# M_n Matrix ring

• Sep 27th 2008, 06:07 PM
roporte
M_n Matrix ring
Let $\displaystyle M_n(A)$ be the ring of matrixes with elements from the ring $\displaystyle A$. $\displaystyle T_n(A) \subset M_n(A)$ the set of matrixes in which all elements below the diagonal are 0. $\displaystyle T'_n(A) \subset T_n(A)$ the set of matrixes with the elements in the principal diagonal are 0. $\displaystyle D_n(A)$ the set of diagonal matrixes.

Prove that,

$\displaystyle T'_n(A) \cong T_n(A)/D_n(A)$
• Sep 27th 2008, 06:20 PM
ThePerfectHacker
Quote:

Originally Posted by roporte
Let $\displaystyle M_n(A)$ be the ring of matrixes with elements from the ring $\displaystyle A$. $\displaystyle T_n(A) \subset M_n(A)$ the set of matrixes in which all elements below the diagonal are 0. $\displaystyle T'_n(A) \subset T_n(A)$ the set of matrixes with the elements in the principal diagonal are 0. $\displaystyle D_n(A)$ the set of diagonal matrixes.

Prove that,

$\displaystyle T'_n(A) \cong T_n(A)/D_n(A)$

I do it for $\displaystyle n=3$ it will be clear.

Define the mapping $\displaystyle \begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{bmatrix} \mapsto \begin{bmatrix} 0 & b & c \\ 0 & 0 & e \\ 0 & 0 & 0 \end{bmatrix}$

Then the kernel of this mapping ($\displaystyle \theta: T_n(A) \to T'_n(A)$) is $\displaystyle D_n$ and range is $\displaystyle T'_n(A)$.

By the fundamental isomorphism theorem,
$\displaystyle T'_n(A) \simeq T_n(A)/D_n(A)$