The diophantine problem for $R[T]$ is solvable iff the diophantine problem for $R$ is solvable

Question

One part of the paper that I am reading is the following:

Let $R$ be a commutative ring with unity and let $R'$ be a subring of $R$.

We say that the diophantine problem for $R$ with coefficients in $R'$ is unsolvable (solvable) if there exists no (an) algorithm to decide whether or not a polynomial equation (in several variables) with coefficients in $R'$ has a solution in $R$.

$$\dots \dots \dots \dots \dots$$

Theorem.

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 $T$.)

$$\dots \dots \dots \dots \dots$$

It is obvious that the diophantine problem for $R[T]$ with coefficients in $\mathbb{Z}$ is solvable if and only if the diophantine problem for $R$ with coefficients in $\mathbb{Z}$ is solvable.

$$$$

Could you explain to me the last sentence?

Why does this stand?

Does the direction $\Leftarrow$ stand because of the following?

We know that there is an algorithm that decides whether or not an equation with coefficients in $\mathbb{Z}$ has a solution in $R$.

We consider this equation as the constant term of a polynomial equation, so there is an algorithm that decides whether or not an equation with coefficients in $\mathbb{Z}$ has a solution in $R[T]$.

Is the justification of this direction correct?

Answer 1

The polynomial equation $$P(x_1,\dots,x_n)=0,\tag{1}$$ where $P$ has integer coefficients, has a solution in $R[T]$ if and only if it has a solution in $R$.

For one direction, note that any solution of (1) in $R$ is in particular a solution in $R[T]$.

For the other direction, suppose that the ordered $n$-tuple $(Q_1(T), \dots, Q_n(T))$ of polynomials is a solution of (1) in $R[T]$. Then the ordered $n$-tuple $(Q_1(0),\dots,Q_n(0))$ is a solution of (1) in $R$.

Thus any algorithm for determining solvability in one of the rings $R$ or $R[T]$ automatically determines solvability in the other.