Exercise A0.V.1.15

Lemma P1.1

Let $R$ be an integral domain. Then $R[x_1,x_2,x_3, \ldots]$ is an integral domain.

Proof

Let $f, g \in R[x_1,x_2,x_3, \ldots]$ be non-zero. Being finitely supported, there exists a subring $S = R[x_{j_1}, x_{j_2}, \ldots, x_{j_n}]$ such that $f, g \in S.$ By III.1.15 we know that $S$ is an integral domain, hence $fg \neq 0.$ Since $f$ and $g$ were arbitrary, we can conclude that $R[x_1,x_2,x_3, \ldots]$ is an integral domain.

$\blacksquare$

Lemma P1.2

Let $R$ be an integral domain and let $u, v \in R[x_1,x_2,x_3,\ldots]$ be non-zero. Let $\Xi(f)$ denote the set of indeterminates appearing in $f$ with non-zero coefficients. Then $\Xi(uv) = \Xi(u) \cup \Xi(v).$

Proof

For any indeterminate $x_k$ we can consider $u, v$ as elements of $S[x_i]$ where $S = R[\{x_1,\ldots\} \setminus \{x_i\}].$ By P1.1 we know that $S$ is an integral domain, hence we can use the degree formula (III.1.14):

$$\deg_{x_i}(uv) = \deg_{x_i}(u) + \deg_{x_i}(v).$$

Whenever $x_i$ is present in $u$ or $v$, its degree is non-zero in $uv,$ and whenever it is present in $uv,$ its degree must be non-zero in $u$ or $v.$ Since

$$x_i \in \Xi(u) \iff \deg_{x_i}(u) \neq 0,$$

this suffices to establish equality of sets.

$\blacksquare$

Proposition

Let $R = \mathbb{Z}[x_1,x_2,x_3,\ldots]$ and let $S = \mathbb{Z}[x_1,\ldots,x_n]$ be a subring of $R.$ Let $f \in S$ be non-zero and let $g \in R$ be such that $(f) \subseteq (g).$ Then $g \in S.$

Proof

Since $(f) \subseteq (g)$ there exists $u \in R$ such that $ug = f.$ By P1.2 we know that $\Xi(f) = \Xi(ug) = \Xi(u) \cup \Xi(g).$ So in particular, $\Xi(g) \subseteq \Xi(f).$ This allows us to express $g$ in terms of the indeterminates $x_1, \ldots, x_n,$ and hence as an element of $S.$

$\blacksquare$