let $b$ be a prime number, and $a$ be any quadratic non-residue(mob $b$), i.e. $x^2 \equiv a \ (mod \ b)$ has no solutions in integers. Then the algebra $A=( \frac{a,b} { \mathbb{Q}})$ is a division algebra.
I have seen the proof of this in the book Fuchsian Groups, Svetlana Katok, at the page 116-117.But I could not understand it .Would you please give me some easy prove such that I can understand this.
The central simple algebra $$A=\left(\frac{a,b}{\Bbb Q}\right)$$ is isomorphic to $M_2(\Bbb Q)$ iff the quadratic form $Q(x,y,z)=x^2-ay^2-bz^2$ is isotropic (that is there is a solution $(x,y,z)\ne(0,0,0)$ to $(x,y,z)=0$ to over $\Bbb Q$) and is a division algebra otherwise. So for $A$ to be a division algebra, one requires $a$ to be a non-square over $\Bbb Q$, and moreover $b$ not to be a norm of an element of the quadratic field $\Bbb Q(\sqrt a)$. (These are the conditions that $Q(x,1,0)=0$ and $Q(x,y,1)$ have no solutions over $\Bbb Q$.)
If $a=p$ is an odd prime, then $a$ is certainly not a square of a rational. Moreover if $b$ is a quadratic non-residue of $p$, then there is no solution to $x^2-py^2=b$ over $\Bbb Q$. If there is a solution with $p$ dividing the denominator of $x$ or $y$, then $p$ must divide the denominators of $x^2$ and $py^2$ the same number of times; impossible on parity grounds. Otherwise $x$ and $y$ are $p$-integers, and then $x^2-py^2\equiv b\pmod p$ has a solution; impossible as $b$ is a quadratic non-residue modulo $p$.
But if $a=2$ we can have $A\cong M_2(\Bbb Q)$, say if $b\equiv\pm1\pmod 8$ is a prime.