Let $R$ be a commutative ring, $D$ be a finite dimensional associative $R$-algebra with unit.
Then for $\alpha\in D$, left multiplication of $\alpha$, $L_{\alpha}: D\to D$ is $R$-linear map, and we get a norm $N(\alpha):=\det L_{\alpha}\in R$.
If $\alpha\in D$ is invertible (i.e. there exists $\beta\in D$ s.t. $\alpha\beta=\beta\alpha=1$), then $N(\alpha)\in R$ is invertible.
Now is the above converse true?
I know it's true if $D=M_n(R)$ of a matrix group over $R$. But is it true for generality?
Say that $N(\alpha) \neq 0$. Then $\text{det } L_\alpha \neq 0$, so $L_\alpha^{-1}$ exists as a linear transformation. Now the Cayley-Hamilton theorem tells us that $L_\alpha^{-1}$ can be written as a polynomial in $L_\alpha$. That is, for some polynomial $p \in R[x]$, we know $L_\alpha^{-1} = p \left ( L_\alpha \right )$.
Now it's easy to see that $p(\alpha) \in D$ should be $\alpha^{-1}$. Indeed one can check that $\alpha \cdot p(\alpha) = 1 = p(\alpha) \cdot \alpha$ (because it's true for $L_\alpha$) so that $\alpha^{-1} = p(\alpha) \in D$.
I hope this helps ^_^