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I know that the prime ideals of $\mathbb Z[x]$ are either the ideals of the form $(p,f(x))$, for a prime $p\in \mathbb Z$ and a polynomial irreducible in $\mathbb F_p[x]$, or the ones of the form $(f(x))$ for any irreducible polynomial in $\mathbb Z[x]$. In particular, the prime ideals of this last form are in bijection with the prime ideals of $\mathbb Q[x]$. This means exactly that a polynomial irreducible in $\mathbb Z[x]$ is also irreducible in $\mathbb Q[x]$, right?

Dr. Scotti
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    The short answer is: No. On the one hand, the zero ideal is missing (which is also prime as $\mathbb{Z}[X]$ is an integral domain. On the other hand, you should rather speak of prime ideals $(r)$ generated by an irreducible element $r \in \mathbb{Z}[X]$. This could also be a prime number $p$. I.e., $(p) \subseteq \mathbb{Z}[X]$ is a prime ideal, but $p$ is a unit in $\mathbb{Q}$, so does not generate a prime ideal. – Algebrus Nov 25 '21 at 17:11
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    In fact, for a UFD R, let Q be its fraction field, then f(x) is irreducible in Q[x] if and only if f* is irreducible in R[x], where f* is primitive. – ALe0 Nov 25 '21 at 17:12
  • However, Gauss's lemma says that a non-constant polynomial $f \in \mathbb{Z}[X]$ is irreducible in $\mathbb{Z}[X]$ if and only if the gcd of its coefficients is $1$ and it is irreducible in $\mathbb{Q}[X]$; see here: https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomials) – Algebrus Nov 25 '21 at 17:13
  • Also, $\langle p\rangle$ is a prime ideal, where $p\in \mathbb Z$ is prime. I’d have put those in $\langle f(x)\rangle,$ except for your comment about $\mathbb Q[x].$ – Thomas Andrews Nov 25 '21 at 17:13
  • I suppose if you addd the $0$ polynomial to the irreducibles, then you’ve covered $\langle p\rangle$ and the zero ideal… – Thomas Andrews Nov 25 '21 at 17:17
  • Generally if $\rm,D,$ is a domain with fraction field $,\rm K,$ then $$\rm f,\ is\ prime\ in\ D[x]\iff f,\ is\ prime (= irreducible)\ in\ K[x]\ and,\ f,\ is\ superprimitive $$ $$\rm where,\ f,\ is\ {\bf superprimitive}\ in\ D[x],\ :=,\ d,|,cf, \Rightarrow, d,|,c,\ \ for\ all,\ c,d\in D^*$$ – Bill Dubuque Nov 25 '21 at 17:36
  • See here for generalizations of Gauss's Lemma and related results. – Bill Dubuque Nov 25 '21 at 17:41
  • I see that is not true that a polynomial irreducible in $\mathbb Z[x]$ is irreducible in $\mathbb Q[x]$. I just don't understand where is the mistake in this arguement: "$\mathbb Z[x]\to \mathbb Q[x]$ induces a bijection between $\operatorname{Spec}\mathbb Q[x]$ and the prime ideals in $\mathbb Z[x]$ of the form $(f)$ for $f$ irreducible in $\mathbb Z[x]$. If I take a prime ideal $(f)$ in $\mathbb Z[x]$, under this bijection it corresponds to the ideal generated by $f$ in $\mathbb Q[x]$. The fact that $(f)$ in $\mathbb Q[x]$ is still prime imply that $f$ is irreducible in $\mathbb Q[x]$." – Dr. Scotti Nov 25 '21 at 19:20
  • $f$ always assumed to be nonconstant – Dr. Scotti Nov 25 '21 at 19:30
  • Plus Gauss lemma says that if $f$ is a non constant polynomial irreducible in $\mathbb Z[x]$, hence primitive, then is irreducible in $\mathbb Q[x]$ (that is what I wanted to prove with my argument). Am I wrong also in the interpretation of the lemma? – Dr. Scotti Nov 25 '21 at 19:43
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    @Dorian: It is correct that any irreducible non-constant polynomial $f \in \mathbb{Z}[X]$ is irreducible in $\mathbb{Q}[X]$ (by Gauss's lemma), hence induces a prime ideal $(f) \subseteq \mathbb{Q}[X]$. However, it is not true that any principal (i.e., generated by one element) prime ideal in $\mathbb{Z}[X]$ has the shape $(f)$ for a non-constant irreducible polynomial $f \in \mathbb{Z}[X]$: You are missing $(0)$ and the principal ideals generated by prime numbers $p \in \mathbb{Z}$. – Algebrus Nov 26 '21 at 05:58
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    I.e., the description of $\mathrm{Spec}(\mathbb{Z}[X])$ as proposed in your question is incomplete. – Algebrus Nov 26 '21 at 06:00

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