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Thread: abstract algebra:zero divisors

  1. #1
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    abstract algebra:zero divisors

    Suppose p is a prime. Find all the zero divisors in $\displaystyle Z_p$$\displaystyle Z_{p^2}$
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  2. #2
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    Any member of $\displaystyle Z_p\times Z_{p^2}$ is of the form (a, b) where a is in $\displaystyle Z_p$, b is in $\displaystyle Z_{p^2}$ and multiplication is coordinate wise. That is, to be a zero divisor we must have $\displaystyle (a, b)(c, d)= (ac, bd)= (np, mp^2)$ for n and m integers. Since p is prime, at least one of a or c is a multiple of p and so is congruent to 0 modulo p. Either one of b or d is a multiple of $\displaystyle p^2$, and so congruent to 0 modulo $\displaystyle p^2$ or both b and d are multiples of p.
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    yes but the first set has no zero divisors. That is all elements excluding zero have inverses, so i thought the solution would be the empty set for the first element and 5 from the second set
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    I don't know what you mean by "the first set" and "the second set". There is only one set to be found here: the zero divisors of $\displaystyle Z_p\times Z_{p^2}$. Yes, since p is a prime, $\displaystyle Z_p$ has no zero-divisors. So any zero-divisor of $\displaystyle Z_p\times Z_{p^2}$ must be of the form (0, a) where a is a zero-divisor of $\displaystyle Z_{p^2}$. What are the zero-divisors of that?
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  5. #5
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    then the only possible zero divisor is the elemnet 5.
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  6. #6
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    Quote Originally Posted by ulysses123 View Post
    then the only possible zero divisor is the elemnet 5.

    5 is not even an element of the ring you're dealing with. Hallsofivy already wrote you what a member of this ring looks like: it is an ordered pair.
    You have first to understand this to continue trying to solve this problem.

    Tonio
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