How to find an $R$-order $\Lambda$, such that $A=k\Lambda$?

Question

Let $R$ be a Noetherian integral domain with quotient field $k$.

An $R$-order $\Lambda$ is a finitely generated torsionless non-zero $R$-module, which is at the same time an $R$-algebra.

Why is there an $R$-order for every finite-dimensional $k$-algebra $A$, i.e. why is there for every arbitrary, but fixed $k$-algebra $A$ an $R$-order $\Lambda$, such that $A=k\Lambda$?

Thanks for the help!

Answer 1

Choose a $k$-basis $\{1=a_1,a_2,\dots,a_d\}$ of $A$ and let $V=\sum_iRa_i$ be the $R$-submodule of $A$ generated by $\{a_i\}$.

Let $$\Lambda=\left\{a\in A \vert aV\subseteq V\right\}.$$

Then $\Lambda$ is clearly a subring of $A$ and an $R$-algebra, and since $1\in V$, $\Lambda\subseteq V$.

For any $a\in A$ let $aa_i=\sum_j\mu_{ij}a_j$, and let $\mu\in R$ be the product of the denominators of all the $\mu_{ij}$. Then $\mu aa_i\in V$ for all $i$, and so $\mu a\in\Lambda$. So $k\Lambda=A$.

Since $\Lambda$ is an $R$-submodule of the finitely generated $R$-module $V$, and $R$ is Noetherian, $\Lambda$ is also finitely generated as an $R$-module.

So $\Lambda$ is an $R$-order of $A$.