give \(\sigma(\sqrt2)^2=\sigma(2)=2\). This contradicts step 2.