number of divisers of element in UFD

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March 13th 2011, 09:38 PM
student2011

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
March 14th 2011, 04:40 AM
tonio

Quote:

Originally Posted by student2011

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
March 14th 2011, 08:00 AM
student2011

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
March 14th 2011, 11:24 AM
tonio

Quote:

Originally Posted by student2011

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