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Math Help - Abstract Alegbra - Zero Divisors and Units

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    Abstract Alegbra - Zero Divisors and Units

    Z2 ◊ Z4 is a ring under component-wise addition and multiplication:


    (a, b) + (c, d) = (a + c, b + d)

    and

    (a, b) ∑ (c, d) = (ac, bd).


    Classify each element of Z2 ◊ Z4 as a zero divisor, a unit, or neither.


    If the element is a zero divisor, find a nonzero element whose product with the first element is (0,0).
    If the element is a unit, find its multiplicative inverse.

    My Answer:

    Z2 x Z4 = { (0,0), (0,1), (0,2), (0,3), (1,0), (1,1), (1,2), (1,3) }

    Zero Divisors:
    (0,0)*(0,0) = (0,0)
    (0,1)*(1,0) = (0,0)
    (0,2)*(0,2) = (0,0)
    (0,3)*(1,0) = (0,0)
    (1,0)*(1,0) = (0,0)
    (1,2)*(0,2) = (0,0)

    Unit:
    (1,1)

    Neither:
    (0,3)
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    Re: Abstract Alegbra - Zero Divisors and Units

    Zero Divisors:
    (0,0), (0,1), (0,2), (0,3), (1,0), (1,2)

    Units:
    (1,1), (1,3)

    (1,1)*(1,1) = (1,1), so (1,1)-1 = (1,1)
    (1,3)*(1,3) = (1,1), so (1,3)-1 = (1,3)

    Neither:
    None
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    Re: Abstract Alegbra - Zero Divisors and Units

    I didn’t see the method by looking at the answer. So I made the addition and multiplication tables, and then I saw the method (ab is shortcut for (a,b)) :

    Divisors of 0 (ab=0): What do you multiply by to get 00
    00 01 02 03 10 11 12 13
    00 10 12 10 01 .... 02 ..

    Additive inverse (a+b=0) : What do you add to get 00
    00 01 02 03 10 11 12 13
    00 03 02 01 10 13 12 13

    Units (ab=1): What do you multiply by to get 11
    00 01 02 03 10 11 12 13
    ......................11..... 13

    Definition:
    1) amodn is remainder of a/n
    5mod4=1
    2) (a+b)=(a+b)modn, (a*b)=(a*b)modn
    2+2=4mod4=0, 3*3=9mod4=1
    3) Zn: 0,1,2,3,…n with above rules for addition and multiplication.

    For the present example, each component satisfies the Definitions:
    (1,3)+(1,3)=(2mod2,6mod4)=(0,2)
    (1,3)*(1,3)=(1mod2,9mod4)=(1,1)
    Last edited by Hartlw; April 17th 2014 at 08:51 AM.
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    Re: Abstract Alegbra - Zero Divisors and Units

    One can always try trial-and-error, but that is not the most efficient approach, usually.

    Let's break the elements of $\Bbb Z_2 \times \Bbb Z_4$ into 4 types:

    1: (0,0)
    2: (1,0)
    3: (0,a), a = 1,2,3
    4: (1,a), a = 1,2,3

    It should be obvious that types 1-3 are all zero divisors. The zero of a ring is (obviously) always a zero-divisor. For those elements where 0 is one of the coordinates, multiplying by an element where 0 is in the OTHER coordinate will always gives you 0, and we can always pick the SAME coordinate in the second element to be non-zero, that is:

    type 2 times type 3 = (0,0), so these are all zero divisors.

    That leaves 3 elements to check: (1,1),(1,2), and (1,3). (1,1) is a unit, being the ring identity (which is always a unit), so we only need to look at (1,2) and (1,3).

    An observation:

    (1,2)(a,b) = (a,2b). If (1,2) were a unit, we would have to have: a = 1 (mod 2),2b = 1 (mod 4). Similarly, if (1,2) were a zero divisor, we would have to have a = 0 (mod 2), 2b = 0 (mod 4), and since (0,b) cannot be (0,0), we need a non-zero such b.

    Similar considerations hold for (1,3), which is a unit iff there exists b with 3b = 1 (mod 4), and is a zero divisor iff there exists non-zero b with 3b = 0 (mod 4).

