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The reason is : We may assume the prime ideal with norm $p^n$. Note $(p^n)=(p)^n$, and the prime ideal $I$ with norm $p^n$ divides $(P^n)$, thus $Ip_1...p_m=(p^n)=(p)^n$, thus as $I$ is a prime ideal, we get $I$ divides $(p)$. However, this is impossible since $N((p))=p^2$. Thus $I$ doesn't exist.

But I don't think then conclusion is right, so can you tell me where I went wrong?

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    Your question is unclear. – 2017-02-18
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    @Starfall, sorry about that, I mean in my first paragraph, it shows a prime ideal with norm $p^n$ does not exist. I wonder if this proof is right. Does that make sense to you now? – 2017-02-18
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    No, it does not. What ring are we talking about? Where is the prime ideal supposed to exist? What is $ P $? What are the $ p_i $? What is $ I $? – 2017-02-18
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    @Starfall the ring is a ring of integers of some number field, $p$ is a prime number in $\mathbb{Z}$. The prime ideal is supposed to exist in the ring of integers – 2017-02-18
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    You have three more questions to answer... If we take your statement verbatim, *any* prime ideal in *any* number ring has norm which is a power of a prime. I have no clue what your question is trying to say; or what your argument actually is. – 2017-02-18
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    @Starfall $I$ is a prime ideal with norm $p^n$, $p_i$ is a prime ideal that $Ip_1...p_m$ is the unique factorisation of the integral ideal $(p^n)$. – 2017-02-18
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53871/discussion-between-starfall-and-danny). – 2017-02-18

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