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?

  • 1
    $\begingroup$ Do you require associativity? $\endgroup$ – Alex Becker Apr 19 '12 at 6:23
  • $\begingroup$ @Alex Yes, I do. $\endgroup$ – Makoto Kato Apr 19 '12 at 6:42
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Apr 19 '12 at 7:52
  • $\begingroup$ @Jyrki Thank you so much. $\endgroup$ – Makoto Kato Apr 24 '12 at 10:33
  • $\begingroup$ If $M$ is a simple $R$-module, then $\mathrm{End}_R(M)$ is a division ring, so a division $\Bbb Z$-algebra. $\endgroup$ – Watson Jan 28 '17 at 14:47

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?

| cite | improve this answer | |
  • 6
    $\begingroup$ +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. $\endgroup$ – Jyrki Lahtonen Apr 19 '12 at 7:55
  • 1
    $\begingroup$ There you go: someone who actually knew this stuff was needed :D $\endgroup$ – Mariano Suárez-Álvarez Apr 19 '12 at 7:56
  • 1
    $\begingroup$ @Mariano Thanks! $\endgroup$ – Makoto Kato Apr 19 '12 at 22:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.