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I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).
I need help with Exercise 1.1.4 (ii) (Chapter 1: Basics, page 13) concerning matrix rings ... ...
Exercise 1.1.4 (ii) (page 13) reads as follows:
View attachment 2984
I need help with Exercise (ii) above ... indeed I have not been able to progress beyond defining the terms of the problem as follows:Now, the centre of a matrix ring $$M_n (R)$$ over a ring $$R$$ is as follows:
$$ Z(M_n (R)) = \{ A \in M_n (R) \ | \ AM = MA \ \ \forall \ M \in M_n (R) \}$$
We need to show that
$$A \in Z(M_n (R)) \Longleftrightarrow A = aI \ \ \forall \ M \in M_n (R)$$
So, to begin, assume $$A \in Z(M_n (R))$$ ...
Then
$$A \in Z(M_n (R)) \Longrightarrow AM= MA$$
Now we know $$A$$ can be written
$$A = \sum_{h,i} a_{hi} e_{hi}$$ ...
BUT ... where to from here ...
Can someone please help?
***EDIT***
I have been reflecting on this exercise and the proof "the other way" seems easier - that is to show that:
$$ A = aI \text{ with a } \in Z(R) \Longrightarrow A \in Z(M_n (R)) $$
Assume $$A = aI$$ with $$a \in Z(R)$$
Then
$$A = aI \Longrightarrow AM = aIM = aM$$ for any $$M \in M_n (R)$$
Now $$aM = \begin{pmatrix} a m_{11} & a m_{12} & ... & ... & a m_{1n} \\ a m_{21} & a m_{22} & ... & ... & a m_{2n} \\ ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... \\ a m_{n1} & a m_{n2} & ... & ... & a m_{nn} \end{pmatrix}$$
But $$am_{ij} = m_{ij}a$$ since$$ a \in Z(R) $$
Thus $$aM = \begin{pmatrix} m_{11}a & m_{12}a & ... & ... & m_{1n} a \\ m_{21} a & m_{22} a & ... & ... & m_{2n} a \\ ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... \\ m_{n1} a & m_{n2} a & ... & ... & m_{nn} a \end{pmatrix} = Ma$$
Thus we have $$AM = aIM = aM = Ma = MIa $$
Can someone please critique this proof of the fact that:
$$ A = aI with a \in Z(R) \Longrightarrow A \in Z(M_n (R)) $$
Hope someone can help ... ...
Peter
I need help with Exercise 1.1.4 (ii) (Chapter 1: Basics, page 13) concerning matrix rings ... ...
Exercise 1.1.4 (ii) (page 13) reads as follows:
View attachment 2984
I need help with Exercise (ii) above ... indeed I have not been able to progress beyond defining the terms of the problem as follows:Now, the centre of a matrix ring $$M_n (R)$$ over a ring $$R$$ is as follows:
$$ Z(M_n (R)) = \{ A \in M_n (R) \ | \ AM = MA \ \ \forall \ M \in M_n (R) \}$$
We need to show that
$$A \in Z(M_n (R)) \Longleftrightarrow A = aI \ \ \forall \ M \in M_n (R)$$
So, to begin, assume $$A \in Z(M_n (R))$$ ...
Then
$$A \in Z(M_n (R)) \Longrightarrow AM= MA$$
Now we know $$A$$ can be written
$$A = \sum_{h,i} a_{hi} e_{hi}$$ ...
BUT ... where to from here ...
Can someone please help?
***EDIT***
I have been reflecting on this exercise and the proof "the other way" seems easier - that is to show that:
$$ A = aI \text{ with a } \in Z(R) \Longrightarrow A \in Z(M_n (R)) $$
Assume $$A = aI$$ with $$a \in Z(R)$$
Then
$$A = aI \Longrightarrow AM = aIM = aM$$ for any $$M \in M_n (R)$$
Now $$aM = \begin{pmatrix} a m_{11} & a m_{12} & ... & ... & a m_{1n} \\ a m_{21} & a m_{22} & ... & ... & a m_{2n} \\ ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... \\ a m_{n1} & a m_{n2} & ... & ... & a m_{nn} \end{pmatrix}$$
But $$am_{ij} = m_{ij}a$$ since$$ a \in Z(R) $$
Thus $$aM = \begin{pmatrix} m_{11}a & m_{12}a & ... & ... & m_{1n} a \\ m_{21} a & m_{22} a & ... & ... & m_{2n} a \\ ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... \\ m_{n1} a & m_{n2} a & ... & ... & m_{nn} a \end{pmatrix} = Ma$$
Thus we have $$AM = aIM = aM = Ma = MIa $$
Can someone please critique this proof of the fact that:
$$ A = aI with a \in Z(R) \Longrightarrow A \in Z(M_n (R)) $$
Hope someone can help ... ...
Peter
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