Let \(\mathbf F\) be a field containing \(\mathbb Q\) such that \([\mathbf F:\mathbb Q]=2\). Prove that there exists a unique integer \(m\) such that \(m\) has no multiple prime factors and \(\mathbf F\) is isomorphic to \(\mathbb Q(\sqrt m)\).
Solution
1. There is an element \(\alpha\in\mathbf F\setminus\QQ\) such that\[ \mathbf F=\QQ(\alpha), \]and the minimal polynomial of \(\alpha\) over \(\QQ\) has the form\[ x^2+rx+s \]with \(r,s\in\QQ\).
Choose any \(\alpha\in\mathbf F\setminus\QQ\). Then \[ 1<[\QQ(\alpha):\QQ]\leq[\mathbf F:\QQ]=2. \] Hence \([\QQ(\alpha):\QQ]=2\) and therefore \(\QQ(\alpha)=\mathbf F\). The monic minimal polynomial of \(\alpha\) over \(\QQ\) consequently has degree \(2\), so it has the displayed form.
2. If\[ \Delta=r^2-4s, \]then \(\Delta\in\QQ^\times\) is not a square in \(\QQ\) and\[ \mathbf F=\QQ(\sqrt{\Delta}). \]
From \[ \alpha^2+r\alpha+s=0 \] one obtains \[ (2\alpha+r)^2=r^2-4s=\Delta. \] Also \[ \alpha=\frac{(2\alpha+r)-r}{2}, \] so \[ \QQ(\alpha)=\QQ(2\alpha+r). \] Thus \(\mathbf F\) is generated by a square root of \(\Delta\).
1. Therefore \(\Delta\) is a nonzero nonsquare in \(\QQ\).
3. There exists a squarefree integer \(m\) such that\[ \mathbf F\cong\QQ(\sqrt m). \]
Write \[ \Delta=\frac{a}{b} \] with relatively prime nonzero integers \(a,b\) and \(b>0\). If \(\gamma^2=\Delta\), then \[ (b\gamma)^2=ab, \] and conversely \(\gamma=(b\gamma)/b\). Hence \[ \QQ(\sqrt{\Delta})=\QQ(\sqrt{ab}). \]
By unique factorization in \(\ZZ\), write \[ ab=mc^2 \] with \(c\in\ZZ_{>0}\) and \(m\) a squarefree integer, with the sign of \(m\) equal to the sign of \(ab\). Then \[ \QQ(\sqrt{ab})=\QQ(c\sqrt m)=\QQ(\sqrt m). \] Combining this with step 2 gives the required existence.
4. The squarefree integer \(m\) in step 3 is unique.
Suppose \(m\) and \(n\) are squarefree integers and \[ \QQ(\sqrt m)\cong\mathbf F\cong\QQ(\sqrt n). \] Composing the two field isomorphisms gives an isomorphism \[ \varphi:\QQ(\sqrt m)\longrightarrow\QQ(\sqrt n). \] Every field isomorphism between fields of characteristic \(0\) fixes \(\QQ\), so for some \(u,v\in\QQ\), \[ \varphi(\sqrt m)=u+v\sqrt n. \] Squaring and using \(\varphi(m)=m\) gives \[ m=u^2+v^2n+2uv\sqrt n. \] Since \(\QQ(\sqrt n)\) has degree \(2\) over \(\QQ\), one has \(\sqrt n\notin\QQ\), and therefore \[ uv=0. \] The case \(v=0\) would give \(\varphi(\sqrt m)\in\QQ\), impossible because \(\varphi\) is an isomorphism and \(\sqrt m\notin\QQ\). Thus \(u=0\), so \[ \frac{m}{n}=v^2 \] is a square in \(\QQ\).
For every prime \(\ell\), squarefreeness gives \[ \nu_\ell(m),\nu_\ell(n)\in\{0,1\}. \] Since \(m/n\) is a rational square, \(\nu_\ell(m)-\nu_\ell(n)\) is even. It lies in \(\{-1,0,1\}\), so it must be \(0\). Thus \(m\) and \(n\) have the same prime divisors. Moreover \(m/n=v^2>0\), so they have the same sign. Therefore \(m=n\).
5. Q.E.D.
Step 3 proves existence and step 4 proves uniqueness.