    In short, we have to look at the units and zero-divisors of $\Bbb Z_4$.

    Well, in $\Bbb Z_4$ we have:

    2*2 = 0 (mod 4), so 2 is a zero-divisor. this means that (1,2) is likewise a zero-divisor, explicitly: (1,2)*(0,2) = (0,0), and (0,2) is not (0,0).

    In $\Bbb Z_4$, 3 is a unit, since 3*3 = 1 (mod 4). this means that (1,3) is likewise a unit, explicitly: (1,3)*(1,3) = (1,1).

    IN GENERAL, for $\Bbb Z_m \times \Bbb Z_n$, we have:

    (0,b) and (a,0) are zero divisors for ANY a,b.

    (a,b) is a zero divisor if either a OR b is a zero-divisor (or both).

    (a,b) is a unit if BOTH a and b are units in their respective rings.

    ********

    It is possible for elements of a ring $R \times S$ to be neither, although that does not happen if $R,S$ are finite cyclic rings. An example is given by $R = \Bbb Z, S = \Bbb Z_2$, where the element $(2,1)$ is neither:

    (2,1)*(a,b) = (0,0) implies (2a,b) = (0,0), which implies b = 0, and 2a = 0. The only integer a for which 2a = 0, is 0, so (2,1) is not a zero-divisor.

    (2,1)*(a,b) = (1,1) implies (2a,b) = (1,1), which implies b = 1, and 2a = 1. There is NO integer a for which 2a = 1 (1/2 is NOT an integer), so (2,1) is not a unit.
    Thanks from romsek and Hartlw
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    Re: Abstract Alegbra - Zero Divisors and Units

    Quote Originally Posted by Hartlw View Post

    Divisors of 0 (ab=0): What do you multiply by to get 00
    00 01 02 03 10 11 12 13
    00 10 12 10 01 .... 02 ..

    Additive inverse (a+b=0) : What do you add to get 00
    00 01 02 03 10 11 12 13
    00 03 02 01 10 13 12 13

    Units (ab=1): What do you multiply by to get 11
    00 01 02 03 10 11 12 13
    ......................11..... 13
    Deveno, why is your description shorter or clearer than simply writing out the elements of (a,b) and underneath, by inspection, respectively listing the divisors of 0, additive inverse, and units. That is not trial and error, it is a clear, simple, systematic, way to do it. I would think you donít have to belabor the fact that, for example, if you have (0,b) or (a,0), there is nothing you can multiply 0 by to get 1.
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    Re: Abstract Alegbra - Zero Divisors and Units

    Quote Originally Posted by Hartlw View Post
    Deveno, why is your description shorter or clearer than simply writing out the elements of (a,b) and underneath, by inspection, respectively listing the divisors of 0, additive inverse, and units. That is not trial and error, it is a clear, simple, systematic, way to do it. I would think you don’t have to belabor the fact that, for example, if you have (0,b) or (a,0), there is nothing you can multiply 0 by to get 1.
    Deveno uses the qualifier "usually" when stating that writing out every element (a,b) is not the most efficient approach. In this case, the ring is small enough that it is not terribly difficult to write out a complete multiplication table. However, if the ring were \mathbb{Z}/2^{50}\mathbb{Z} \times \mathbb{Z}/2^{100}\mathbb{Z}, you would have a ring with 2^{150} elements. Deveno's method of computing the zero divisors and units still works while trying to write out the 2^{300} possible multiplications of the ring would take quite a long time.

    So, Deveno attempted to give the OP some theory that could help understand future problems that do not use such small groups. The idea being that this knowledge could be useful in solving more complex problems of a similar nature. However, in this example, I agree with you that either method works. Still, there is no harm in imparting a little extra knowledge, is there?
    Last edited by SlipEternal; April 21st 2014 at 06:34 AM.
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    Re: Abstract Alegbra - Zero Divisors and Units

    The first part of Deveno’s post is simply a belaboring of the obvious, it was done in post #3 by inspection.

    Of the three rules given at the end of Deveno’s post, the first two are wrong, and there is no ”theory.”

