Generalised Quaternion Algebra over K - Dauns Section 1-5 no 19

[Math Amateur]

In Dauns book "Modules and Rings", Exercise 19 in Section 1-5 reads as follows: (see attachment)

Let K be any ring with [FONT=MathJax_Main]1[FONT=MathJax_Main]∈[FONT=MathJax_Math]K whose center is a field and $$ 0 \ne x, 0 \ne y \in $$ center K are any elements.

Let I, J, and IJ be symbols not in K.

Form the set K[I, J] = K + KI + KJ + KIJ of all K linear combinations of {1, I, J, IJ}.

The following multiplication rules apply: (These also apply in my post re Ex 18!)

$$ I^2 = x, J^2 = y, IJ = -JI, cI = Ic, cIJ = JIc $$ for all $$ c \in K $$

Prove that the ring K[I, J] is isomorphic to a ring of $$ 2 \times 2 $$ matrices as follows:

$$ a + bJ \rightarrow \begin{pmatrix} a & by \\ \overline{b} & \overline{a} \end{pmatrix} $$ for all $$ a,b \in K $$

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I am not sure how to go about this ... indeed I am confused by the statement of the problem. My issue is the following:

Elements of K[I, J] are of the form r = a + bI + cJ + dIJ so we would expect an isomorphism of K[I, J] to specify how elements of this form are mapped into another ring, but we are only told how elements of the form s = a + bJ are mapped. ?

Can someone please clarify this issue and help me to get started on this exercise?

Peter

 

[Deveno]

K[I,J] is naturally isomorphic to K[J]:

a + bI + cJ + dIJ <--> (a+bI) + (c+dI)J

 

[Math Amateur]

K[I,J] is naturally isomorphic to K[J]:

a + bI + cJ + dIJ <--> (a+bI) + (c+dI)J

Thank you Deveno ... but still thinking

Mind you I (stupidly) missed the point that $$ a, b \in K $$ which your post has made me highly aware of - will go back to this now.

Peter

 

[Math Amateur]

Thank you Deveno ... but still thinking

Mind you I (stupidly) missed the point that $$ a, b \in K $$ which your post has made me highly aware of - will go back to this now.

Peter

Based on Deveno's help I have now worked through a good deal of the problem ... and I am satisfied I now understand it ...

For members of MHB interested in this exercise, I have attached a check of the fact $$ \phi (x, y) = \phi (x) \phi (y) $$ 'works' for $$ {element}_{11} $$ of the isomorphism $$ \phi $$ ...

I must say that John Dauns' is not afraid to set his students at Tulane University a fair amount of symbol manipulation ... :)

Peter

 

[Math Amateur]

Based on Deveno's help I have now worked through a good deal of the problem ... and I am satisfied I now understand it ...

For members of MHB interested in this exercise, I have attached a check of the fact $$ \phi (x, y) = \phi (x) \phi (y) $$ 'works' for $$ {element}_{11} $$ of the isomorphism $$ \phi $$ ...

I must say that John Dauns' is not afraid to set his students at Tulane University a fair amount of symbol manipulation ... :)

Peter

I just looked up Jon Dauns on the Internet and discovered that, very sadly, John Dauns passed away in 2009.

It is a sad loss to algebra ...

Peter