A Question Of Degree

We have intermediate fields $\mathbb{Q} \,\le\, \mathbb{Q}(a)\,,\,\mathbb{Q}(b)\,,\,\mathbb{Q}(ab) \,\le\, \mathbb{Q}(a,b)$ , with $[\mathbb{Q}(a):\mathbb{Q}] = 2$ and $[\mathbb{Q}(b):\mathbb{Q}] = 3$ . Since the minimum polynomial of $a$ over $\mathbb{Q}(b)$ must divide the minimum polynomial of $a$ over $\mathbb{Q}$ , it follows that $[\mathbb{Q}(a,b):\mathbb{Q}(b)] \,\le\, [\mathbb{Q}(a):\mathbb{Q}] = 2$ , and hence $[\mathbb{Q}(a,b):\mathbb{Q}] \; = \; [\mathbb{Q}(a,b):\mathbb{Q}(b)] \,[\mathbb{Q}(b):\mathbb{Q}] \; \le \; 6 \;,$ while both $2 = [\mathbb{Q}(a):\mathbb{Q}]$ and $3 = [\mathbb{Q}(b):\mathbb{Q}]$ divide $[\mathbb{Q}(a,b):\mathbb{Q}]$ . Thus $[\mathbb{Q}(a,b):\mathbb{Q}] = 6$ , and so it follows that $[\mathbb{Q}(ab):\mathbb{Q}]$ divides $6$ .

If $[\mathbb{Q}(ab):\mathbb{Q}] \le 2$ then, since $[\mathbb{Q}(a,b):\mathbb{Q}(ab)] \,\le\, [\mathbb{Q}(a):\mathbb{Q}] \,=\, 2$ , we deduce that $[\mathbb{Q}(a,b):\mathbb{Q}] = [\mathbb{Q}(a,b):\mathbb{Q}(ab)]\,[\mathbb{Q}(ab):\mathbb{Q}]$ would have to be one of $1$ , $2$ or $4$ , which is not the case. Thus it follows that $[\mathbb{Q}(ab):\mathbb{Q}] \,\ge\, 3$ , and so $[\mathbb{Q}(ab):\mathbb{Q}]$ can be either $3$ or $6$ .

Cue the standard example of $a = \omega$ , $b = \sqrt[3]{2}$ to see that a degree of $3$ is possible.

First we give an example where the degree of $ab$ is 3. Let $a=e^{2\pi i/3}$ and $b=\sqrt[3]{2}$ , with minimal polynomials $x^2+x+1$ and $x^3-2$ , respectively. Then $ab=e^{2\pi i/3}\sqrt[3]{2}$ is another root of $x^3-2$ .

Now $[\mathbb{Q}(a,b):\mathbb{Q}(b)]=2$ so $[\mathbb{Q}(a,b):\mathbb{Q}]=6$ so $[\mathbb{Q}(a,b):\mathbb{Q}(a)]= [\mathbb{Q}(ab,a):\mathbb{Q}(a)]= 3$ so $[\mathbb{Q}(ab):\mathbb{Q}]\geq 3$ since the minimal polynomial of $ab$ over $\mathbb{Q}$ may factor over $\mathbb{Q}(a)$ .

Thus the least possible degree of $ab$ is $\boxed{3}$