## Tag `068P`

Chapter 15: More on Algebra > Section 15.27: Koszul regular sequences

Lemma 15.27.16. 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 10.62.19.

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 $R$. $\square$

The code snippet corresponding to this tag is a part of the file `more-algebra.tex` and is located in lines 6337–6343 (see updates for more information).

```
\begin{lemma}
\label{lemma-make-nonzero-divisor}
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.
\end{lemma}
\begin{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 $R$.
\medskip\noindent
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 \ref{algebra-lemma-zero-at-ass-zero}.
\medskip\noindent
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 $R$.
\end{proof}
```

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