Lemma 10.25.1. Let $\mathfrak p$ be a minimal prime of a ring $R$. Every element of the maximal ideal of $R_{\mathfrak p}$ is nilpotent. If $R$ is reduced then $R_{\mathfrak p}$ is a field.

## 10.25 Zerodivisors and total rings of fractions

The local ring at a minimal prime has the following properties.

**Proof.**
If some element $x$ of ${\mathfrak p}R_{\mathfrak p}$ is not nilpotent, then $D(x) \not= \emptyset $, see Lemma 10.17.2. This contradicts the minimality of $\mathfrak p$. If $R$ is reduced, then ${\mathfrak p}R_{\mathfrak p} = 0$ and hence it is a field.
$\square$

Lemma 10.25.2. Let $R$ be a reduced ring. Then

$R$ is a subring of a product of fields,

$R \to \prod _{\mathfrak p\text{ minimal}} R_{\mathfrak p}$ is an embedding into a product of fields,

$\bigcup _{\mathfrak p\text{ minimal}} \mathfrak p$ is the set of zerodivisors of $R$.

**Proof.**
By Lemma 10.25.1 each of the rings $R_\mathfrak p$ is a field. In particular, the kernel of the ring map $R \to R_\mathfrak p$ is $\mathfrak p$. By Lemma 10.17.2 we have $\bigcap _{\mathfrak p} \mathfrak p = (0)$. Hence (2) and (1) are true. If $x y = 0$ and $y \not= 0$, then $y \not\in \mathfrak p$ for some minimal prime $\mathfrak p$. Hence $x \in \mathfrak p$. Thus every zerodivisor of $R$ is contained in $\bigcup _{\mathfrak p\text{ minimal}} \mathfrak p$. Conversely, suppose that $x \in \mathfrak p$ for some minimal prime $\mathfrak p$. Then $x$ maps to zero in $R_\mathfrak p$, hence there exists $y \in R$, $y \not\in \mathfrak p$ such that $xy = 0$. In other words, $x$ is a zerodivisor. This finishes the proof of (3) and the lemma.
$\square$

The total ring of fractions $Q(R)$ of a ring $R$ was introduced in Example 10.9.8.

Lemma 10.25.3. Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset consisting of nonzerodivisors. Then $Q(R) \cong Q(S^{-1}R)$. In particular $Q(R) \cong Q(Q(R))$.

**Proof.**
If $x \in S^{-1}R$ is a nonzerodivisor, and $x = r/f$ for some $r \in R$, $f \in S$, then $r$ is a nonzerodivisor in $R$. Whence the lemma.
$\square$

We can apply glueing results to prove something about total rings of fractions $Q(R)$ which we introduced in Example 10.9.8.

Lemma 10.25.4. Let $R$ be a ring. Assume that $R$ has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$, and that $\mathfrak q_1 \cup \ldots \cup \mathfrak q_ t$ is the set of zerodivisors of $R$. Then the total ring of fractions $Q(R)$ is equal to $R_{\mathfrak q_1} \times \ldots \times R_{\mathfrak q_ t}$.

**Proof.**
There are natural maps $Q(R) \to R_{\mathfrak q_ i}$ since any nonzerodivisor is contained in $R \setminus \mathfrak q_ i$. Hence a natural map $Q(R) \to R_{\mathfrak q_1} \times \ldots \times R_{\mathfrak q_ t}$. For any nonminimal prime $\mathfrak p \subset R$ we see that $\mathfrak p \not\subset \mathfrak q_1 \cup \ldots \cup \mathfrak q_ t$ by Lemma 10.15.2. Hence $\mathop{\mathrm{Spec}}(Q(R)) = \{ \mathfrak q_1, \ldots , \mathfrak q_ t\} $ (as subsets of $\mathop{\mathrm{Spec}}(R)$, see Lemma 10.17.5). Therefore $\mathop{\mathrm{Spec}}(Q(R))$ is a finite discrete set and it follows that $Q(R) = A_1 \times \ldots \times A_ t$ with $\mathop{\mathrm{Spec}}(A_ i) = \{ q_ i\} $, see Lemma 10.24.3. Moreover $A_ i$ is a local ring, which is a localization of $R$. Hence $A_ i \cong R_{\mathfrak q_ i}$.
$\square$

## 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.

## Comments (4)

Comment #6046 by Shurui Liu on

Comment #6047 by Johan on

Comment #8084 by Miles Reid on

Comment #8090 by Stacks Project on