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

Question

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.

Answer 1

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.