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The definition of an algebra

  1. Jun 12, 2013 #1
    "Let R be a commutative ring. We say that M is an algebra over R, or that M is an R-algebra if M is an R-module that is also a ring (not necessarily commutative), and the ring and module operations are compatible, i.e., [tex]r(xy) = (rx)y = x(ry)[/tex] for all [itex]x, y \in M[/itex] and [itex]r \in R[/itex]."

    I'm not really sure why the second equality is true, because it implies commutativity and the definition tells us that an R-module is not necessarily commutative, right?

    Thank you in advance
     
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  3. Jun 12, 2013 #2

    HallsofIvy

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    No, it does NOT imply commutativity because x and y are not commuted. What is says is that it doesn't matter if you multiply the real number "r" by x or by y before you multiply the two module members.
     
  4. Jun 12, 2013 #3
    Thanks, but I cannot see any ring/module operation that would permit that. We know that associativity and distributivity hold (and of course others too, but they aren't related to this case). But I can't see how these basic operations would imply what you said.
     
  5. Jun 12, 2013 #4

    micromass

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    The definition does that ##rx = xr##, but that's for ##r\in R## and ##x\in M##. The definition does not state that ##xy = yx## for ##x,y\in M##. So ##M## is not commutative in that sense.
     
  6. Jun 13, 2013 #5
    Thanks a lot.
     
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