Why does this equivalence stand?

Question

I am reading the proof of the following theorem:

THEOREM A.

Let $R$ be an integral domain of characteristic zero; then the diophantine problem for $R[T]$ with coefficients in $\mathbb{Z}T]$ is unsolvable.

($R[T]$ denotes the ring of polynomials over $R$, in one variable $T$.)

at a paper.

We write $V \sim W$ to denote that the polynomials $V$ and $W$ in $R[T]$ take the same value at $T=1$.

$$Imt(Y) \leftrightarrow Y \in R[T]\land \exists X\in R[T]: X^2-(T^2-1)Y^2=1$$

Lemma 2.3.

We have:

(i) The relation $Imt(Y)$ is diophantine over $R[T]$ with coefficients in $\mathbb{Z}[T]$.

(ii) If $Y$ satisfies $Imt(Y)$, then there exists an integer $m$ such that $Y \sim m$.

(iii) For every integer $m$ there exists a polynomial $Y$ satisfying $Imt(Y)$ and $Y \sim m$.

Could you take a look at the proof:

and explain to me the following?

$$\exists z_1, \dots , z_n\in \mathbb{Z}: P(z_1, \dots , z_n)=0 \leftrightarrow \exists Z_1, \dots , Z_n\in R[T]: (Imt(Z_1) \land \dots \land Imt(Z_n) \land P(Z_1, \dots , Z_n) \sim 0)$$

Why does this equivalence stand?

Answer 1

Firstly I think the formula in the paper was typeset misleadingly and should have read: $\def\zz{\mathbb{Z}}$ $\def\eq{\Leftrightarrow}$

$\exists z_1,\cdots,z_n \in \zz\ ( P(z_1,\cdots,z_n) = 0 )$

$\ \eq \exists Z_1,\cdots,Z_n \in R[T]\ ( Imt(Z_1) \land \cdots \land Imt(Z_n) \land P(Z_1,\cdots,Z_n) \sim 0 )$

For the forward implication, for each $k \in [1..n]$ let $Z_k \in R[T]$ such that $Imt(Z_k)$ and $Z_k \sim z_k$ by Lemma 2.3(iii), and then $P(Z_1,\cdots,Z_n) \sim P(z_1,\cdots,z_n) = 0$.

For the backward implication, let $z_k \in \zz$ such that $Z_k \sim z_k$ by Lemma 2.3(ii), and then $P(z_1,\cdots,z_n) \sim P(Z_1,\cdots,Z_n) \sim 0$. But $P(z_1,\cdots,z_n)$ is a constant in $R[T]$ and hence $P(z_1,\cdots,z_n) = 0$.