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Wikipedia defines an order $\mathcal O$ of a finite type $\Bbb Q$-algebra $A$ to be a subring of $A$ satisfying the following properties. Here, by finite type $\Bbb Q$-algebra, I mean that $A=\Bbb Q[x_1,...,x_n]/I$.

  1. $\Bbb Q \mathcal O = A$
  2. $\mathcal O$ is a $\Bbb Z$-lattice in $A$.

But then the definition of a $\Bbb Z$-lattice is that $\mathcal O$ is a finitely generated, torsion-free $\Bbb Z$-module in some $FF(\Bbb Z)$-vector space (i.e. in a $\Bbb Q$-vectorspace). I take that the only reasonable vector space to parse here would be $\Bbb Q[e_1,...,e_n]$, where the basis vectors $e_i$ are identified with the $x_i$ generating $A$.

My issue is that Wikipedia goes on to paraphrase:

The last two conditions condition can be stated in less formal terms: Additively, $\mathcal{O}$ is a free abelian group generated by a basis for $A$ over $\mathbb{Q}$.

But we only required that $\mathcal O$ is a finitely generated module that spans $A$, over $\Bbb Q$! We never said that it was generated by a basis of $A$. For all I know, it could be generated by (finitely) many more elements.

Also, in the lattice entry, Wikipedia seems to have the wrong paraphrase of what is torsion-free. They say that torsion-free $R$-module $M$ means that no element of $R$ annihilates $M$. But this is a weaker definition of what is torsion-free, that is only equivalent the correct following definition in the case $R$ is a PID, right? A torsion free module means that $0$ is the only element of $M$ annihilated by a regular element of $R$.

Rodrigo
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1 Answers1

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It turns out that if $\mathcal{O} = \mathbb{Z}^n$, then $\mathbb{Q}\mathcal{O} = \mathbb{Q} \otimes_{\mathbb{Z}} \mathcal{O} = \mathbb{Q}^n$, so a $\mathbb{Z}$-basis for $\mathcal{O}$ is always a $\mathbb{Q}$-basis for $\mathbb{Q}\mathcal{O}$.

  • Yes, but that's not what I intended to say. I meant that a basis of $A$ that lies in $\mathcal O$ does not necessarily generate $\mathcal O$ – Rodrigo Mar 02 '14 at 19:35
  • That is true. But there are bases of $A$ that do span $\mathcal{O}$. –  Mar 02 '14 at 19:46
  • Okay, what about the comment that I made about the lattice entry. The paraphrase of "torsion-free" there is wrong, right? – Rodrigo Mar 02 '14 at 19:54
  • It's not true for PIDs either: e.g. no integer annihilates the $\mathbb{Z}$-module $\mathbb{Q}/\mathbb{Z}$. –  Mar 02 '14 at 20:08