## 10.25 Zerodivisors and total rings of fractions

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

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.

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

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

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

3. $\bigcup _{\mathfrak p\text{ minimal}} \mathfrak p$ 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 $A_ i \cong R_{\mathfrak q_ i}$. $\square$

Comment #6046 by Shurui Liu on

Lemma 00EW (3) doesn't need R to be reduced and hence lemma 02LX needn't require $\mathfrak q_1 \cup \ldots \cup \mathfrak q_ t$ to be the set of zerodivisors of R (since it is automatically true).

Comment #6047 by on

@#6046. No. I suggest trying to make a counterexample.

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