next up previous
下一頁: The Hamilton quaternions 上一頁: 一些 Noncommutative Ring 前一頁: 一些 Noncommutative Ring

Matrix ring M2(R)

令 R 是一個 commutative ring with 1. 考撰隻X M2(R) 是所有係數在 R 的 2×2 矩骨狾赤熄隻X, 也就是說 M2(R) 中的元素都是

$\displaystyle \left(\vphantom{\begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} a & b \\ c & d \\ \end{array}$$\displaystyle \left.\vphantom{\begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right)$

這種形式其中 a, b, c, d $ \in$ R. 因為 R 是一個 ring 我們可以定 M2(R) 中的加法和乘法就是一般矩高漸[法和乘法, 即:

$\displaystyle \left(\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} a & b \\ c & d \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right)$ + $\displaystyle \left(\vphantom{ \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}%% }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{cc} a+a' & b+b' \\ c+c' & d+d' \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} a+a' & b+b' \\ c+c' & d+d' \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a+a' & b+b' \\ c+c' & d+d' \\ \end{array}%% }\right)$    和

$\displaystyle \left(\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} a & b \\ c & d \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right)$ . $\displaystyle \left(\vphantom{ \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}%% }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{cc} a\cdot a'+b\cdot c' & a\cdot ... ... d' \\ c\cdot a'+d\cdot c' & c\cdot b'+d\cdot d' \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} a\cdot a'+b\cdot c' & a\cdot b'+b\cdot d' \\ c\cdot a'+d\cdot c' & c\cdot b'+d\cdot d' \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a\cdot a'+b\cdot c' & a\cdot ... ... d' \\ c\cdot a'+d\cdot c' & c\cdot b'+d\cdot d' \\ \end{array}%% }\right)$.

因為 R 是一個 ring with 1, 不難發畦H上的加法和乘法使得 M2(R) 成為一個 ring, 而且 $ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$ 和 $ \left(\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}%% }\right.$$ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}%% }\right)$ 分別是 M2(R) 的 0 和 1. 所以說 M2(R) 是一個 ring with 1. 不過即使 R 是 commutative, M2(R) 也不會是 commutative ring. 這可由以下的例子看出:

$\displaystyle \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}%% }\right)$ . $\displaystyle \left(\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}%% }\right)$    但是    $\displaystyle \left(\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$ . $\displaystyle \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}%% }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$.

(注意: 當我們要說明一個 ring R 是 commutative 時, 我們必須證明對任意的 a, b $ \in$ R 皆有 a . b = b . a. 不過若要說明 R 是 noncommutative 時, 只要找到一組 a, b $ \in$ R 使得 a . b$ \ne$b . a 即可.)

從上面的式子我們知道 $ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}%% }\right)$ 和 $ \left(\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$ 是 M2(R) 的 zero-divisor. 同時在這個例子裡我們也發畢b一個 noncommutative ring 中是有可能發生 a . b = 0 但 b . a$ \ne$ 0 的眸H.

接下來我們想找到 M2(R) 中所有的 zero-divisor 和 unit. 首先觀察以下的式子:

$\displaystyle \left(\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}<tex2html_comment_mark>86 }\right.$$\displaystyle \begin{array}{cc} a & b \\ c & d \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}<tex2html_comment_mark>86 }\right)$ . $\displaystyle \left(\vphantom{ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}<tex2html_comment_mark>88 }\right.$$\displaystyle \begin{array}{cc} d & -b \\ -c & a \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}<tex2html_comment_mark>88 }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{cc} a\cdot d-b\cdot c & 0 \\ 0 & a\cdot d-b\cdot c \\ \end{array}<tex2html_comment_mark>90 }\right.$$\displaystyle \begin{array}{cc} a\cdot d-b\cdot c & 0 \\ 0 & a\cdot d-b\cdot c \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a\cdot d-b\cdot c & 0 \\ 0 & a\cdot d-b\cdot c \\ \end{array}<tex2html_comment_mark>90 }\right)$. (5.2)

要注意我們需要 R 是 commutative 式子 (5.2) � 會對. 大家應該對 a . d - b . c 這個值不陌生, 它是 $ \left(\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right.$$ \begin{array}{cc} a & b \\ c & d \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right)$ 的 determinant. 通常給一矩� A $ \in$ M2(R) 我們用 det(A) 表示其 determinant. 由於 R 是一個 ring, 所以對任意的 A $ \in$ M2(R), 我們都可得 det(A) $ \in$ R. Determinant 還有以下這個重要的性質:

det(A . B) = det(A) . det(B),    $\displaystyle \forall$ A, B $\displaystyle \in$ M2(R). (5.3)

到底 M2(R) 中有哪些 zero-divisor 呢? 同學可能想到 determinant 為 0 的元素. 沒錯, 當 A = $ \left(\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right.$$ \begin{array}{cc} a & b \\ c & d \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right)$$ \ne$$ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$ 但 det(A) = 0 時, 由於 $ \left(\vphantom{ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}%% }\right.$$ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}%% }\right)$$ \ne$$ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$, 由式子 (5.2) 知 A 是一個 zero-divisor.

