Matrix Rings - Exercise 1.1.4 (ii) - Berrick and Keating (B&K) - page 12

[Math Amateur]

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

 

[Deveno]

It should be clear that if $A = aI$ that:

$AM = (aI)M = a(IM) = a(MI) = (aM)I$

now since $a \in Z(R)$, we have:

$\displaystyle aM = a\left(\sum_{h,i} a_{hi}e_{hi}\right) = \sum_{h,i} a(a_{hi}e_{hi})$

$\displaystyle = \sum_{h,i} (a_{hi}e_{hi})a = \left(\sum_{h,i} a_{hi}e_{hi}\right)a = Ma$.

Hence: $(aM)I = (Ma)I = M(aI) = MA$.

So the center of the matrix ring clearly contains all such matrices.

Of course, this is the "easy part".

To do the other way, I find it helpful to consider two cases.

Case 1: $A$ is diagonal, but not equal to $aI$ for any $a \in Z(R)$.

Case 1a: $A = bI$ for some $b \not \in Z(R)$. Then there is some $c \in R$ with $bc \neq cb$.

Then $A(cI) = (bI)(cI) = (bc)I \neq (cb)I = (cI)(bI) = (cI)A$, so $A$ does not commute with $cI$.

Case 1b: $A = \text{diag}(a_1,a_2,\dots,a_n)$ with $a_i \neq a_j$ for some $i \neq j$.

In this case, we have (I am going to use capital letters for the $e$ matrices to underscore that these are matrices, not ring-elements):

$A(E_{ij}) = a_jE_{ij}$ whereas:

$(E_{ij})A = a_iE_{ij}$, and these are unequal, so $A$ does not commute with $E_{ij}$.

Case 2: $A$ is not diagonal. Suppose that we have $a_{ij} \neq 0$ for $i \neq j$.

If we denote $A(E_{jk})$ by $B$, we have:

$\displaystyle b_{ik} = \sum_n a_{in}e_{nk} = a_{ij} \neq 0$.

Similarly, if we denote $(E_{jk})A$ by $C$, we have:

$\displaystyle c_{ik} = \sum_n e_{in}a_{nk} = 0$, since $e_{in} = 0$ for all $i \neq j$ and any $n$.

Since $B$ and $C$ differ in the $i,k$-entry, they cannot be equal. Hence $A$ does not commute with $E_{jk}$.

So ONLY those matrices of the form $aI$ are in the center (for every other matrix, we have at least one matrix it doesn't commute with).

Since the mapping $Z(R) \to Z(M_n(R))$ given by $a \mapsto aI$ is a ring-monomorphism, it is common practice to identify these two objects, and say $Z(R)$ IS the center $Z(M_n(R))$. This essentially allows us to obtain $M_n(R)$ as an extension ring of $Z(R)$, something that becomes important when $R$ is a field.

 

[Math Amateur]

It should be clear that if $A = aI$ that:

$AM = (aI)M = a(IM) = a(MI) = (aM)I$

now since $a \in Z(R)$, we have:

$\displaystyle aM = a\left(\sum_{h,i} a_{hi}e_{hi}\right) = \sum_{h,i} a(a_{hi}e_{hi})$

$\displaystyle = \sum_{h,i} (a_{hi}e_{hi})a = \left(\sum_{h,i} a_{hi}e_{hi}\right)a = Ma$.

Hence: $(aM)I = (Ma)I = M(aI) = MA$.

So the center of the matrix ring clearly contains all such matrices.

Of course, this is the "easy part".

To do the other way, I find it helpful to consider two cases.

Case 1: $A$ is diagonal, but not equal to $aI$ for any $a \in Z(R)$.

Case 1a: $A = bI$ for some $b \not \in Z(R)$. Then there is some $c \in R$ with $bc \neq cb$.

Then $A(cI) = (bI)(cI) = (bc)I \neq (cb)I = (cI)(bI) = (cI)A$, so $A$ does not commute with $cI$.

Case 1b: $A = \text{diag}(a_1,a_2,\dots,a_n)$ with $a_i \neq a_j$ for some $i \neq j$.

In this case, we have (I am going to use capital letters for the $e$ matrices to underscore that these are matrices, not ring-elements):

$A(E_{ij}) = a_jE_{ij}$ whereas:

$(E_{ij})A = a_iE_{ij}$, and these are unequal, so $A$ does not commute with $E_{ij}$.

Case 2: $A$ is not diagonal. Suppose that we have $a_{ij} \neq 0$ for $i \neq j$.

If we denote $A(E_{jk})$ by $B$, we have:

$\displaystyle b_{ik} = \sum_n a_{in}e_{nk} = a_{ij} \neq 0$.

Similarly, if we denote $(E_{jk})A$ by $C$, we have:

$\displaystyle c_{ik} = \sum_n e_{in}a_{nk} = 0$, since $e_{in} = 0$ for all $i \neq j$ and any $n$.

Since $B$ and $C$ differ in the $i,k$-entry, they cannot be equal. Hence $A$ does not commute with $E_{jk}$.

So ONLY those matrices of the form $aI$ are in the center (for every other matrix, we have at least one matrix it doesn't commute with).

Since the mapping $Z(R) \to Z(M_n(R))$ given by $a \mapsto aI$ is a ring-monomorphism, it is common practice to identify these two objects, and say $Z(R)$ IS the center $Z(M_n(R))$. This essentially allows us to obtain $M_n(R)$ as an extension ring of $Z(R)$, something that becomes important when $R$ is a field.

Thanks Deveno ... Appreciate your help ...

Will now be working through the details of your post ...

Thanks again,

Peter