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Thread: Center of a matrix ring

  1. #1
    Junior Member Greg98's Avatar
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    Center of a matrix ring

    Hello,
    the task is to prove that center of a matrix ring $\displaystyle R=Mat_2(\mathbb{R})$ is the scalar matrices $\displaystyle \alpha I, \alpha \in \mathbb{R}$.

    I found two more general proofs:
    Center of an Algebra Abstract Nonsense
    The center of a matrix ring over a commutative ring is precisely the scalar matrices Project Crazy Project

    However I didn't understand those totally, so I coudn't reduce them to this simpler case. But I think the main point is:
    Let's assume that $\displaystyle M \in Z(Mat_2(\mathbb{R}))$ and$\displaystyle E=\left(\begin{array}{cc}1&0\\0&0\end{array}\right )$.
    Matrix multiplication $\displaystyle ME$ is commutative ($\displaystyle ME=EM$) iff $\displaystyle M$ is in form $\displaystyle \left(\begin{array}{cc}a_1&0\\0&a_2\end{array}\rig ht)$.

    I don't know how to continue etc., so any help is appreciated. Thanks!
    Last edited by Greg98; Mar 28th 2011 at 09:59 AM.
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  2. #2
    Junior Member
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    [QUOTE=Greg98;634437]Hello,
    the task is to prove that center of a matrix ring $\displaystyle R=Mat_2(\mathbb{R})$ is the scalar matrices $\displaystyle \alpha I, \alpha \in \mathbb{R}$.


    If $\displaystyle M=\left(\begin{array}{cc}m_{1}&m_{2}\\m_{3}&m_{4}\ end{array}\right) \in Z(Mat_2(\mathbb{R}))$ then $\displaystyle M \cdot E = E\cdot M, \forall E \in Mat_2(\mathbb{R})$.

    If you take $\displaystyle E=\left(\begin{array}{cc}1&0\\0&0\end{array}\right )$ then you have: $\displaystyle M \cdot E = \left(\begin{array}{cc}m_{1}&0\\m_{3}&0\end{array} \right)$ and $\displaystyle E \cdot M= \left(\begin{array}{cc}m_{1}&m_{2}\\0&0\end{array} \right)$.

    So, $\displaystyle m_{2} = m_{3} = 0$.

    If $\displaystyle E=\left(\begin{array}{cc}0&1\\0&0\end{array}\right )$ then you have $\displaystyle m_{1} = m_{4} =m$ $\displaystyle \Rightarrow$ $\displaystyle M=\left(\begin{array}{cc}m&0\\0&m\end{array}\right ) = m \cdot I.$
    Last edited by zoek; Mar 28th 2011 at 11:48 AM.
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