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Math Help - Algebra, Problems For Fun (13)

  1. #1
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    Algebra, Problems For Fun (13)

    Definition: An element a in a ring is called an idempotent if a^2=a.

    Question: Given an integer n \geq 2, how many idempotents are there in \mathbb{Z}/n\mathbb{Z} ?
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    If there is an integer a such that 1<a<n and n=a^2-a then there are 3 idempotents, namely 0, 1 and a. Otherwise 0 and 1 are the only idempotents.

    Is that right?
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    Quote Originally Posted by TheAbstractionist View Post
    If there is an integer a such that 1<a<n and n=a^2-a then there are 3 idempotents, namely 0, 1 and a. Otherwise 0 and 1 are the only idempotents.

    Is that right?
    unfortunately it's not right! for example the ring \mathbb{Z}/6\mathbb{Z} has 4 idempotents. i suggest you first solve the problem for n=p^k, where p is a prime.
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    Argh! I was thinking along the line that if 1<a<n is an idempotent then n=a^2-a. It should be n\mid a^2-a rather. Stupid, stupid.

    Okay, back to the beginning.
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  5. #5
    Senior Member TheAbstractionist's Avatar
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    Okay, this is actually very simple. I can’t believe I’ve taken such a long time to twig it.

    It is clear that a is an idempotent of \mathbb Z/n\mathbb Z if and only if n divides a^2-a=a(a-1). Let n=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r} where p_1,p_2,\ldots,p_r are the distinct prime factors of n. If a is an idempotent then each prime p_i must divide a or a-1, but not both – otherwise it would divide a-(a-1)=1. So whichever factor it divides, the whole lot p_i^{k_i} will divide that factor. And since each p_i either divides a or a-1 (2 choices), it emerges that the number of idempotents a is precisely 2^r.

    Hence the number of idempotents of \mathbb Z/n\mathbb Z where n\ge2 is 2^r where r is the number of distinct prime factors of n.

    This is one of the most wonderful problems I’ve ever tried – thank you ever so much, NonCommAlg. It’s a shame I can only thank you once – I would thank you more times if I could.
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  6. #6
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    that's correct. you could also use the isomoprhism \mathbb{Z}/n\mathbb{Z} \cong \bigoplus_{j=1}^r \mathbb{Z}/p_j^{k_j}\mathbb{Z} and this simple fact that each \mathbb{Z}/p_j^{k_j}\mathbb{Z} has exactly two idempotents.
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