Exhibit infinitely many quadratic extensions of \(\mathbb Q\) and prove that they are pairwise nonisomorphic.
Solution
For each prime number \(p\), let \[ K_p\coloneqq\QQ(\sqrt p). \]
1. Each \(K_p\) is a quadratic extension of \(\QQ\).
A prime number \(p\) is not a square in \(\QQ\). Hence \[ x^2-p \] is irreducible over \(\QQ\), so \[ [K_p:\QQ]=2. \]
2. Every field isomorphism between two fields \(K_p\) and \(K_q\) fixes\(\QQ\) pointwise.
A field isomorphism sends \(1\) to \(1\), hence fixes the prime field \(\QQ\).
3. If \(p\) and \(q\) are distinct primes, then \(K_p\) and \(K_q\) are notisomorphic.
1. Thus \(a=0\), and therefore\[ \frac pq=b^2. \]But the prime factorization of \(p/q\) has exponent \(1\) at \(p\) and exponent\(-1\) at \(q\), whereas every rational square has even exponent at everyprime. This is impossible.
4. There are infinitely many pairwise nonisomorphic quadratic extensionsof \(\QQ\).
There are infinitely many prime numbers. By step 1 each prime \(p\) produces a quadratic extension \(K_p\), and step 3 shows that extensions coming from distinct primes are nonisomorphic.
5. Q.E.D.
Step 4 gives the required infinite family.