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Could anyone please show me an example of finite dimensional noncommutative associative division algebra over the field of rational numbers $Q$ other than quaternion algebras?

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Do you require associativity? –  Alex Becker Apr 19 '12 at 6:23
    
@Alex Yes, I do. –  Makoto Kato Apr 19 '12 at 6:42
    
For examples (from the same scheme as in Mariano's answer) see also this or this question. Matt E gave an example of a division algebra with center $\mathbf{Q}$. My example had a larger center. –  Jyrki Lahtonen Apr 19 '12 at 7:52
    
@Jyrki Thank you so much. –  Makoto Kato Apr 24 '12 at 10:33
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1 Answer

up vote 5 down vote accepted

There a many constructions. A simple one is the following:

Let $K/F$ be a cyclic extension of fields of degree $n$ and let $\sigma$ be a generator of the Galois group $G=Gal(K/F)$. Let $N_{K/F}(K^\times)$ be the subgroup of the multiplicative group $F^\times$ of norms of non-zero elements of $K$, and let $\alpha\in F^\times$ be an element of order $n$ modulo $N_{K/F}(K^\times)$. Then the $K$-vector space $$A=K\oplus xK\oplus x^2K\oplus\cdots\oplus x^{n-1}K$$ with basis $\{1,x,\dots,x^{n-1}\}$ is an $F$-algebra with multiplication extending that of $K$ and such that $$kx=x\sigma(k), \qquad\forall k\in K$$ and $$x^n=\alpha$$ is an associative division $F$-algebra.

So we can do this general contruction with $F=\mathbb Q$. For example, the polynomial $f=X^3-3X-1\in\mathbb Q[X]$ has discriminant $81$, which is a square, so its Galois group $G$ has only even permutations: since $f$ is irreducible, we see that $G$ is cyclic of order $3$.

Let $\zeta$ be a root of $f$ in $\mathbb Z$, and let $K=\mathbb Q[\zeta]$. Then $K/\mathbb Q$ is a cyclic extension of degree $3$. A generator $\sigma$ of the Galois group maps $\zeta$ to $2-\zeta^2$. The norm of the element $a+b\zeta+c\zeta^2$ is $$N_{K/F}(a+b\zeta+c\zeta^2)=a^3+6 a^2 c-3 a b^2-3 a b c+9 a c^2+b^3-3 b c^2+c^3.$$ Now we need an number $\alpha\in\mathbb Q$ such that its cube is in the image on $N_{K/\mathbb Q}$ but which itself isn't there. I think $2$ works (its cube is $N(-2\zeta-2\zeta^2)$, but one has to check that $2$ is not a norm...) The above construction, then, provides us with a rational division algebra of dimension $9$.

Can someone check that $2$ is not a norm?

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+1: $f$ is irreducible modulo $2$, so $2$ is inert in the extension $K/\mathbb{Q}$. Therefore it cannot be a norm of an element, as the fractional ideal generated by the norm of any element will have $(2)$ as a factor with a multiplicity divisible by three. –  Jyrki Lahtonen Apr 19 '12 at 7:55
    
There you go: someone who actually knew this stuff was needed :D –  Mariano Suárez-Alvarez Apr 19 '12 at 7:56
    
@Mariano Thanks! –  Makoto Kato Apr 19 '12 at 22:58
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