還有沒有其他的 zero-divisor 呢? 其實當 det(A) 是 R 的 zero-divisor 時, A 也會是 M2(R) 的 zero-divisor. 這是因為如果 A = $ \left(\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right.$$ \begin{array}{cc} a & b \\ c & d \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right)$$ \ne$$ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$ 且 det(A) = $ \alpha$ 是 R 的一個 zero-divisor. 設 $ \beta$$ \ne$ 0 是 R 中一元素滿足 $ \alpha$ . $ \beta$ = 0. 有以下兩種可能發生:

(1) a . $ \beta$, b . $ \beta$, c . $ \beta$ 和 d . $ \beta$ 都等於 0: 此時令 B = $ \left(\vphantom{ \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}%% }\right.$$ \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}%% }\right)$, 如此一來因 $ \beta$$ \ne$ 0, 所以 B$ \ne$$ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$, 但是

A . B = $\displaystyle \left(\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} a & b \\ c & d \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right)$ . $\displaystyle \left(\vphantom{ \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}%% }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{cc} a\cdot\beta & b\cdot\beta \\ c\cdot\beta & d\cdot\beta \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} a\cdot\beta & b\cdot\beta \\ c\cdot\beta & d\cdot\beta \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a\cdot\beta & b\cdot\beta \\ c\cdot\beta & d\cdot\beta \\ \end{array}%% }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$.

因此在這個情形時, A 是 M2(R) 的一個 zero-divisor.

(2) a . $ \beta$, b . $ \beta$, c . $ \beta$ 和 d . $ \beta$ 不全為 0: 則我們考�

B = $\displaystyle \left(\vphantom{ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} d & -b \\ -c & a \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}%% }\right)$ . $\displaystyle \left(\vphantom{ \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}%% }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{cc} d\cdot\beta & -b\cdot\beta \\ -c\cdot\beta & a\cdot\beta \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} d\cdot\beta & -b\cdot\beta \\ -c\cdot\beta & a\cdot\beta \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} d\cdot\beta & -b\cdot\beta \\ -c\cdot\beta & a\cdot\beta \\ \end{array}%% }\right)$$\displaystyle \ne$$\displaystyle \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$.

然而

A . B = $\displaystyle \left(\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} a & b \\ c & d \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right)$ . $\displaystyle \left(\vphantom{ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} d & -b \\ -c & a \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}%% }\right)$ . $\displaystyle \left(\vphantom{ \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}%% }\right)$,

由式子 (5.2) 知

A . B = $\displaystyle \left(\vphantom{ \begin{array}{cc} \alpha & 0 \\ 0 & \alpha \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} \alpha & 0 \\ 0 & \alpha \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} \alpha & 0 \\ 0 & \alpha \\ \end{array}%% }\right)$ . $\displaystyle \left(\vphantom{ \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} \beta & 0 \\ 0 & \beta \\ \end{array}%% }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{cc} \alpha\cdot\beta & 0 \\ 0 & \alpha\cdot\beta \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} \alpha\cdot\beta & 0 \\ 0 & \alpha\cdot\beta \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} \alpha\cdot\beta & 0 \\ 0 & \alpha\cdot\beta \\ \end{array}%% }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$.

所以在這個情況 A 還是 M2(R) 的一個 zero-divisor.

