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 $0$.

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    $\begingroup$ 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. $\endgroup$ – Berci May 27 at 0:40
  • $\begingroup$ Why does it want (1-ra),(1-tc) to be units? $\endgroup$ – 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*}

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  • $\begingroup$ In the paragraph before the end you write $(1-ra)(1-tc) \in \mathbb{Z}_{63}^*$. Do you mean $\in \mathbb{Z}_{63}^X$ $\endgroup$ – Group23 May 27 at 10:30
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    $\begingroup$ Both $R^\times$ and $R^*$ are notations for the group of units of $R$. $\endgroup$ – Richard D. James May 27 at 16:28
  • $\begingroup$ So when you used both you meant the same thing? $\endgroup$ – Group23 May 27 at 19:49
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    $\begingroup$ Yes, I did. I'll change them all to $\times$ to avoid confusion. $\endgroup$ – Richard D. James May 27 at 19:52

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