If \(F = \mathbb{Q}(\sqrt 2, \sqrt 3)\), compute \([F: \mathbb{Q}]\) and find a basis of \(F/\mathbb{Q}\).
Do the same for \(\mathbb{Q}(i, \sqrt 3, \zeta_3)\) where \(\zeta_3\) is a complex third root of 1.
Solution
We use the tower law in the following form: if \([L:K]=m\) with basis \(\{x_i\}\) and \([M:L]=n\) with basis \(\{y_j\}\), then \([M:K]=mn\) with basis \(\{x_iy_j\}\).
1. (1) \([\QQ(\sqrt 2, \sqrt 3) : \QQ] = \boxed{4}\), with basis \(\boxed{\{1, \sqrt 2, \sqrt 3, \sqrt 6\}}\).
1.1. \([\QQ(\sqrt 2) : \QQ] = 2\) with basis \(\{1,\sqrt2\}\).
\(\sqrt2\) is irrational, so \(X^2-2\) has no rational root and is the minimal polynomial of \(\sqrt2\).
1.2. \(\sqrt 3 \notin \QQ(\sqrt 2)\), so \([\QQ(\sqrt 2, \sqrt 3) : \QQ(\sqrt 2)] = 2\) with basis \(\{1,\sqrt3\}\).
Suppose \(\sqrt 3 = a + b\sqrt 2\) with \(a, b \in \QQ\). Squaring gives \(3=a^2+2b^2+2ab\sqrt2\), so \(ab=0\) because \(\sqrt2\) is irrational. If \(b=0\), then \(\sqrt3=a\in\QQ\); if \(a=0\), then \(\sqrt6=2b\in\QQ\). Both contradict irrationality of \(\sqrt3\) and \(\sqrt6\). Hence \(X^2-3\) has no root in \(\QQ(\sqrt2)\) and is the minimal polynomial of \(\sqrt3\) over it.
1.3. Q.E.D.
By the tower law and steps 1 and 2, the degree is \(2\cdot2=4\) and the products \(1,\sqrt2,\sqrt3,\sqrt2\sqrt3=\sqrt6\) form a basis.
2. (2) \([\QQ(i, \sqrt 3, \zeta_3) : \QQ] = \boxed{4}\), with basis \(\boxed{\{1, i, \sqrt 3, i\sqrt 3\}}\).
2.1. \(\QQ(i, \sqrt 3, \zeta_3) = \QQ(i, \sqrt 3)\).
The complex third roots of \(1\) other than \(1\) are \(\frac{-1 \pm i\sqrt 3}{2}\in\QQ(i,\sqrt3)\), and \(\zeta_3=1\) also lies in \(\QQ\).
2.2. \([\QQ(i) : \QQ] = 2\) with basis \(\{1,i\}\), and \([\QQ(i, \sqrt 3) : \QQ(i)] = 2\) with basis \(\{1,\sqrt3\}\).
The polynomial \(X^2 + 1\) has no rational root, so it is the minimal polynomial of \(i\). Every element \(a+bi\) of \(\QQ(i)\) with \(a,b\in\QQ\) is real only when \(b=0\), so \(\QQ(i)\cap\RR=\QQ\). Since \(\sqrt3\) is real and irrational, \(\sqrt 3 \notin \QQ(i)\), and \(X^2-3\) is the minimal polynomial of \(\sqrt3\) over \(\QQ(i)\).
2.3. Q.E.D.
By step 1 and the tower law applied to step 2, the degree is \(2\cdot2=4\) and the products \(1,i,\sqrt3,i\sqrt3\) form a basis.