那麼當 A = $ \left(\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right.$$ \begin{array}{cc} a & b \\ c & d \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right)$ 且 det(A) = $ \alpha$ 不是 R 的 zero-divisor 時又會怎樣呢? 假設存在 B = $ \left(\vphantom{ \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}%% }\right.$$ \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}%% }\right)$$ \ne$$ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$ (也就是說 a', b', c' 和 d' 不全為 0) 滿足 A . B = $ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$. 此時考� C = $ \left(\vphantom{ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}%% }\right.$$ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} d & -b \\ -c & a \\ \end{array}%% }\right)$, 則由式子 (5.2) 知

(C . A) . B = $\displaystyle \left(\vphantom{ \begin{array}{cc} \alpha & 0 \\ 0 & \alpha \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} \alpha & 0 \\ 0 & \alpha \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} \alpha & 0 \\ 0 & \alpha \\ \end{array}%% }\right)$ . $\displaystyle \left(\vphantom{ \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} a' & b' \\ c' & d' \\ \end{array}%% }\right)$ = $\displaystyle \left(\vphantom{ \begin{array}{cc} \alpha\cdot a' & \alpha\cdot b' \\ \alpha\cdot c' & \alpha\cdot d' \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} \alpha\cdot a' & \alpha\cdot b' \\ \alpha\cdot c' & \alpha\cdot d' \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} \alpha\cdot a' & \alpha\cdot b' \\ \alpha\cdot c' & \alpha\cdot d' \\ \end{array}%% }\right)$.

因為 $ \alpha$ 不是 zero-divisor 且 a', b', c' 和 d' 不全為 0, 所以知 $ \alpha$ . a', $ \alpha$ . b', $ \alpha$ . c' 和 $ \alpha$ . d' 不全為 0. 也就是說 (C . A) . B$ \ne$$ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$. 這和

(C . A) . B = C . (A . B) = $\displaystyle \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$

相矛盾, 所以不可能找到 B$ \ne$$ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$ 滿足 A . B = $ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$. 同理可知不可能找到 B$ \ne$$ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$ 滿足 B . A = $ \left(\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right.$$ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array}%% }\right)$. 所以 A 絕對不會是 M2(R) 的一個 zero-divisor. 因此我們得證:

Proposition 5.5.1   若 R 是一個 commutative ring 且 A $ \in$ M2(R), 則 A 是 M2(R) 的一個 zero-divisor 若且唯若 det(A) = 0 或 det(A) 是 R 的一個 zero-divisor.

由 Proposition 5.5.1 我們知在 M2($ \mathbb {Z}$) 和 M2($ \mathbb {Q}$) 中 determinant 為 0 的矩健|是 zero-divisor, 而 determinant 不為 0 的矩偌N不會是 zero-divisor.

當 R 是 commutative ring with 1 時 M2(R) 會有哪些 unit 呢? 我們有以下的結果:

Proposition 5.5.2   若 R 是一個 commutative ring with 1 且 A $ \in$ M2(R), 則 A 是 M2(R) 的一個 unit 若且唯若 det(A) 是 R 的一個 unit.

証 明. 假設 A 是 M2(R) 的一個 unit, 則存在 B $ \in$ M2(R) 滿足

A . B = B . A = $\displaystyle \left(\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}%% }\right)$.

利用式子 (5.3) 得

det(A) . det(B) = det(B) . det(A) = 1.

然而 det(A), det(B) $ \in$ R, 故得 det(A) 是 R 的一個 unit.

反之, 若 A = $ \left(\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right.$$ \begin{array}{cc} a & b \\ c & d \\ \end{array}$$ \left.\vphantom{ \begin{array}{cc} a & b \\ c & d \\ \end{array}%% }\right)$ 且 det(A) = $ \alpha$ 是 R 的一個 unit, 則考�

B = $\displaystyle \left(\vphantom{ \begin{array}{cc} \alpha^{-1}\cdot d & \alpha^{... ...-b) \\ \alpha^{-1}\cdot(-c) & \alpha^{-1}\cdot a \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} \alpha^{-1}\cdot d & \alpha^{-1}\cdot (-b) \\ \alpha^{-1}\cdot(-c) & \alpha^{-1}\cdot a \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} \alpha^{-1}\cdot d & \alpha^{... ...-b) \\ \alpha^{-1}\cdot(-c) & \alpha^{-1}\cdot a \\ \end{array}%% }\right)$.

因 $ \alpha^{-1}_{}$ $ \in$ R, 我們知 B $ \in$ M2(R). 利用式子 (5.2), 可得

A . B = B . A = $\displaystyle \left(\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}%% }\right.$$\displaystyle \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}%% }\right)$.

因此 A 是 M2(R) 的一個 unit. $ \qedsymbol$

由 Proposition 5.5.2 知在 M2($ \mathbb {Z}$) 中惟有 determinant 是 ±1 的矩陘 會是 unit, 而在 M2($ \mathbb {Q}$) 中所有 determinant 不是 0 的矩偃ㄦ|是 unit.

next up previous
下一頁: The Hamilton quaternions 上一頁: 一些 Noncommutative Ring 前一頁: 一些 Noncommutative Ring
Administrator 2005-06-18