Analogy between the quaternion ring and extensions of the rationals

Question

I've started studing fields and their extensions. As an exercise I proved that $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]=4$ by showing that $B=(1,\sqrt2,\sqrt3,\sqrt6)$ is a base for the extension field as a vector space over $\mathbb{Q}$.

Now, I couldn't of not noticed the analogy between this ring and the quaternion ring. Actually it feels more like there is an anolgy between how $\mathbb{Q}(\sqrt2,\sqrt3)$ relate to $\mathbb{Q}$ and how the quaternion ring relates to $\mathbb{R}$. Indeed: $\sqrt2^2,\sqrt3^2,\sqrt6^2\in 1\mathbb{Q}$ and $\sqrt2\cdot\sqrt3\in\sqrt6\mathbb{Q}, \sqrt2\cdot\sqrt6\in\sqrt3\mathbb{Q}, \sqrt3\cdot\sqrt6\in\sqrt2\mathbb{Q}$ - similar to the quaternions.

By this analogy it seems like the linear independence of $B$ is implied by the fact that the quaternions are a ring and hence have a unique $0$ element. Showing that $Sp(B)=\mathbb{Q}(\sqrt2,\sqrt3)$ becomes very easy by the fact that the quaternions are a division ring.

How would you formally express this analogy (if you agree that is a good analogy). Clearly the same idea can be applied to the how the complex field relates to the reals and how $\mathbb{Q}(\sqrt2)$ related to $\mathbb{Q}$ Thanks

Answer 1

Instead of looking an analogy with the real quaternions $\Bbb H$, you should consider the quaternion algebra $$ D=\Bbb Q\oplus\Bbb Qi\oplus\Bbb Qj\oplus\Bbb Qk $$ with $i^2=2$, $j^2=3$ and $ij=k=-ji$. They are both 4-dimensional over $\Bbb Q$ and admit a basis of elements whose squares are in $\Bbb Q$. Yet one is commutative, the other is not.

One can define plenty of these quaternion algebras over $\Bbb Q$, just fix $a$ and $b$ in $\Bbb Q$ such that $i^2=a$, $j^2=b$. Classifying them up to isomorphism is an interesting exercise (!) in Number Theory (Hint: read the 1st part of Serre's Course d'Arithmetique first). Sometimes they are division algebras, sometimes they are not.

Answer 2

Suppose $A$ is an $F$-algebra with a $F$-basis $B$. If the $F$-lines generated by $B$, $\{xF:x\in B\}$, form a group under multiplication, we shall call this the turtle group $G=G(A,B,F)$. Equivalently, the turtle group is the image of $B$ under $A^\times\to A^\times/F^\times$, if it is indeed a group.

Your observation comes down to the fact

$$G\left(\Bbb Q(\sqrt{a},\sqrt{b}),\{1,\sqrt{a},\sqrt{b},\sqrt{ab}\},\Bbb Q\right) \cong G\left(\Bbb H,\{1,i,j,k\},\Bbb R\right)$$

are the same turtle group, the Klein four group $C_2\times C_2$.