Question about the definition of Diophantine sets

Question

I am currently reading "A Course in Mathematical Logic for Mathematicians" by Manin. The book defines Diophantine sets as follows:

The projections of the level sets of a special kind of primitive recursive functions polynomials with coefficients in $\mathbb{Z}^+$—are called Diophantine sets.

It appears to me that all Diophantine sets are of the form $E\times \mathbb{Z}^i$ where $E$ is a finite subset of $\mathbb{Z}^n$.

Proof: Let $f:(\mathbb{Z}^+)^{n+m}\rightarrow \mathbb{Z}^+$ be polynomial with coefficients in $\mathbb{Z}^+$. I will write the polynomial $f$ as $f(x_1,x_2,...,x_n,y_1,...,y_m)$. Where $y_1,...,y_m$ are the variables that $f$ don't depend on.

Claim: Let $c\in \mathbb{Z}^{+}$, If $(a_1,a_2,...,a_n,b_1,...,b_m)\in f^{-1}(c)$ , then $ \land_{i=1}^n(1 \leq a_i \leq c)$

Proof: Lets prove this for $a_1$. Since $f$ depends on $a_1$, therefore $$f(a_1,...,b_m)=da_1\times(\text{positive function in the rest of the variables})+\text{another positive function in the rest of the variables}=c$$

Since the postive functions in the equation bove are positive integers (at least 1), thus: $$da_1\leq c$$ Since $d$ is a positive integer, therefore

$a_1\leq c$

$\square$

Now it follows that $f^{-1}(c)=E\times (\mathbb{Z^+})^m$ for some finite subset $E$ of $\{1,2,...,c\}^n$. It follows easily as well that every projection of $f^{-1}(c)$ will be of the form $F\times (\mathbb{Z}^+)^i$ where $F$ is finite.

Question: What is wrong with my argument ?

Thank you

Answer 1

You are right !

We have that a polynomial is :

$g \in \mathbb Z [x_0, . . . , x_n]$

and define a set $E$ which coincides with the set of positive values of $g$ at points in $(\mathbb Z^+)^{n+1}$.

This means that when the $(n+1)$-uple $(x_0, . . . , x_n)$ "spans over" $(\mathbb Z^+)^{n+1}$ the value of the polynomial $g$ "describes" the set $E$.

In more formal words :

$g : (\mathbb Z^+)^{n+1} \rightarrow E \subset \mathbb Z^+$.

Consider the polynomial $g := x + y$; when the "point" $(x,y)$ "moves" into the space $(\mathbb Z^+)^2$ the values og $g$ are "unlimited".

From page 192 :

Recall that the level set at $m$ (or simply the $m$-level) of a function $f$ from $(\mathbb Z^+)^n$ to $\mathbb Z^+$ is the set $E ⊂ D(f)$ such that : $x \in E ⇔ f(x) = m$.

Thus, in our example with $g := x + y$, the level set at $1$ is the empty set, because there are no positive integers $x,y$ such that $x+y = 1$.

With $m=2$ we have only : $x=y=1$; with $m=3$ we have two "solutions". And so on.

For each $m$ the number of "solutions" is finite.