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In First Course in Noncommutative rings of T.Y.Lam (p.210), the author stated that "It is known that for each $n$, there exists a $\mathbb{Q}$-division algebra $A_n$ of dimension $p_n^2$, with $Z(A_n)=\mathbb{Q}$." Here $p_n$ is a prime number and $Z(A_n)$ is the center of $A_n$.

How to construct $A_n$?

Thanks!

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The recipe is the following:

  • Locate an extension field $K$ such that $[K:\Bbb{Q}]=p=p_n$, and $K/\Bbb{Q}$ is Galois with a cyclic Galois group generated by an automorphism $\sigma$. This latter requirement is automatic as a group of prime order is necessarily cyclic, but if you want non-prime index then you need to ascertain this also.
  • Locate an element $\gamma\in\Bbb{Q}$ such that $\gamma\neq N(z)$ for all $z\in K$. Here $N:K\to\Bbb{Q}$ is the norm map: $$N(z)=\prod_{i=0}^{p-1}\sigma^i(z).$$
  • Given all this let $A_p$ be the $p$-dimensional vector space over $K$ with basis $1,u,u^2,\ldots, u^{p-1}$. Here $u$ is an auxiliary generator. $A_p$ becomes a $\Bbb{Q}$-algebra when we define a multiplication by the rules $u^p=\gamma$, and $ux=\sigma(x)u$ for all $x\in K$.
  • The relation $ux=\sigma(x)u$ really forces the center of $A_p$ to be $\Bbb{Q}$ only. Among other things it implies that if an element $x\in K$ commutes with $u$, we need $\sigma(x)=x$, i.e. $x\in\Bbb{Q}$.
  • The requirement $\gamma\neq N(x)$ for all $x\in K$ turns out be a sufficient criterion for $A_p$ not to have zero divisors, hence proving that $A_p$ is a division algebra. This fact (and the previous bullet) are not at all obvious, see e.g. Jacobson's Basic Algebra II, chapter 8, for more details.

A few remarks:

  • You can find the required field $K$ as a subfield of a suitable cyclotomic extension. In fact, by Kronecker-Weber, this is the only way to find such a $K$.
  • Finding a $\gamma$ takes a bit of algebraic number theory. There are many choices that work (I like Lord Shark the Unknown's suggestion of using an inert prime). Two choices $\gamma$ and $\gamma'$ yield isomorphic division algebras, iff $\gamma/\gamma'=N(x)$ for some $x\in K$.
  • You can realize $A_p$ as matrices of the form $$\left(\begin{array}{ccccc} x_0&\sigma(x_{p-1})&\sigma^2(x_{p-2})&\cdots&\sigma^{p-1}(x_1)\\ \gamma x_1&\sigma(x_0)&\sigma^{2}(x_{p-1})&\cdots&\sigma^{p-1}(x_2)\\ \gamma x_2&\gamma\sigma(x_1)&\sigma^2(x_0)&\cdots&\sigma^{p-1}(x_3)\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \gamma x_{p-1}&\gamma\sigma(x_{p-2})&\gamma\sigma^2(x_{p-2})&\cdots&\sigma^{p-1}(x_0) \end{array}\right).$$ Here $x_0,x_1,\ldots,x_{p-1}$ are arbitrary elements of $K$, $\gamma$ is the prescribed non-norm element and appears as a factor in all the entries below the diagonal. The powers of $\sigma$ increase column-by-column from left to right, and the variables $x_i,i=0,1,\ldots,p-1$, appear in a circulant way.
  • If you want a non-prime index $n$, then the extra requirement is that you need to check that the lowest power $\gamma^\ell$, $\ell>0$, that is a norm of some element occurs only at $\ell=n$.
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  • $\begingroup$ Thanks for your answer. Can you give me some ways to find $\gamma$? If $\gamma$ is an inert prime, then how to prove it is not contained in $N(K^*)$ ? $\endgroup$ – Huỳnh Việt Khánh Apr 2 '18 at 9:22
  • $\begingroup$ @HuỳnhViệtKhánh If $\gamma=q$ is an inert prime, then the ideal of the integers $\mathcal{O}_K$ of the field $K$ it generates remains a prime ideal, call it $\mathfrak{q}$. If the norm of an element $x\in K$ were equal to $q$, then the fractional ideal $\mathcal{O}_Kx$ must have $\mathfrak{q}$ as factor (with positive multiplicity). But, the same then holds for all the conjugates of $x$, so the norm will be divisible by $\mathfrak{q}^p$, i.e. with $q^p$. Here $p>1$ so the norm cannot be equal to $q$. $\endgroup$ – Jyrki Lahtonen Apr 2 '18 at 9:35
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Let $p$ be a prime. Then there is a prime $q\equiv1\pmod p$, and so the $q$-th cyclotomic field will have a subfield $K$ of degree $p$ over $\Bbb Q$ with cyclic Galois group generated by $\sigma$. There will be a prime $\ell$ which remains inert in $K$.

One now constructs $A$ as a crossed product of $K$. It is a free left $K$ module $A=K\oplus Ku\oplus Ku^2\cdots \oplus Ku^{p-1}$ where we define multiplication by $u^p=\ell$ and $u\alpha=\sigma(\alpha)u$ for $\alpha\in K$.

This is a central simple algebra by the theory of crossed product. Tensoring with $\Bbb Q_\ell$ will give a division algebra over $\Bbb Q_\ell$, so $A$ must be a division algebra.

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