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I'm reading the proof here: Classification of prime ideals of $\mathbb{Z}[X]$

The part I am having trouble with is the following.

We are considering the case where $P$ is an ideal in $\mathbb{Z}[x]$, and $P \cap \mathbb{Z} = (p)$ for some prime $p$. Then we consider the image of $P$ by the isomorphism $\mathbb{Z}[x]/(p) \cong F_p[x]$. we are concerned with the case where $p(x)$ is a polynomial in $\mathbb{Z}[x]$ that reduces to irreducible polynomial in $F_p[x]$.

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What I do not get is that we assume $p(x)$ is monic, but I have no idea how to justify that assumption.

If someone could clear up my misunderstanding, I would be grateful.

Thanks.

Phil
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    $p(x)$ can be any polynomial reducing to $q(x) \bmod p$, and you can choose such a polynomial to be monic, since $q(x)$ is monic. (And $q(x)$ is monic WLOG because $\mathbb{F}_p$ is a field.) – Qiaochu Yuan Jan 19 '21 at 01:14
  • @QiaochuYuan say $3x+1$, that reduces to x+1 in $F_2$. What would "make" $3x+1$ monic? – Phil Jan 19 '21 at 01:48
  • Nothing is "making" $3x + 1$ monic. It's just that in the argument $p(x)$ can be any polynomial reducing to $q(x) \bmod p$, as in we are free to choose it, and the answerer chooses it so that it's monic. In your example we can choose $p(x) = x + 1$ instead. – Qiaochu Yuan Jan 19 '21 at 01:49
  • @QiaochuYuan ah yeah, I missed the quantify. That makes sense, thanks! – Phil Jan 19 '21 at 01:53

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