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This is a particular case of [Corollary, McCoy]

Lemma 15.30.16. Let $R$ be a ring. Let $a_1, \ldots , a_ n \in R$ be elements such that $R \to R^{\oplus n}$, $x \mapsto (xa_1, \ldots , xa_ n)$ is injective. Then the element $\sum a_ i t_ i$ of the polynomial ring $R[t_1, \ldots , t_ n]$ is a nonzerodivisor.

Proof. If one of the $a_ i$ is a unit this is just the statement that any element of the form $t_1 + a_2 t_2 + \ldots + a_ n t_ n$ is a nonzerodivisor in the polynomial ring over $R$.

Case I: $R$ is Noetherian. Let $\mathfrak q_ j$, $j = 1, \ldots , m$ be the associated primes of $R$. We have to show that each of the maps

\[ \sum a_ i t_ i : \text{Sym}^ d(R^{\oplus n}) \longrightarrow \text{Sym}^{d + 1}(R^{\oplus n}) \]

is injective. As $\text{Sym}^ d(R^{\oplus n})$ is a free $R$-module its associated primes are $\mathfrak q_ j$, $j = 1, \ldots , m$. For each $j$ there exists an $i = i(j)$ such that $a_ i \not\in \mathfrak q_ j$ because there exists an $x \in R$ with $\mathfrak q_ jx = 0$ but $a_ i x \not= 0$ for some $i$ by assumption. Hence $a_ i$ is a unit in $R_{\mathfrak q_ j}$ and the map is injective after localizing at $\mathfrak q_ j$. Thus the map is injective, see Algebra, Lemma 10.63.19.

Case II: $R$ general. We can write $R$ as the union of Noetherian rings $R_\lambda $ with $a_1, \ldots , a_ n \in R_\lambda $. For each $R_\lambda $ the result holds, hence the result holds for $R$. $\square$

Comments (2)

Comment #2778 by on

This is also a particular case of McCoy's theorem, which states that if a polynomial is a zero-divisor, then all its coefficients are annihilated by a nonzero scalar (simultaneously). This is well known for polynomials in 1 variable (e.g., and countless math.stackexchange threads); the case of n variables can be reduced to 1 variable by induction.

There are also:

  • 4 comment(s) on Section 15.30: Koszul regular sequences

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