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

Question

Let $(R,m)$ be a local commutative integral domain. Are the residue field $k= R/m$ and the quotient field of $R$ the same?

I would think they are since $m$ is the only maximal ideal. What if $R$ is not an integral domain?

Answer 1

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 $$S^{-1} R = {\left\{\frac r s : r \in R, \, s \in S, \text{ and } \frac r s = \frac{r'}{s'} \text{ iff } \exists \, s'' \in S \text{ s.t. } s''(s'r - r's) = 0 \right\}},$$ where $S = R \setminus \{r \in R : rs = 0 \text{ for some nonzero } s \in R \}.$ Observe that $S$ is the multiplicatively closed set consisting of all elements 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 (algebraic) properties that distinguish these two rings (as algebraic objects) in general.

1.) Of course, the residue field is always a field, but the total ring of fractions need not be a field. If $R$ 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 $R,$ hence $\mathfrak m$ is precisely the set of zero divisors of $R.$) Consequently, we have that $S = R \setminus \mathfrak m$ 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 $R = \mathbb Z / 4 \mathbb Z.$ Observe that $2$ is the only zero divisor of $R,$ hence we have $S = \{1, 3 \}.$ But these are units in $R,$ so inverting them accomplishes nothing, i.e., $Q(R) = R = \mathbb Z / 4 \mathbb Z$ in which $2$ is nilpotent.

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 $0 = ab = 1(1 \cdot ab - 0 \cdot 1)$ so that $\frac a 1 \cdot \frac b 1 = \frac{ab} 1 = \frac 0 1.$

If $R$ is a domain, then these are all non-issues. Explicitly, we have that $S = R \setminus \{0\}$ so that $Q(R)$ coincides with the field of fractions $\operatorname{Frac}(R).$ On the other hand, 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 $R / \mathfrak m \cong \operatorname{Frac}(R).$

Answer 2

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.