In this question, a ring is not supposed to be commutative.
Let $K=\mathbb{Q}[i]$, and $A$ a subalgebra of $\mathcal{M}_n(K)$ (the algebra of $n \times n$ matrices). We suppose $A$ is a division ring. Is $A$ necessarily commutative ?
More generally, if $K$ is an algebraic extension of $\mathbb{Q}$, where $-1$ is a square, is $A$ necessarily commutative ?
If $-1$ is not the sum of three square in $K$, we can construct a division ring, that is an algebraic extension of $K$. The construction is similar to the construction of quaternions over $\mathbb{R}$. So, $A$ is not necessarily commutative in this case.
Thanks in advance.
The answer to your question is no. More generally, if $K$ is any number field, then for any $n > 1$ there exists a division algebra $D/K$ with center $K$ and $\operatorname{dim}_K D = n^2$. In particular $D$ is not commutative.
If $K$ is not formally real then you can't use the easy trick of choosing a quaternion algebra $\left( \frac{a,b}{K} \right)$ with $a,b$ both negative. But you can still do the $p$-adic analogue of this.
Here first is a general recipe: let $K/\mathbb{Q}$ be any number field, and choose a prime number $p$ which splits completely in $K$ (the set of such primes has positive density). Now choose a division algebra $D_0$ over $\mathbb{Q}$ such that $D_0 \otimes_{\mathbb{Q}} \mathbb{Q}_p$ is still a division algebra. Since $p$ splits completely in $K$, $K$ embeds in $\mathbb{Q}_p$ and thus $D = D_0 \otimes_{\mathbb{Q}} K$ is still a division algebra. (In yet more abstract terms, the Brauer group of any number field is infinite.)
Here is a concrete example: take $D_0$ to be the quaternion algebra $\left( \frac{-2,-5}{\mathbb{Q}} \right)$. This algebra is ramified at $\infty$ and $5$, and $5$ splits in $K =\mathbb{Q}(\sqrt{-1})$, so $\left( \frac{-2,-5}{\mathbb{Q(\sqrt{-1})}} \right)$ is still a division algebra.