The generic fiber of a morphism of schemes

Question

Let $m$ be a non-zero integer and $f: \operatorname{Spec}\mathbb{Z}[T_1,T_2]/\langle T_1T_2^2-m\rangle \to \operatorname{Spec}\mathbb{Z}$.

Why is the generic fiber $\operatorname{Spec} \mathbb{Q}[T_1,T_2]/\langle T_1T_2^2-m\rangle=\operatorname{Spec}\mathbb{Q}[T_2,\frac{1}{T_2}]$?

Suppose that, instead of $\mathbb{Q}$, we have $\mathbb{F}_p$. What are the conditions according to which $X_p$ is integral?

Answer 1

Here is an answer to your first question:

Note that the morphism $f$ is induced by the injective homomorphism $$\mathbb{Z} \to \mathbb{Z}[T_1,T_2]/(T_1 T_2^2 - m).$$ As MooS wrote, the generic fibre of $f$ is given by those prime ideals of $\mathbb{Z}[T_1,T_2]/(T_1 T_2^2 - m)$ that restrict to $(0)$ in $\mathbb{Z}$. These are precisely the prime ideals of the localization $(\mathbb{Z}\setminus \{0\}) ^{-1}(\mathbb{Z}[T_1,T_2]/(T_1 T_2^2 - m)) = \mathbb{Q}[T_1,T_2]/(T_1 T_2^2 - m)$.

For the second part of the first question it suffices to construct an isomorphism between $\mathbb{Q}[T_1, T_2]/(T_1 T_2^2 - m)$ and $\mathbb{Q}[T_2, 1/T_2]$. For this, consider $$\mathbb{Q}[T_2, 1/T_2] \to \mathbb{Q}[T_1,T_2]/(T_1 T_2^2 - m)$$ defined by sending $T_2$ to the class of $T_2$. Note that this is well-defined, since the class of $T_2$ in $\mathbb{Q}[T_1, T_2]/(T_1 T_2^2 - m)$ is invertible with inverse the class of $T_1 T_2/m$.

Conversely, consider the $\mathbb{Q}$-homomorphism $$\mathbb{Q}[T_1,T_2]/(T_1 T_2^2 - m) \to \mathbb{Q}[T_2, 1/T_2]$$ defined by sending the class of $T_2$ to $T_2$ and the class of $T_1$ to $m/T_2^2$ (by definition this is well-defined).

These two homomorphisms are inverse to each other.

Answer 2

Per definition the generic fibre consists of primes $\mathfrak p \in Spec ~\mathbb Z[T_1,T_2]/(T_1T_2^2-m)$ with $\mathfrak p \cap \mathbb Z=(0)$.

These are precisely the primes, which survive localisation of $\mathbb Z[T_1,T_2]/(T_1T_2^2-m)$ at $S := \mathbb Z \setminus \{0\}$, hence equal to $Spec ~ \mathbb Q[T_1,T_2]/(T_1T_2^2-m)$.

Furthermore we can calculate $\mathbb Q[T_1,T_2]/(T_1T_2^2-m) = \mathbb Q[T_2,\frac{m}{T_2^2}] = \mathbb Q[T_2,\frac{1}{T_2^2}] = \mathbb Q[T_2,\frac{1}{T_2}]$.

Answer 3

For the sake of completeness (You have already solved it and i believe you :)), let us consider the 2nd question. If $X$ is a scheme over $\mathbb F_p$, the fibre over $p$ of the unique morphism $X \to Spec ~\mathbb Z$ is $X$ itself. So we just need to check for which integers $m$ the polynomial $T_1T_2^2-m$ is prime (=irreducible) in $\mathbb F_p[T_1,T_2]$, which is an easy exercise.