The Stacks project

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., https://ysharifi.wordpress.com/tag/mccoy-theorem/ and countless math.stackexchange threads); the case of n variables can be reduced to 1 variable by induction.

There are also:

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

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 068P. Beware of the difference between the letter 'O' and the digit '0'.