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Generalised Quaternion Algebra over K - Dauns Section 1-5 no 18

In Dauns book "Modules and Rings", Exercise 18 in Section 1-5 reads as follows: (see attachment)

Let K be any ring with 1∈*K* 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}.

Prove that the subring K[I] = K + KI has an automorphism of order 2 defined by $\displaystyle a = a_1 + a_2I \rightarrow \overline{a} = a_1 - a_2I $ for $\displaystyle a_1, a_2 \in K $ which extends to an inner automorphism of order two of all of K[I,J], where $\displaystyle q \rightarrow \overline{q} = J_{-1}qJ = (1/y)JqJ $ for $\displaystyle q = a + bJ, a, b \in K[I] $. Show that $\displaystyle \overline{q} = \overline{a} + \overline{b}J $.

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I can show that K[I] = K + KI is an automorphism, but I am unsure of the meaning of an automorphism having order two - let alone proving it! So I would appreciate some help on showing that the automorphism has order two.

I cannot however show that it 'extends' to the inner automorphism of order two that is then defined.

I would appreciate help with these parts of the exercise.

Peter

[This has also been posted on MHB]

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Re: Generalised Quaternion Algebra over K - Dauns Section 1-5 no 18

OK, I have just been checking the order of an automorphism and it is actually quite straightforward - definition is as follows:

If the automorphism is $\displaystyle f: a \rightarrow \overline{a} $ as in the exercise that we are focused on, then the order is the smallest $\displaystyle n \ge 1 $ such that $\displaystyle f^n = E $ where E is the identity function.

OK so in the exercise we have that $\displaystyle f: a \rightarrow \overline{a} $ is obviously of order 2.

However I am still unable to show that K[I] = K + KI 'extends' (?) to the inner automorphism specified.

I would very much appreciate some help.

Peter

[This has also been posted on MHB]