    By Definition (as given in my post #3):
    1) (a,b) is a divisor of 0 if a and b are divisors of 0.
    2) (a,b) is a unit if a and b are units.
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    Re: Abstract Alegbra - Zero Divisors and Units

    Quote Originally Posted by Hartlw View Post
    The first part of Deveno’s post is simply a belaboring of the obvious, it was done in post #3 by inspection.

    Of the three rules given at the end of Deveno’s post, the first two are wrong, and there is no ”theory.”

    By Definition (as given in my post #3):
    1) (a,b) is a divisor of 0 if a and b are divisors of 0.
    2) (a,b) is a unit if a and b are units.
    Perhaps what Deveno wrote is not connecting with you, but it may be helpful to others. Ring theory is "theory" so I don't understand why you dislike my use of that word. And I don't understand why you say the first two rules Deveno posted are "wrong". For example, you showed that (1,2) is a zero divisor, but 1 is not a zero divisor in \Bbb{Z}/2\Bbb{Z}. So, your "rule 1" seems inaccurate.
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    Re: Abstract Alegbra - Zero Divisors and Units

    Quote Originally Posted by SlipEternal View Post
    Perhaps what Deveno wrote is not connecting with you, but it may be helpful to others. Ring theory is "theory" so I don't understand why you dislike my use of that word. And I don't understand why you say the first two rules Deveno posted are "wrong". For example, you showed that (1,2) is a zero divisor, but 1 is not a zero divisor in \Bbb{Z}/2\Bbb{Z}. So, your "rule 1" seems inaccurate.
    I wasn't objecting to your use of the word ring "theory." What's the point? We are obviously talking about zeros and units of the ring ZnXZm as defined in OP.

    Z is irrelevant. The subject is Zn (finite).
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    Re: Abstract Alegbra - Zero Divisors and Units

    Quote Originally Posted by Hartlw View Post
    Z is irrelevant. The subject is Zn (finite).
    The OP originally thought one of the elements of the ring was neither a zero-divisor nor a unit. Deveno brought up \Bbb{Z} to discuss such elements by pointing out that every element of any finite product of finite cyclic rings is either a zero-divisor or a unit. Then, to expand upon that, he gave an example of a ring containing an element that is neither a zero-divisor nor a unit. Exposure to concrete examples is frequently useful to students learning ring theory for the first time.
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    Re: Abstract Alegbra - Zero Divisors and Units

    Quote Originally Posted by Deveno View Post

    .......
    It is possible for elements of a ring $R \times S$ to be neither, although that does not happen if $R,S$ are finite cyclic rings. An example is given by $R = \Bbb Z, S = \Bbb Z_2$, where the element $(2,1)$ is neither:

    (2,1)*(a,b) = (0,0) implies (2a,b) = (0,0), which implies b = 0, and 2a = 0. The only integer a for which 2a = 0, is 0, so (2,1) is not a zero-divisor.

    (2,1)*(a,b) = (1,1) implies (2a,b) = (1,1), which implies b = 1, and 2a = 1. There is NO integer a for which 2a = 1 (1/2 is NOT an integer), so (2,1) is not a unit.
    What is that all about?

    By Definition (as given in my post #3):
    1) (a,b) is a divisor of 0 if a and b are divisors of 0.
    2) (a,b) is a unit if a and b are units.

    In ZXZ2, The only unit of Z is1, so (2,1) is not a unit by definition. The only unit is (1,1)
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    Re: Abstract Alegbra - Zero Divisors and Units

    Quote Originally Posted by Hartlw View Post
    What is that all about?

    By Definition (as given in my post #3):
    1) (a,b) is a divisor of 0 if a and b are divisors of 0.
    2) (a,b) is a unit if a and b are units.
    By your (1), in (\Bbb{Z}/2\Bbb{Z})\times (\Bbb{Z}/4\Bbb{Z}), (1,0) is not a zero-divisor since 1 is not a zero-divisor in \Bbb{Z}/2\Bbb{Z}. However, in post #3, you show the multiplication that proves that (1,0) is a zero divisor. Therefore, (1) is not correct.

