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Math Help - Integral rings

  1. #1
    r45
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    Integral rings

    Hi there, I'm stuck on a couple of questions about integral rings:

    1. State the difference(s) between an integral ring and an integral domain.

    2. Suppose R is an integral ring, and x, y, z are 3 elements in R. Furthermore, suppose x \neq 0 and xy = xz. Show that y = z.

    3. Again, suppose R is an integral ring, and x is a non-zero element in R. Is the map of x into ax an injective mapping of R into R? Why/why not?

    For 1, is an integral domain simply an integral ring in which every element is commutative? Or is there something else that differentiates the two?

    For 2, I am confused, at first glance it seems like an obvious result but I've been struggling to show it. The fact that R is an integral ring (as opposed to an integral domain) means that x, y, z may not have a multiplicative inverse. So I cannot just multiply xy = xz on the left by x^{-1} to get y = z. I've tried using the associativity axiom of rings etc but I haven't been able to get the desired result, what am I missing?!

    And for 3, I think the answer is yes (that's the intuitive answer, at least - although by now I've realised that by no means does that mean that it's the right answer!) - but why, can you show why it is an injective mapping?

    Many thanks for any help!
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  2. #2
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    Quote Originally Posted by r45 View Post
    Hi there, I'm stuck on a couple of questions about integral rings:

    1. State the difference(s) between an integral ring and an integral domain.

    2. Suppose R is an integral ring, and x, y, z are 3 elements in R. Furthermore, suppose x \neq 0 and xy = xz. Show that y = z.

    3. Again, suppose R is an integral ring, and x is a non-zero element in R. Is the map of x into ax an injective mapping of R into R? Why/why not?

    For 1, is an integral domain simply an integral ring in which every element is commutative? Or is there something else that differentiates the two?

    For 2, I am confused, at first glance it seems like an obvious result but I've been struggling to show it. The fact that R is an integral ring (as opposed to an integral domain) means that x, y, z may not have a multiplicative inverse. So I cannot just multiply xy = xz on the left by x^{-1} to get y = z. I've tried using the associativity axiom of rings etc but I haven't been able to get the desired result, what am I missing?!

    And for 3, I think the answer is yes (that's the intuitive answer, at least - although by now I've realised that by no means does that mean that it's the right answer!) - but why, can you show why it is an injective mapping?

    Many thanks for any help!

    It'd help if you add the definitions you're working with: as far as I know, integral ring ALWAYS goes with respect to another ring: a ring R is integral OVER ANOTHER ring A if every element in r satisfies a monic pol. over A.
    Apparently you people are using "integral ring" in another sense...perhaps a non commutative ring without proper zero divisors, or perhaps a non-unitary ring w/o proper zero divisors...?

    Tonio
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  3. #3
    r45
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    The definition for integral ring in my textbook is:

    "A commutative ring without divisors of zero, and such that 1 \neq 0 is called an INTEGRAL RING"

    An alternative definition I have from my lecture notes is:

    "A ring is an INTEGRAL RING if whenever ab=0, either a=0 or b=0. Furthermore, if the ring is commutative then it is an INTEGRAL DOMAIN"

    Does this help?
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    Quote Originally Posted by r45 View Post
    The definition for integral ring in my textbook is:

    "A commutative ring without divisors of zero, and such that 1 \neq 0 is called an INTEGRAL RING"

    An alternative definition I have from my lecture notes is:

    "A ring is an INTEGRAL RING if whenever ab=0, either a=0 or b=0. Furthermore, if the ring is commutative then it is an INTEGRAL DOMAIN"

    Does this help?

    What textbook\lecture notes/course are those?! In the first definition it's being explicitly required that the ring is commutative and, besides, that it has a unit (otherwise what would they require 1\neq 0?), but in the second definition from the lecture notes they don't require neither commutativity nor unit and then they add that if the condition of commutativity exists then it is called an integral domain...so what exactly is an integral ring according to that course of yours, for Euclides' sake?!??

    Tonio
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  5. #5
    r45
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    The textbook is Undergraduate Algebra by Lang. The questions I am trying to answer come from the textbook, so I suppose it is those definitions that should be obeyed, i.e.

    "A commutative ring without divisors of zero, and such that is called an INTEGRAL RING"
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  6. #6
    r45
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    OK here's what I've come up with for 2.

    (xy)z = (xy)z
    (xy)z = (yx)z by commutativity
    (xy)z = y(xz) by transitivity
    (xz)z = y(xz) since xy = xz from the question

    Now if xz = 0 then z = 0 since we are told in the question that x is non-zero. Similarly if xz = 0 then xy = 0 also, and so y = 0 and y = z as required.

    If xz is nonzero, then we have:

    (xz)z = y(xz)

    But from here what can I do? I can't just "divide" through by xz can I? And I can't multiply on each side by (xz)^-1 since it's not guaranteed to have a multiplicative inverse, is it?
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  7. #7
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    Quote Originally Posted by r45 View Post
    OK here's what I've come up with for 2.

    (xy)z = (xy)z
    (xy)z = (yx)z by commutativity
    (xy)z = y(xz) by transitivity
    (xz)z = y(xz) since xy = xz from the question

    Now if xz = 0 then z = 0 since we are told in the question that x is non-zero. Similarly if xz = 0 then xy = 0 also, and so y = 0 and y = z as required.

    If xz is nonzero, then we have:

    (xz)z = y(xz)

    But from here what can I do? I can't just "divide" through by xz can I? And I can't multiply on each side by (xz)^-1 since it's not guaranteed to have a multiplicative inverse, is it?
    Easier: xy=xz\Longrightarrow x(y-z)=0 , and since x\neq 0 and we're in an integral ring then y-z=0\Longleftrightarrow y=z

    This must also be enough to solve (3) .

    Tonio
    Last edited by tonio; March 24th 2010 at 08:38 PM. Reason: additional info.
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  8. #8
    r45
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    Ah yeah, how did I miss that! Thanks for your help.
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