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Let $A$ be a Baer $*$-ring. Let us denote $B$ by the space of all upper triangular matrices $\left(\begin{array}{cc} a_1& a_2 \\ 0 & a_4 \end{array}\right)$ where $a_i$'s are in $A$. Is $B$ a Baer *-ring too?

As for the involution on $B$, I mean any involution that makes the mapping $$a\to \left(\begin{array}{cc} a& 0 \\ 0 & 0 \end{array}\right)$$ forms an embedding.

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There is not such involution on B. In general, there exist two involution on B which are not satisfy in this condition.

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    $\begingroup$ Could you please explain a bit more? $\endgroup$ – Ali Bagheri Aug 3 '18 at 8:52
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There exist two involutions on ring $B$ as follows:

$\begin{pmatrix} a&b\\ 0&c \end{pmatrix}^{\ast_1}= \begin{pmatrix} c^{\ast}&b^{\ast}\\ 0&a^{\ast} \end{pmatrix},$ and $\begin{pmatrix} a&b\\ 0&c \end{pmatrix}^{\ast_2}= \begin{pmatrix} c^{\ast}&-b^{\ast}\\ 0&a^{\ast} \end{pmatrix}$.

An involution is proper if $aa^{\ast}=0$, then $a=0$. If the $\ast$-ring $R$ is Baer $\ast$, then $\ast$ is a involution proper.

The involutions $\ast_1$ and $\ast_2$ are not proper, because $\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}^{\ast_1}=\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}^{\ast_2}=\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}\begin{pmatrix} 0&0\\ 0&1 \end{pmatrix}=0$ and $\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}\not=0$. Hence $B=T_2(A)$ is not a Baer $\ast$-ring.

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  • $\begingroup$ This does not explain why $M_2(B)$ is or is not a Baer ring $\endgroup$ – Yemon Choi Aug 4 '18 at 20:32

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