# What is the relationship between the residue field and the quotient field of a local ring?

Let $$(R,m)$$ be a local commutative integral domain. Are the residue field $$k= R/m$$ and the quotient field of $$R$$ are the same? I would think they are since $$m$$ is the only maximal ideal. What if $$R$$ is not an integral domain?

Consider a commutative local ring $$(R, \mathfrak m).$$ By definition, the residue field of $$R$$ is given by $$k = R / \mathfrak m.$$ On the other hand, the total ring of fractions $$Q(R)$$ is given by $$Q(R) = S^{-1} R = \biggl\{\frac r s \,\bigg|\, r \in R, \, s \in S, \text{ and } \frac r s = \frac{r'}{s'} \text{ iff } \exists \, s'' \in S \text{ s.t. } s''(s'r - r's) = 0 \},$$ where we have that $$S = R - \{r \in R \,|\, rs = 0 \text{ for some nonzero } s \in R \}.$$ Observe that $$S$$ is the multiplicatively closed set that consists of the elements of $$R$$ that are neither zero divisors nor $$0.$$

Purely in terms of their definitions, these two rings are not the same as sets: the residue field consists of equivalence classes of elements modulo the unique maximal ideal $$\mathfrak m,$$ and the total ring of fractions $$Q(R)$$ consists of fractions of elements in $$R$$ whose denominator is neither a zero divisor nor $$0.$$ But there are more properties that distinguish these two rings in general.

1.) Of course, the residue field is always a field, but the total ring of fractions need not be a field. Given that $$R$$ is Artinian (i.e., $$R$$ is Noetherian and $$\dim R = 0$$) but not a field, the maximal ideal $$\mathfrak m$$ consists of all zero divisors of $$R.$$ (Indeed, every prime ideal of $$R$$ is maximal, hence $$\mathfrak m$$ is the unique prime ideal of $$R$$ so that $$\mathfrak m$$ is the unique minimal prime ideal of $$R,$$ from which it follows that $$\mathfrak m$$ is precisely the set of zero divisors of $$R.$$) Consequently, we have that $$S = R - \mathfrak m$$ so that $$Q(R) = S^{-1} R =$$ $$R_{\mathfrak m} \cong R$$ is not a field by assumption.

2). Every field is reduced (i.e., the only nilpotent element of a field is $$0$$); however, the total ring of fractions need not be reduced. Consider the ring $$R = \mathbb Z / 4 \mathbb Z.$$ Observe that $$2$$ is the only zero divisor of $$R,$$ hence we have $$S = \{0, 2 \}.$$ We have therefore that $$Q(R) = \bigl \{0, 1, 2, 3, \frac 1 3, \frac 2 3 \bigr \}$$ is not reduced, as $$\frac 2 3$$ is nilpotent. Particularly, we have that $$\bigl(\frac 2 3 \bigr)^2 = 0.$$

3.) Every field is an integral domain; however, the total ring of fractions need not be a domain. Particularly, the zero divisors of $$R$$ give rise to zero divisors of $$Q(R).$$ Explicitly, given that $$ab = 0$$ in $$R,$$ we have that $$0 = ab = 1(1 \cdot ab - 0 \cdot ss'),$$ from which it follows that $$\frac a s \cdot \frac b {s'} = \frac{ab}{ss'} = \frac 0 1.$$

Given that $$R$$ is a domain, these are non-issues. Explicitly, we have that $$S = R - \{0\}$$ so that $$Q(R)$$ coincides with the field of fractions $$\operatorname{Frac}(R)$$; that immediately does away with points (1.) and (3.). Further, if $$R$$ is reduced, then $$Q(R)$$ is reduced, hence (2.) is irrelevant. But it remains to be seen how the residue field $$k = R / \mathfrak m$$ and the field of fractions $$\operatorname{Frac}(R)$$ relate in this case.

Like you mentioned, there are a few things we can say.

1.) We have a short exact sequence $$R \to k \to 0,$$ i.e., $$R \to k$$ is surjective.

2.) We have a short exact sequence $$0 \to R \to \operatorname{Frac}(R),$$ i.e., $$R \to \operatorname{Frac}(R)$$ is injective.

3.) There exists a commutative local domain $$(R, \mathfrak m)$$ such that $$R / \mathfrak m \cong \operatorname{Frac}(R).$$

I think you may be confusing two concepts. Take the example $$R = \mathbb{Z}_p$$, the $$p$$-adic integers. Then $$m = p\mathbb{Z}_p$$. The residue field $$k = R/m$$ is the field of $$p$$ elements, while the quotient field of $$R$$ is $$\mathbb{Q}_p$$, the $$p$$-adic rationals.

• I see. But it should be the case that there's a surjection from the residue field into $R$ and an injection from $R$ into the quotient field, right? – FearfulSymmetry May 10 '20 at 20:55
• There is a surjection from $R$ onto the residue field (not the other way around as you stated), and an injection from $R$ into the quotient field. – Ted May 10 '20 at 20:56
• Does the residue field inject into the quotient field? – FearfulSymmetry May 10 '20 at 21:00
• Did you think about the above example? Does the field of $p$ elements inject into $\mathbb{Q}_p$ ? – Ted May 11 '20 at 19:40