If $R[s], R[t]$ are finitely generated as $R$-modules, the so is $R[s + t]$.

Question

Let $S \supset R$ as rings with $1 \in R$. Suppose that $s, t \in S$ and that the subrings $R[s], R[t]$ are finitely generated by $\{1, s, \dots, s^k\}$ and $\{1, t, \dots, t^m \}$. Then $R[s + t]$ is also finitely generated.

Let $g$ be a minimal degree monic polynomial for $s$ over $R$.

I'm thinking let $f(X) = X + t. \ $ Then $s + t = f(s). \ $ So $f(X)^n = q(X) g(X) + r(X)$ by the division algo, where $r(X) = 0$ or $\deg{r} \lt \deg{g}$. Then $f(s)^n = (s+t)^n = 0 + r(s), \ $ for some $r(X) \in R[t][X]$.

Now I'm very confused, but this looks like the right direction. Please work within it if you can, but all answers accepted. Thanks.

Answer 1

The ordinary approach is to prove $R[s]$ is a finitely generated $R$-module if and only if $s$ is integral over $R$ and that the elements of $S$ that are integral over $R$ form a subring of $S$. (See, for instance, Atiyah-Macdonald 5.1, 5.3 or Eisenbud 4.2, 4.6).

This settles your problem: $s$ and $t$ are integral over $R$ and since the integral elements of $S$ over $R$ form a ring, so is $s + t$, which shows $R[s+t]$ is finitely generated.

The proof of these results rests on the Cayley-Hamilton Theorem and, in the words of David Eisenbud, "It would be natural to prove [these results] by starting with the equations satisfied by two integral elements and simply writing down the equations satisfied by their sum and product. In a sense this is what we shall do. But in general the necessary polynomials are complicated. The Cayley-Hamilton Theorem gives them implicitly."

Edit. Let me at least quote the crucial theorem (Atiyah-Macdonald 5.1; Eisenbud 4.6); the formulation is more-or-less that from Atiyah-Macdonald.

Theorem. Let $S$ be an $R$-algebra and $s \in S$. Then the following four statements are equivalent:

1. $s$ is integral over $R$;
2. $R[s]$ is a finitely generated $R$-module;
3. $R[s]$ is contained in a subring $T$ of $S$ such that $T$ is a finitely generated $R$-module;
4. there exists a faithful $R[x]$-module $M$ which is finitely generated as an $R$-module. (Faithful means that no non-zero element of $R[x]$ annihilates $M$.)

The crucial implication is $4 \rightarrow 1$, which uses Cayley-Hamilton.

Answer 2

Let $R[s]$ be fin-gen with monic $f$ and $R[t]$ fin-gen with monic $g$.

$(X + t)^k, k \geq 1$, is a polynomial in $R[t][X]$ . Since $s$ is integral over $R$ with $f$ it's also integral over $R[t]$ with $f$ since $R \subset R[t]$. Then $(X + t)^k = q(X) f(X) + r(X)$ for some $r = 0$ or $\deg r \lt \deg f$. This means that if $h(X) = (X + t)^k$, then $h(s) = (s + t)^k$ and $h(s) = r(s)$ for some $r$. So we have that $p \in R[s+t]$ can be expressed as $p = p_0 + p_1 (s+t) + \dots p_n (s+t)^n = p_0 + p_1 r_1(s, t) + \dots + p_n r_n(s,t)$, each $r_i$ having degree $\leq \deg f + \deg g$, and each of which is a polynomial in $\{s^{i}t^{j} : 0 \leq i,j, i+j \leq \deg f + \deg g \} $. This set is finite and clearly generates $R[s + t]$ over $R$ as an $R$-module.

QED