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Let $\mathfrak{p} \subseteq \mathbb{Z}[X_0,\dots,X_n]$ be a homogeneous prime ideal generated by some homogeneous, irreducible polynomials $P_1,\dots,P_r$. Pick a prime number $p$ that does not divide any coefficient of the $P_i$'s. Is it true that the homogeneous ideal $\mathfrak{p} + (p)$ is still prime?

Some background: I am trying to prove that, given a projective, irreducible curve defined over $\mathbb{Q}$, there are only finitely many primes $p$ such that the "corresponding" curve over $\mathbb{F}_p$ is not irreducible. This should be true, but I am not sure whether my argument is correct. Certainly one needs the assumption that $p$ does not divide the coefficients (or something similar), because for instance $XY^2-pZ^3$ is irreducible but $(XY^2-pZ^3,p)=(XY^2,p)$ is not prime. We may assume that $\mathfrak{p}$ does not contain constants, if this helps.

57Jimmy
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    What about $r=1$, $n=1$, $P=X_0^2+X_1^2$ and $p=2$? – user26857 Feb 02 '18 at 20:52
  • @user26857 Is $(X_0^2+X_1^2,2)$ not a prime ideal in $\mathbb{Z}[X_0,X_1]$? – 57Jimmy Feb 03 '18 at 14:42
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    @57Jimmy : no, because $$(X_0^2+X_1^2) - 2 \cdot X_1^2 = X_0^2 - X_1^2 = (X_0-X_1)(X_0+X_1)$$ belongs to this ideal, but neither $X_0-X_1$ nor $X_0 + X_1$ belongs to it. – Watson Feb 03 '18 at 15:18
  • @Watson Ok, got it, thanks. Do you have any hints for my original question then? Given a prime ideal $\mathfrak{p}$ in $\mathbb{Z}[X_0,\dots,X_n]$, how do I determine for which primes the quotient ideal $\mathfrak{p}/(p)$ stays prime? They should be all but finitely many, but I have no idea how to show it – 57Jimmy Feb 03 '18 at 15:50

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Your first question is answered negatively in the comments.

As for your background question, it doesn't work either. Basically, we can mimic the proof that $X^4+1$ is reducible mod $p$ for any prime $p$, but with the homoegenous polynomial $X^4+Y^4$, which is an irreducible element of $\Bbb Z[X,Y]$ (see here). We prove that for any prime $p$, the reduction $X^4+Y^4 \in \Bbb F_p[X,Y]$ is reducible.

  • If $-1$ is a square in $\Bbb F_p$ (which includes the case $p=2$), say $a^2=-1$, then we have $$X^4+Y^4=X^4-a^2Y^4=(X^2+aY^2)(X^2-aY^2).$$

  • If $p$ is odd and $2$ is a square in $\Bbb F_p$, say $2=b^2$, then we have $$X^4+Y^4=(X^2+Y^2)^2-(bXY)^2=(X^2+bXY+Y^2)(X^2-bXY+Y^2). $$

  • If $p$ is odd and neither $-1$ nor $2$ is a square, then their product $-2$ is a square, say $-2=c^2$. (since $\Bbb F_p^\times$ is a cyclic group of even order). Then we have $$ X^4+Y^4=(X^2-Y^2)^2-(cXY)^2=(X^2-cXY-Y^2)(X^2+cXY-Y^2).$$

Watson
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  • Notice that we have for any $f \in \Bbb Z[X_1, ..., X_n]$, the ideal $(f,p) \subset \Bbb Z[X_1, ..., X_n]$ is prime if and only if $\bar f \in \Bbb F_p[X_1, ..., X_n]$ is irreducible. – Watson Feb 03 '18 at 16:53
  • Great, many thanks! – 57Jimmy Feb 03 '18 at 17:19