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

#### Peter

##### Well-known member
MHB Site Helper
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][FONT=MathJax_Main]∈[/FONT][FONT=MathJax_Math]K[/FONT] whose center is a field and $$\displaystyle 0 \ne x, 0 \ne y \in$$ 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!)

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

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

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

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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

Last edited:

#### Deveno

##### Well-known member
MHB Math Scholar
K[I,J] is naturally isomorphic to K[J]:

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

#### Peter

##### Well-known member
MHB Site Helper
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 $$\displaystyle a, b \in K$$ which your post has made me highly aware of - will go back to this now.

Peter

#### Peter

##### Well-known member
MHB Site Helper
Thank you Deveno ... but still thinking

Mind you I (stupidly) missed the point that $$\displaystyle a, b \in K$$ 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 $$\displaystyle \phi (x, y) = \phi (x) \phi (y)$$ 'works' for $$\displaystyle {element}_{11}$$ of the isomorphism $$\displaystyle \phi$$ ...

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

#### Peter

##### Well-known member
MHB Site Helper
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 $$\displaystyle \phi (x, y) = \phi (x) \phi (y)$$ 'works' for $$\displaystyle {element}_{11}$$ of the isomorphism $$\displaystyle \phi$$ ...

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