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I know that there is a division algebra over $\mathbb{Q}$ such that it is algebraic and infinite dimensional over it's center i.e. $\mathbb{Q}$. But for construct this division algebra. we can use prime numbers.

I want to know a general method of construction for every arbitrary field.

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Such objects do not always exist. If $D$ is a division algebra with center $K$, and $z\in D\setminus K$ is some element, then $K[z]$ is an algebraic extension of $K$. It is commutative because powers of $z$ commute with each other, and they obviously commute with $K=Z(D)$. It is a finite dimensional $K$-algebra as $z$ has a minimal polynomial over $K$ (this is how I interpret your phrase "$D$ is algebraic over $K$"). Because $D$ has no zero divisors the f.d. $K$-algebra $K[z]$ has no zero-divisors either, and thus $K[z]$ is a f.d. extension of the field $K$.

But if $K$ is algebraically closed, then it has no non-trivial algebraic extensions, and hence no non-trivial algebraic $D$ either.


OTOH if $K$ is a number field (or a f.d. extension of the field of $p$-adic numbers), then it seems to me that we can construct an infinite dimensional divison algebra of the prescribed type. A construction I have in mind is to build a nested sequence of cyclic division algebras of indices, say powers of two. At least in the $p$-adic case the standard construction seems to work, because we can locate a chain of unramified extension fields of degrees $2^n, n\in\Bbb{N}$, each contained in the next. The nested union is then a division algebra of the prescribed type. Over a number field we need some deepish results describing the Brauer group to do the same.


I would think that Wedderburn's theorem about the commutativity of finite division algebras rules out the possibility of such a division algebra to exist over a finite field, but I cannot wrap my head around an argument at this time. If I'm wrong about this, then that is, of course, much more exciting! Similarly, over real closed fields all those elements $z$ will be quadratic, and this likely makes the construction impossible, but I'm missing a link or five from the argument.

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