3
$\begingroup$

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 $0$.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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*}

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Yes, I did. I'll change them all to $\times$ to avoid confusion. $\endgroup$ – Richard D. James May 27 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.