# Jacobson radical of upper triangular matrix ring

I do not follow this solution for the Jacobson radical of the upper triangular matrix ring $$U_2(\mathbb{Z}_{63})$$.

NB: in the solution, the result that “$$1-ra$$ is a unit for all $$r\in R$$ iff $$a$$ belongs to every maximal left ideal in a ring $$R$$“ is being used

In the last paragraph it says it is sufficient for $$(1-ra),(1-tc)$$ to belong to $$\mathbb{Z}_{63} \setminus \{0\}$$. How can this be the case when $$\mathbb{Z}_{63}$$ is a domain? For example $$7,9 \in \mathbb{Z}_{63}$$ but their product is $$0$$.

• The only mistake in this solution is what you asked for: $\Bbb Z_{63}^*=\Bbb Z_{63}\setminus \{0\}$ is clearly false. But the rest are valid. – Berci May 27 at 0:40
• Why does it want (1-ra),(1-tc) to be units? – Group23 May 27 at 0:41

First, as pointed out in the comments, it is not true that $$\newcommand{\Z}{\mathbb{Z}}(\mathbb{Z}/63\mathbb{Z})^\times = {\mathbb{Z}/63\mathbb{Z}} \setminus \{0\}$$. As you rightly pointed out, $$7$$ and $$9$$ are zero divisors so they can't be units.
However, this doesn't matter for the solution. Basically it applies the result you stated, \begin{align} \label{radical} \tag{1} \DeclareMathOperator{\Jac}{Jac} a \in \Jac(R) \iff 1-ra \in R^\times \, \text{for all r \in R} , \end{align} to the ring of $$2 \times 2$$ upper triangular matrices and then to $$\mathbb{Z}/63\mathbb{Z}$$ itself. (Here $$\Jac(R)$$ is the Jacobson radical of $$R$$.) A square matrix over a commutative ring $$R$$ is invertible iff its determinant is a unit in $$R$$. So assuming the matrix $$\DeclareMathOperator{\R}{\mathcal{R}} M = \begin{pmatrix} 1 - ra & -(rb+sc)\\ 0 & 1 - tc \end{pmatrix}$$ is invertible, then its determinant $$(1 - ra) (1 - tc) \in (\mathbb{Z}/63\mathbb{Z})^\times$$, and you can show that this means $$1 - ra, 1 - tc \in (\mathbb{Z}/63\mathbb{Z})^\times$$ individually, too. Now we apply (\ref{radical}) to $$a$$ and $$c$$: since $$r, t \in \Z/63\Z$$ were arbitrary, what we've shown implies that $$a,c \in \Jac(\Z/63\Z) = 21 \Z$$. Moreover, both the statement about invertible matrices and (\ref{radical}) were equivalences, so the converse holds, too. Thus we've shown $$\Jac(\R) = \begin{pmatrix} 21\Z/63\Z & \Z/63\Z\\ 0 & 21\Z/63\Z \end{pmatrix} \, ,$$ where $$\R$$ is the ring of $$2 \times 2$$ upper triangular matrices over $$\Z/63\Z$$.
As an aside, I find doing both implications of a proof at once a little sneaky, so as a summary: \begin{align*} \begin{pmatrix} a & b\\ 0 & c \end{pmatrix} \in \Jac(\R) &\iff \begin{pmatrix} 1 - ra & -(rb+sc)\\ 0 & 1 - tc \end{pmatrix} \in \R^\times \ \text{for all r,s,t, \in \Z/63\Z}\\ &\iff (1 - ra)(1 - tc) \in (\Z/63\Z)^\times \ \text{for all r,t \in \Z/63\Z}\\ &\iff 1 - ra, 1 - tc \in (\Z/63\Z)^\times \ \text{for all r,t \in \Z/63\Z}\\ &\iff a, c \in \Jac(\Z/63\Z) = 21\Z/63\Z \, . \end{align*}
• In the paragraph before the end you write $(1-ra)(1-tc) \in \mathbb{Z}_{63}^*$. Do you mean $\in \mathbb{Z}_{63}^X$ – Group23 May 27 at 10:30
• Both $R^\times$ and $R^*$ are notations for the group of units of $R$. – Richard D. James May 27 at 16:28
• Yes, I did. I'll change them all to $\times$ to avoid confusion. – Richard D. James May 27 at 19:52