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- Jun 22, 2012

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I am reading R.Y. Sharp: Steps in Commutative Algebra, Chapter 3: Prime Ideals and Maximal Ideals.

Exerise 3.47 on page 52 reads as follows:

=====================================================

Let P be a prime ideal of the commutative ring R.

Show that \(\displaystyle \sqrt (P^n) = P \)

======================================================

Can someone please help me get started on this problem.

Working from definitions of terms we have that

\(\displaystyle p_i \in P^n \Longrightarrow p_i = p_{11}p_{12} ... \ ... p_{1n} + p_{21}p_{22} ... \ ... p_{2n} + p_{i1}p_{i2} ... \ ... p_{in} \)

and then

\(\displaystyle (P^n) = \{ \sum_{i=1}^{n} r_i p_i , r_i \in R, p_i \in P^n \} \) ...

and

\(\displaystyle \sqrt (P^n) = \{ p \in R \ \ | \ \) there exists \(\displaystyle m \in \mathbb{N} \) with \(\displaystyle p^m \in (P^n) \} \)

But how to move beyond a restatement of the definitions ... ... ?

Can someone please help ...

Peter

Exerise 3.47 on page 52 reads as follows:

=====================================================

Let P be a prime ideal of the commutative ring R.

Show that \(\displaystyle \sqrt (P^n) = P \)

======================================================

Can someone please help me get started on this problem.

Working from definitions of terms we have that

\(\displaystyle p_i \in P^n \Longrightarrow p_i = p_{11}p_{12} ... \ ... p_{1n} + p_{21}p_{22} ... \ ... p_{2n} + p_{i1}p_{i2} ... \ ... p_{in} \)

and then

\(\displaystyle (P^n) = \{ \sum_{i=1}^{n} r_i p_i , r_i \in R, p_i \in P^n \} \) ...

and

\(\displaystyle \sqrt (P^n) = \{ p \in R \ \ | \ \) there exists \(\displaystyle m \in \mathbb{N} \) with \(\displaystyle p^m \in (P^n) \} \)

But how to move beyond a restatement of the definitions ... ... ?

Can someone please help ...

Peter

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