    Quote Originally Posted by Hartlw View Post
    In ZXZ2, The only unit of Z is1, so (2,1) is not a unit by definition. The only unit is (1,1)
    That was Deveno's point exactly. I'm glad you agree.
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    Re: Abstract Alegbra - Zero Divisors and Units

    Quote Originally Posted by SlipEternal View Post
    By your (1), in (\Bbb{Z}/2\Bbb{Z})\times (\Bbb{Z}/4\Bbb{Z}), (1,0) is not a zero-divisor since 1 is not a zero-divisor in \Bbb{Z}/2\Bbb{Z}. However, in post #3, you show the multiplication that proves that (1,0) is a zero divisor. Therefore, (1) is not correct.

    That was Deveno's point exactly. I'm glad you agree.
    What is (Z/2Z)X(Z/4Z) and what does it have to do with Z2XZ4?

    (1,0) is not a divisor of zero in Z2/Z4. I made a mistake, as can easily be seen from defintion 1) below. 1 is not a zero of Z2. Without all the unintelligible, undefined blah blah.

    Deveno's point was:

    In ZXZ2:

    "(2,1)*(a,b) = (0,0) implies (2a,b) = (0,0), which implies b = 0, and 2a = 0. The only integer a for which 2a = 0, is 0, so (2,1) is not a zero-divisor.

    (2,1)*(a,b) = (1,1) implies (2a,b) = (1,1), which implies b = 1, and 2a = 1. There is NO integer a for which 2a = 1 (1/2 is NOT an integer), so (2,1) is not a unit.

    My point was:

    (2,1) is neither a unit or divisor of ZXZ2 by definiton. (2 is neither a unit or divisor of 0 in Z)

    By Definition (as given in my post #3):
    1) (a,b) is a divisor of 0 if a and b are divisors of 0.
    2) (a,b) is a unit if a and b are units.

    My point is general and simple. If you don't see the difference, sorry.
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    Re: Abstract Alegbra - Zero Divisors and Units

    Quote Originally Posted by Hartlw View Post
    What is (Z/2Z)X(Z/4Z) and what does it have to do with Z2XZ4?
    \Bbb{Z}/2\Bbb{Z} is always the cyclic ring of order 2. \Bbb{Z}_2 is the notation for the ring of 2-adic integers. I avoid using \Bbb{Z}_2 to avoid confusion. So, if you don't like my notation, that is your preference, and \Bbb{Z}/2\Bbb{Z} = \Bbb{Z}_2 and \Bbb{Z}/4\Bbb{Z} = \Bbb{Z}_4.

    Your definition (as given in post #3) is a different definition from the standard definition of a zero divisor of a ring. The standard definition (Wikipedia - Zero divisor):

    In abstract algebra, an element a of a ring R is called a left zero divisor if there exists a nonzero x such that ax = 0, or equivalently if the map from R to R that sends x to ax is not injective.
    As you showed in post 3, (1,0)\cdot (0,1) = (0,0). Since (0,1) \neq (0,0), this satisfies the definition so (1,0) is a zero-divisor (even though 1 is not a zero-divisor).

    So, I don't know where you got your definitions (as given in your post #3), but they are not standard.
    Last edited by SlipEternal; April 21st 2014 at 01:50 PM.
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    Re: Abstract Alegbra - Zero Divisors and Units

    MY SUMMARY:

    Post 2 answered OP correctly without any explanation.
    Post 3 gave a systematic method for answering OP with general defs of zero divisors and units: ab=0, and ab=1. Missing: a≠0 and b≠0 for ab=0. (but very limited)

    Post 4:
    1) (0,b) and (a,0) are zero divisors for ANY a,b.
    2) (a,b) is a zero divisor if either a OR b is a zero-divisor (or both).
    3) (a,b) is a unit if BOTH a and b are units in their respective rings.

    Comment:
    1) a and b both ≠ 0
    2) if a is a zero divisor: ac=0, a≠0, c≠0. Then
    (a,b)(c,0)=(a,0)(c,b)=(0,0), any b.

    CONCLUSION:
    Post 4 answered the OP, which I realized after I got past the cramped wordiness, which is a put-off.

    Turned out to be an interesting, educational thread, except for the irrelevant obtuse abstract jargon copied from wiki articles.
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