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In his book Introduction to arithmetic groups, Dave Witte Morris implicitly gives a construction of central division algebras of degree 3 over $\mathbb{Q}$ in Proposition 6.7.4. More precisely, let $L/\mathbb{Q}$ be a cubic Galois extension and $\sigma$ a generator of its Galois group.If $p \in \mathbb{Z}^+$ and $p \neq t\sigma(t)\sigma^2(t)$ for all $t \in L$, then $$ D=\left\{ \begin{pmatrix} x & y & z\\ p\sigma(z) & \sigma(x) & \sigma(y)\\ p\sigma^2(y) & p\sigma^2(z) & \sigma^2(x) \end{pmatrix} :(x,y,z)\in L^3 \right\} $$ is a division algebra.

On page 145, just before Proposition 6.8.8, Morris claims that it is known that every division algebra of degree 3 arises in this manner. This should follow from the fact that every central division algebra of degree 3 is cyclic. I could not find this explicit construction in my references (e.g. Pierce - Associative Algebras, though maybe I missed something) and I would like to know if there is a reference or a quick way to see that this exhausts all central division algebras of degree 3 over $\mathbb{Q}$.

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    $\begingroup$ This may not constitute a "construction", but.one can use the Albert–Brauer–Hasse–Noether theorem and the Grunwald-Wang theorem to see that central simple algebras over number fields are cyclic. $\endgroup$ – LeechLattice Apr 13 at 11:49
  • $\begingroup$ @LeechLattice Thank you for the comment. Yes, this was mentioned by Morris and is a big topic in some references I have, but I am specifically interested in the explicit construction of the algebras. So taking the theorems you mention into consideration, the question is: why would the construction above yield all cyclic algebras? $\endgroup$ – Radu T Apr 13 at 12:51
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    $\begingroup$ For any cyclic algebra $C$ of degree $9$ over $\mathbb{Q}$, you can choose a maximal subfield $L$ of $C$ which is cyclic of degree $3$ over $\mathbb{Q}$, and an $L$-basis for $C$. Then the left multiplication action of $C$ on itself gives a map $i : C \hookrightarrow \operatorname{End}(C) \cong \operatorname{GL}_3(L)$. I guess it shouldn't be hard to show, using the definition of cyclic algebra, that the image of $i$ is always of the form that Morris gives (where it may or may not be important how you choose your basis for $C$). $\endgroup$ – RP_ Apr 13 at 14:34
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Looking more carefully in Pierce - Associative algebras, I found the answer I was looking for, which I'm going to describe here for future reference.

The algebra $D$ in the question is a realisation of a special type of crossed product algebra, which gives precisely the central division algebras of degree 3 over $\mathbb{Q}$.

The first step is to use the powerful result mentioned in the comments, which states that all central simple algebras over number fields are cyclic. For degree 3 algebras, this was proved by Wedderburn. Recall that an algebra $A$ over a field $F$ of degree $n$ is cyclic if there is a cyclic Galois extension $E/F$ of degree $n$ such that $E$ is a maximal subfield inside $A$.

Proposition a in Pierce, section 15.1, states that a cyclic algebra $A$ as above, where $\sigma$ generates the Galois group of $E/F$, contains an invertible element $u$ such that

  1. $A=\bigoplus_{0 \leq j <n} u^jE$
  2. $u^{-1}du=\sigma(d)$ for all $d\in E$, and
  3. $u^n=a \in F^\times$.

This is equivalent to saying that $A$ is isomorphic to the crossed product $(E,G,\Phi_a)$, defined in Pierce. The only thing left to do is to check that the algebra $D$ in the question is a construction of this crossed product. You can check this by first embedding $E$ into $M_3(E)$ by $$x \mapsto \text{diag}(x,\sigma(x),\sigma^2(x)). $$ Then take a rational number $p\neq 0$, which will play the role of $a$ in Proposition a, and put $$ u = \begin{pmatrix} & & p\\ 1 & & \\ & 1 & \end{pmatrix}. $$ Then one easily checks that (at least after transposing to ensure the same order of the factors) the conditions in the proposition are satisfied.

The rest of the section in Pierce deals with the condition that $p$ is not a norm.

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    $\begingroup$ Your construction requires that $p$ be a positive integer (I think; I'm not sure what $\mathbb Z^+$ means). How is that arranged? (Oh, I guess just choose a suitable rational multiple of the original $u$ that will clear the denominator, and flip the sign if needed.) $\endgroup$ – LSpice Apr 15 at 13:52
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    $\begingroup$ Just wondering: the algebra $A$ has a "local Brauer invariant" $c_\ell(A) \in \tfrac{1}{3}\mathbf{Z} / \mathbf{Z}$ for each prime $\ell$, and these invariants uniquely determine $A$ up to isomorphism. Can we write this down explicitly in terms of $p$ and $L$? I'm guessing it is something like $v_\ell(p) / 3$ if $\ell$ is unramified in $L$. $\endgroup$ – David Loeffler Apr 15 at 14:07
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    $\begingroup$ @DavidLoeffler it is what you write if $\sigma$ is the Frobenius at $\ell$; it is $-v_\ell(p)/3$ if $\sigma$ is the inverse of $Frob_\ell$, and it is $0$ if $Frob_\ell$ is trivial. And I realise I should have written it this way: if $Frob_\ell = \sigma^k$ then the invariant is $kv_\ell(p)/3$. $\endgroup$ – Aurel Apr 15 at 20:04

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