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

[Math Amateur]

In Dauns book "Modules and Rings", Exercise 18 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}.

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

---------------------------------------------------------------------

I can show that K = 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 MHF]

 

[Math Amateur]

In Dauns book "Modules and Rings", Exercise 18 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}.

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

---------------------------------------------------------------------

I can show that K = 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

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

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

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

Peter

 

[Deveno]

if we send every element of K to it's "I-conjugate":

a+bI --> a-bI

then wouldn't it make sense to also send:

(a+bI) + (c+dI)J --> (a-bI) + (c-dI)J?

all you're being asked to show is three things:

1) q-->q* is an automorphism (check that it's bijective, and is a ring-homomorphism).

2)(q*)* = q (and there exists some q for which q* is not q).

3) if q = a + bI + 0J + 0IJ, then q* = a - bI + 0J + 0IJ (so that q-->q* extends the automorphism of K).

Having shown this, verify by direct computation that for r,s in K:

(r + sJ)* = r* + s*J (use the automorphism property you proved above!).

(hint: what is J*, directly from the definition...isn't it (1/y)J³ = (1/y)(J²​)(J)?).