Is $\mathbb Z_3+\mathbb Z_3\mathbf i+\mathbb Z_3\mathbf j+\mathbb Z_3\mathbf k$ a local ring?

Question

I consider the quaternion division ring on $\mathbb Q_3$: that is $$\mathbb H_{\mathbb Q_3}=\{a+b\mathbf i+c\mathbf j+d\mathbf k \mid a,b,c,d\in\mathbb Q_3\}$$ with $\mathbf i^2=\mathbf j^2=\mathbf k^2=\mathbf i\mathbf j\mathbf k=-1$. Is the subring $$\mathbb Z_3+\mathbb Z_3\mathbf i+\mathbb Z_3\mathbf j+\mathbb Z_3\mathbf k$$ a noncommutative local ring, i.e., does it have a unique maximal left ideal (or equivalently a unique maximal right ideal)?

Thanks in advance for any help.

Answer 1

Since $-1$ is a unit in $\mathbb{Q}_3$, then the Hilbert symbol is trivial: $(-1, -1)_{\mathbb{Q}_3} = 1$. This means that $\mathbb{H}_{\mathbb{Q}_3}$ is split, i.e., $\mathbb{H}_{\mathbb{Q}_3} \cong M_2(\mathbb{Q}_3)$, the ring of $2 \times 2$ matrices. (So in particular it is not a division ring.) Then $\mathbb Z_3+\mathbb Z_3\mathbf i+\mathbb Z_3\mathbf j+\mathbb Z_3\mathbf k \cong M_2(\mathbb{Z}_3)$, which indeed has the unique maximal ideal $M_2(3 \mathbb{Z}_3)$.

For reference, see $\S12.4$ of this book.