- Let \(u=\sqrt{3+\sqrt{2}}\). Is \(\QQ(u)\) the splitting field over \(\QQ\) of the minimal polynomial of \(u\)?
Solution
Set \[ E=\QQ(\sqrt2), \qquad F=\QQ(u)=E(u). \]
1. The minimal polynomial of \(u\) over \(\QQ\) is\[ f(x)=x^4-6x^2+7. \]
Since \[ u^2=3+\sqrt2, \] we have \(\sqrt2=u^2-3\in\QQ(u)\), so \(E\subseteq F\). Also \[ (u^2-3)^2=2, \] which gives \(f(u)=0\).
It remains to show \([F:\QQ]=4\). We have \([E:\QQ]=2\). If \(u\in E\), then \(3+\sqrt2\) would be a square in \(E\). But for \(a\in E\), \[ N_{E/\QQ}(a^2)=N_{E/\QQ}(a)^2 \] is a square in \(\QQ\), whereas \[ N_{E/\QQ}(3+\sqrt2)=(3+\sqrt2)(3-\sqrt2)=7 \] is not a square in \(\QQ\). Thus \(u\notin E\), so \([F:E]=2\) and therefore \([F:\QQ]=4\). Since \(f\) has degree \(4\), it is the minimal polynomial of \(u\).
2. The roots of \(f\) are\[ \pm u, \qquad \pm v, \qquad v:=\sqrt{3-\sqrt2}. \]
The equation \(f(x)=0\) is \[ (x^2-3)^2=2, \] so \[ x^2=3\pm\sqrt2. \] Taking square roots gives exactly the four displayed roots. Also \[ uv=\sqrt{(3+\sqrt2)(3-\sqrt2)}=\sqrt7. \]
3. One has \(\sqrt7\notin F\).
Because \(F/E\) is quadratic with basis \(1,u\), write any element of \(F\) uniquely as \(a+bu\) with \(a,b\in E\). Suppose \[ (a+bu)^2=7. \] Then \[ a^2+b^2u^2+2ab\,u=7. \] The first two terms lie in \(E\), while \(1,u\) are \(E\)-linearly independent, so \(2ab=0\). Hence either \(a=0\) or \(b=0\).
If \(b=0\), then \(a^2=7\) in \(E=\QQ(\sqrt2)\). Write \(a=r+s\sqrt2\) with \(r,s\in\QQ\). Comparing the \(\sqrt2\) coefficient gives \(2rs=0\). If \(s=0\), then \(r^2=7\); if \(r=0\), then \(2s^2=7\). Neither equation has a rational solution.
If \(a=0\), then \[ b^2=\frac7{u^2}=\frac7{3+\sqrt2}=3-\sqrt2. \] But \[ N_{E/\QQ}(3-\sqrt2)=7 \] is not a square in \(\QQ\), whereas the norm of \(b^2\) would be a square. This is again impossible. Thus \(\sqrt7\notin F\).
4. The root \(v=\sqrt{3-\sqrt2}\) does not lie in \(F\).
If \(v\in F\), then by 2, \[ \sqrt7=uv\in F, \] contradicting 3.
5. Therefore \(F=\QQ(u)\) is not the splitting field of the minimal polynomial of \(u\) over \(\QQ\).
By <1>1–<1>2, the minimal polynomial is \(f(x)=x^4-6x^2+7\) and one of its roots is \(v\). By 4, \(v\notin F\), so \(f\) does not split over \(F\).