# Every order in a finite-dimensional $\mathbb{Q}$-algebra is contained in a maximal order

Definition. A lattice in a finite-dimensional $$\mathbb{Q}$$-algebra $$V$$ is a finitely generated $$\mathbb{Z}$$-submodule $$\mathcal{L} \subset V$$ such that $$\mathcal{L}\mathbb{Q}=V$$ (i.e., $$\mathcal{L}$$ spans $$V$$ over $$\mathbb{Q}$$).

We have the following useful characterization of lattices in finite-dimensional algeberas over $$\mathbb{Q}$$:

Let $$V$$ be a finite-dimensional vector space over $$\mathbb{Q}$$. $$\mathcal{L} \subset V$$ is a lattice if and only if $$\mathcal{L}=\mathbb{Z} x_1 \oplus \ldots \oplus \mathbb{Z} x_n$$ where $$x_1,\ldots,x_n$$ is a $$\mathbb{Q}$$-basis for $$V$$.

Definition. An order $$\mathcal{O}$$ in a finite-dimensional $$\mathbb{Q}$$-algebra $$B$$ is a lattice that is also a subring having $$1\in B$$. $$\mathcal{O}$$ is called maximal if it's not contained in another order in $$B$$.

I need to prove the following statement:

Every order of a finite-dimensional $$\mathbb{Q}$$-algebra is contained in a maximal order.

Attempt: Let $$O$$ be any order of a finite-dimensional $$\mathbb{Q}$$-algebra $$B$$ and define $$S=\lbrace M \subset B \text{ an order containing } O \rbrace$$. Indeed, $$S\neq \emptyset$$ because $$O\in S$$. Let $$\lbrace M_i \rbrace_{i\in I}$$ be a chain of orders in $$S$$ and set $$N=\bigcup_{i\in I}M_i$$. Indeed, $$N$$ is a subring of $$B$$ having $$1$$. How can I prove that $$N$$ is a lattice so that $$N$$ turns to be an order and we can therefore apply Zorn's Lemma ?!.

I appreciate any help in completing this proof or even in beginning a new one. Thanks in advance.

• Interestingly, the union of an ascending chain of lattices is not necessarily a lattice. So somehow we have to use that they are all subrings with the same $1$. Probably, Noetherian-ness of $\mathbb Z$ will be used too ... Mar 20 at 23:46

The statement is not correct in this level of generality: it is necessary (and also sufficient) that $$B$$ is a semi-simple $$\mathbb{Q}$$-algebra, not just any finite-dimensional algebra.

Indeed, suppose that $$B$$ is not semi-simple (but is a finite-dimensional $$\mathbb{Q}$$-algebra), and let $$O\subset B$$ be an order. We will construct an order $$O'\subset B$$ strictly containing $$O$$ (therefore there is no maximal order in $$B$$).

Since $$B$$ is not semi-simple, it has a non-zero nilpotent two-sided ideal $$J$$ (namely its Jacobson radical). Let us write $$J^{n+1}=0$$ for some $$n>0$$ with $$J^n\neq 0$$. Then let $$O_k=O\cap J^k$$ for any $$k\in \mathbb{N}$$. Clearly, $$O_k$$ is a lattice in the $$\mathbb{Q}$$-vector space $$J^k$$, with in particular $$O_0=O$$, and $$O_k\cdot O_{k'}\subset O_{k+k'}$$. Now write $$O' = \sum_{k=0}^n 2^{-k}O_k\subset B.$$ It is clear that $$O'$$ is a subring of $$B$$, and since the sum is finite it is a sublattice in $$B$$, therefore it is an order. Furthermore, by construction $$O\subset O'$$, but also $$2^nO'\subset O$$, so since $$O$$ is a lattice we cannot have $$O'=O$$.

To see that on the other hand when $$B$$ is semi-simple, then indeed any order is contained in a maximal one, then you can finish your proof by noticing that every element in $$N$$ is integral over $$\mathbb{Z}$$, and that $$N$$ generates $$B$$ as a $$\mathbb{Q}$$-vector space.

In fact, first prove that if $$Tr: B\to \mathbb{Q}$$ is the usual trace of a finite-dimensional algebra, then $$Tr(N)\subset \mathbb{Z}$$. Then let $$(x_i)$$ be a basis of $$B$$ as a $$\mathbb{Q}$$-vector space with $$x_i\in N$$, and write $$d=\det(Tr(x_ix_j))$$. The fact that $$B$$ is semi-simple (in characteristic $$0$$) implies that the trace form is non-degenerate, and therefore $$d\neq 0$$. Then try to prove that $$N\subset d^{-1}\bigoplus \mathbb{Z}x_i$$, and that therefore $$N$$ is a lattice.

• Excuse me. Does the proof of this fact exist in some paper or some book?!. I'm writing a scientific paper and I just want to cite the final conclusion. @Captain Lama May 18 at 22:16
• Yes, everything I said is done in Maximal Orders by Reiner. May 19 at 7:14
• @CaptainLama that is a nearly 400-page book. It would be nice to include in your answer an indication o where the argument can be found in that book.
– KCd
May 19 at 21:33
• Could you please mention the name of the book, the chapter, the number of the result?!. Thanks for your help. @CaptainLama May 20 at 19:25