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*-rings and involution

  1. Apr 8, 2008 #1
    Someone, please help me solve this problem:

    First if R is a ring and * is an involution, then U(R, *):= {x \in R|x* · x = 1}

    (an involution * is an antihomomorphism such that a** = a for any a)

    Now the problem. Find two rings (R, S) with involutions (*, ^) such that U(R, *) is homomorphic to (S, ^). and R and S are not homomorphic.

    My first problem is that i do not know of any involutions except for conjugation and transposition for matrixes.
     
  2. jcsd
  3. Apr 8, 2008 #2

    morphism

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    I'm assuming you want to find an example where U(R,*) and U(S,^) are isomorphic (as groups), while R and S aren't isomorphic (as rings). Correct?

    Maybe this example of a *-ring will help you: The ring of polynomials with complex coefficients [itex]\mathbb{C}[z][/itex] (viewed as functions on [itex]\mathbb{C}[/itex]), with [itex]p(z)^* = \overline{p(\bar{z})}[/itex], where the bar denotes complex conjugation. What's the unitary group of ([itex]\mathbb{C}[z][/itex],*)?
     
    Last edited: Apr 9, 2008
  4. Apr 9, 2008 #3
    Thank you for another example. The unitary group would be all the complex numbers a: |a|= 1.
    I have also thought of Examples with the unitary groups isomprphis to Z\2 and Z\4, but didn't help to solve he problem...

    You are correct on the remark, the first isomorphism is an isomorphism of miltiplicative groups.
     
  5. Apr 9, 2008 #4
    Actually, now that I think about it: an example would be complex numbers (with conjugation = *) and polynomials. Their unitary groups are the same while they are not homomorphic as rings!! I hope I'm right!! Thanks again for the example!
     
  6. Apr 9, 2008 #5
    By the way the unitary group is +1 and -1 (not what i have said before, because it is true iff there was only one conjugation over the argument)and it is also isomorphic to Z2.. I guess that is another example along with matrices 2*2 (with elements = 0, 1) with * = transposition.
     
  7. Apr 9, 2008 #6

    morphism

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    Why isn't U(C[z],*) identifiable with the set of complex numbers with absolute value 1 (i.e. the unit circle)? I actually agree with what you said in post #4; that's the example I had in mind.
     
  8. Apr 10, 2008 #7
    yeah, what i said later isn't true, sorry
     
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