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Math Help - number of divisers of element in UFD

  1. #1
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    number of divisers of element in UFD

    Hello;
    I need your help is proving this statement;

    Let R be a unique factorization domain and let a be non zero element of R with a=p_1^(α1)p_2^(α2)...P_n^(αn ), where pi are prime elements and αi are positive integers. Show that the number of divisors of a is the product of (1+αi ) where i=1,2,...,n

    May be we can proceed by induction, but how can we start.


    Thank you in advance
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  2. #2
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    Quote Originally Posted by student2011 View Post
    Hello;
    I need your help is proving this statement;

    Let R be a unique factorization domain and let a be non zero element of R with a=p_1^(α1)p_2^(α2)...P_n^(αn ), where pi are prime elements and αi are positive integers. Show that the number of divisors of a is the product of (1+αi ) where i=1,2,...,n

    May be we can proceed by induction, but how can we start.


    Thank you in advance

    Exactly, by induction: if there's only one prime in the decomposition of a=p^a , then is divisors are

    1,\,p,\,p^2,\,\ldots,\,p^\alpha ,and thus \alpha +1 divisors. Continue from here.

    Tonio
    Last edited by tonio; March 14th 2011 at 11:22 AM.
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  3. #3
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    Thaaaaaank you very much for your help. I will show you my solution:
    So we use induction on the number of prime numbers appears in a;
    If we have only one prime, then a_1=p_1^{\alpha_1} and the divisors of a_1 are 1,p_1,p_1^{2},....,p_1^{\alpha_1}. Hence the number of divisors of a_1 is 1+\alpha_1.

    Inductive Step: Suppose that a_{n-1} contains n-1 prime numbers, and the statement is true for a_{n-1}. i.e a_{n-1}=p_1^{\alpha_1}p_2^{\alpha_2}....p_{n-1}^{\alpha_{n-1}}and the number of divisors of a_{n-1} is \prod_{i=1}^{n-1}{1+\alpha_i}

    Our goal is to show that the statement is true for a_n which contain n prime numbers.

    We have a_n=p_1^{\alpha_1}p_2^{\alpha_2}....p_{n-1}^{\alpha_{n-1}}.p_n^{\alpha_n}. Hence a_n=a_{n-1}.p_n^{\alpha_n} By inductive hypothesis, the number of divisors of a_{n-1} is \prod_{i=1}^{n-1}{1+\alpha_i} and we know that the number of divisors of p_n^{\alpha_n} is 1+\alpha_n. So, the number of divisors of a_n is \prod_{i=1}^{n-1}{1+\alpha_i}.(1+\alpha_n) which is \prod_{i=1}^{n}{1+\alpha_i
    This is my solution. If there is any mistakes or comments please guide me. Thank you again
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  4. #4
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    Quote Originally Posted by student2011 View Post
    Thaaaaaank you very much for your help. I will show you my solution:
    So we use induction on the number of prime numbers appears in a;
    If we have only one prime, then a_1=p_1^{\alpha_1} and the divisors of a_1 are 1,p_1,p_1^{2},....,p_1^{\alpha_1}. Hence the number of divisors of a_1 is 1+\alpha_1.

    Inductive Step: Suppose that a_{n-1} contains n-1 prime numbers, and the statement is true for a_{n-1}. i.e a_{n-1}=p_1^{\alpha_1}p_2^{\alpha_2}....p_{n-1}^{\alpha_{n-1}}and the number of divisors of a_{n-1} is \prod_{i=1}^{n-1}{1+\alpha_i}

    Our goal is to show that the statement is true for a_n which contain n prime numbers.

    We have a_n=p_1^{\alpha_1}p_2^{\alpha_2}....p_{n-1}^{\alpha_{n-1}}.p_n^{\alpha_n}. Hence a_n=a_{n-1}.p_n^{\alpha_n} By inductive hypothesis, the number of divisors of a_{n-1} is \prod_{i=1}^{n-1}{1+\alpha_i} and we know that the number of divisors of p_n^{\alpha_n} is 1+\alpha_n. So, the number of divisors of a_n is \prod_{i=1}^{n-1}{1+\alpha_i}.(1+\alpha_n) which is \prod_{i=1}^{n}{1+\alpha_i
    This is my solution. If there is any mistakes or comments please guide me. Thank you again

    Nice. Just put parentheses after the Pi product sign: \displaystyle{\prod\limits^n_{i=1}(\alpha_i + 1) .

    Tonio
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