MHB Prime and Maximal Ideals .... Bland -AA - Theorem 3.2.16 .... ....

[Math Amateur]

I am reading The Basics of Abstract Algebra by Paul E. Bland ...

I am focused on Section 3.2 Subrings, Ideals and Factor Rings ... ...

I need help with the proof of Theorem 3.2.16 ... ... Theorem 3.2.16 and its proof reads as follows:
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In the above proof of $$(3) \Longrightarrow (1)$$ by Bland, we read the following:

" ... ... Then $$n_1 n_2 = p \in p \mathbb{Z}$$, so either $$n_1 \in p \mathbb{Z}$$ or $$n_2 \in p \mathbb{Z}$$. ... ... Can someone please explain how/why exactly ... $$n_1 n_2 = p \in p \mathbb{Z}$$ implies that either $$n_1 \in p \mathbb{Z}$$ or $$n_2 \in p \mathbb{Z}$$. ... ... Peter
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***NOTE***

It may help readers to have access to Bland's definition of a prime ideal ... so I am providing the same as follows:
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Sorry about the legibility ... but Bland shades his definitions ...Peter

 

[steenis]

The answer to your question lies in the definition: if $P$ is a prime ideal and $xy \in P$, then eiter $x \in P$ or $y \in P$.

It is given that $p \mathbb{Z}$ is aprime ideal, of course $p=p1 \in p \mathbb{Z}$. So, if $p=n_1 n_2 \in p \mathbb{Z}$, then either $n_1 \in p \mathbb{Z}$ or $n_2 \in p \mathbb{Z}$ by definition.

 

[Math Amateur]

The answer to your question lies in the definition: if $P$ is a prime ideal and $xy \in P$, then eiter $x \in P$ or $y \in P$.

It is given that $p \mathbb{Z}$ is aprime ideal, of course $p=p1 \in p \mathbb{Z}$. So, if $p=n_1 n_2 \in p \mathbb{Z}$, then either $n_1 \in p \mathbb{Z}$ or $n_2 \in p \mathbb{Z}$ by definition.

Thanks Steenis ...

Peter