Is the following the proof in Atiyah & Macdonald? I find it constructive:
The $A$-module $M$ is the quotient of $A^n$ for some natural number $n$.
There exists a morphism $\Phi: A^n \to A^n$ covering $\phi: M \to M$.
Since $A$ is a commutative ring, determinant makes sense and one can define the characteristic polynomial $p_\Phi(t) := \det(\Phi-t\,\mathrm{Id})$.
By the Cayley–Hamilton Theorem, $p_\Phi(\Phi)=0: A^n \to A^n$.
Observe that this morphism covers $p_\Phi(\phi): M \to M$.
Since the morphism $A^n \to M$ is surjective, it follows that $p_\Phi(\phi)=0$ also.
For a given $\phi$, I don't know anything about the uniqueness the polynomial $p_\Phi(t)$.
Already posted this on mathoverflow.net.
So we solved this today, in a way similar to how the Hilbert basis theorem might be proved. First show the following lemma, then the claim that $D^n(B)$ is artinian as a module over $B[t_1,\dotsc,t_n]$ is a direct consequence.
Lemma If $M$ is an artinian $R$-module, then $D^1 \otimes M$ is an artinian $R[t]$ module.
Proof: We can think of elements in $D^1 \otimes M$ as polynomials with coefficients in $M$. So each element $f$ has a degree $\deg f$ and a leading coefficient $lc(f)$. Now consider a descending chain
$$ N_1 \supset N_2 \supset \dots$$
an set $M_{k, l } := \{ lc(f) : f \in N_k, \deg(f) = l \} \subset M$. Clearly
$M_{k,l}\supset M_{k+1,l}$ and $M_{k,l}\supset M_{k,l+1}$.
Now observe that $N_k = N_{k+1}$ if and only if $M_{k,l} = M_{k+1,l}$ for all $l$. "$\Rightarrow$" is clear, so for "$\Leftarrow$" consider $f \in N_k \setminus N_{k+1}$ with minimal degree. Then there is some $g \in N_{k+1}$ with $lc(f) = lc(g)$ and $\deg(f) = \deg(g)$. So $f - g \in N_k\setminus N_{k+1}$ has lower degree, which is a contradiction.
Next consider the diagram
\begin{align*}
\vdots && \vdots && \vdots \\
\cap && \cap && \cap\\
M_{3,1} &\supset& M_{3,2} &\supset& M_{3,3} &\supset& \dots\\
\cap && \cap && \cap\\
M_{2,1} &\supset& M_{2,2} &\supset& M_{2,3} &\supset& \dots\\
\cap && \cap && \cap \\
M_{1,1} &\supset& M_{1,2} &\supset& M_{1,3} &\supset& \dots
\end{align*}
Now the sequence $M_{1,1} \supset M_{2,2} \supset M_{3,3} \supset \dots$ has to stabilize at some point, because $M$ is artinian, say at $M_{k,k}$. Now it is easy to see that $M_{n,m} = M_{k,k}$ for all $n,m\geq k$. Thus there are only finitely many rows in the diagram which are not yet equal, but they will equal after finitely many steps. So the criterion above applies and we see that the sequence $N_k$ has to stabilize, proving the claim.
This directly proves the following
Corollary If $B$ is artinian, then $D^n(B)$ is artinian as a module over $B[t_1,\dotsc,t_n]$.
Proof: Apply the lemma inductively.
Best Answer
Yes, you are correct, it is a polynomial ring on infinitely many generators. Given any set $T=\{t_1,t_2,\ldots\}$, we can form the free $A$-algebra generated by $T$, denoted by $A[t_1,t_2,\ldots]$ or just $A[T]$, which is the polynomial ring with coefficients in $A$ and with the elements of $T$ acting as indeterminates. This construction does not depend on $T$ being finite or infinite. In fact we could take $T$ to be uncountable. An important thing to keep in mind is that while there could be infinitely many elements of $T$, any single element $f(t_i)\in A[T]$, i.e. a polynomial in the symbols $t_i$ with coefficients in $A$, will only use finitely many of the $t_i$.
When $B$ is any $A$-algebra, $B$ is a quotient of $A[\{t_b\}_{b\in B}]$ by sending, for example, $a_1t_{b}+a_2t_{3b}^2$ to $a_1b+a_2(3b)^2=a_1b+9a_2b^2\in B$.
For example, the ring $\mathbb{C}$ is a $\mathbb{Q}$-algebra. We map $\mathbb{Q}[\{t_{\alpha}\}_{\alpha\in\mathbb{C}}]$ to $\mathbb{C}$ by sending each $\mathbb{Q}$-polynomial $f(t_{\alpha},t_{\beta},\ldots)$ to its evaluation when we replace $t_\alpha$ with $\alpha\in\mathbb{C}$, $t_\beta$ with $\beta\in\mathbb{C}$, etc., so that $\mathbb{C}$ is isomorphic to the quotient of $\mathbb{Q}[\{t_{\alpha}\}_{\alpha\in\mathbb{C}}]$ by the kernel of the evaluation map.