The Stacks project

10.107 Epimorphisms of rings

In any category there is a notion of an epimorphism. Some of this material is taken from [Autour] and [Mazet].

Lemma 10.107.1. Let $R \to S$ be a ring map. The following are equivalent

  1. $R \to S$ is an epimorphism,

  2. the two ring maps $S \to S \otimes _ R S$ are equal,

  3. either of the ring maps $S \to S \otimes _ R S$ is an isomorphism, and

  4. the ring map $S \otimes _ R S \to S$ is an isomorphism.

Proof. Omitted. $\square$

Lemma 10.107.2. The composition of two epimorphisms of rings is an epimorphism.

Proof. Omitted. Hint: This is true in any category. $\square$

Lemma 10.107.3. If $R \to S$ is an epimorphism of rings and $R \to R'$ is any ring map, then $R' \to R' \otimes _ R S$ is an epimorphism.

Proof. Omitted. Hint: True in any category with pushouts. $\square$

Lemma 10.107.4. If $A \to B \to C$ are ring maps and $A \to C$ is an epimorphism, so is $B \to C$.

Proof. Omitted. Hint: This is true in any category. $\square$

This means in particular, that if $R \to S$ is an epimorphism with image $\overline{R} \subset S$, then $\overline{R} \to S$ is an epimorphism. Hence while proving results for epimorphisms we may often assume the map is injective. The following lemma means in particular that every localization is an epimorphism.

Lemma 10.107.5. Let $R \to S$ be a ring map. The following are equivalent:

  1. $R \to S$ is an epimorphism, and

  2. $R_{\mathfrak p} \to S_{\mathfrak p}$ is an epimorphism for each prime $\mathfrak p$ of $R$.

Proof. Since $S_{\mathfrak p} = R_{\mathfrak p} \otimes _ R S$ (see Lemma 10.12.15) we see that (1) implies (2) by Lemma 10.107.3. Conversely, assume that (2) holds. Let $a, b : S \to A$ be two ring maps from $S$ to a ring $A$ equalizing the map $R \to S$. By assumption we see that for every prime $\mathfrak p$ of $R$ the induced maps $a_{\mathfrak p}, b_{\mathfrak p} : S_{\mathfrak p} \to A_{\mathfrak p}$ are the same. Hence $a = b$ as $A \subset \prod _{\mathfrak p} A_{\mathfrak p}$, see Lemma 10.23.1. $\square$


Lemma 10.107.6. Let $R \to S$ be a ring map. The following are equivalent

  1. $R \to S$ is an epimorphism and finite, and

  2. $R \to S$ is surjective.

Proof. (This lemma seems to have been reproved many times in the literature, and has many different proofs.) It is clear that a surjective ring map is an epimorphism. Suppose that $R \to S$ is a finite ring map such that $S \otimes _ R S \to S$ is an isomorphism. Our goal is to show that $R \to S$ is surjective. Assume $S/R$ is not zero. The exact sequence $R \to S \to S/R \to 0$ leads to an exact sequence

\[ R \otimes _ R S \to S \otimes _ R S \to S/R \otimes _ R S \to 0. \]

Our assumption implies that the first arrow is an isomorphism, hence we conclude that $S/R \otimes _ R S = 0$. Hence also $S/R \otimes _ R S/R = 0$. By Lemma 10.5.4 there exists a surjection of $R$-modules $S/R \to R/I$ for some proper ideal $I \subset R$. Hence there exists a surjection $S/R \otimes _ R S/R \to R/I \otimes _ R R/I = R/I \not= 0$, contradiction. $\square$

Proof. This is clear from Lemma 10.107.1 part (3) as the map $S \to S \otimes _ R S$ is the map $R \to S$ tensored with $S$. $\square$

Lemma 10.107.8. If $k \to S$ is an epimorphism and $k$ is a field, then $S = k$ or $S = 0$.

Proof. This is clear from the result of Lemma 10.107.7 (as any nonzero algebra over $k$ is faithfully flat), or by arguing directly that $R \to R \otimes _ k R$ cannot be surjective unless $\dim _ k(R) \leq 1$. $\square$

Lemma 10.107.9. Let $R \to S$ be an epimorphism of rings. Then

  1. $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is injective, and

  2. for $\mathfrak q \subset S$ lying over $\mathfrak p \subset R$ we have $\kappa (\mathfrak p) = \kappa (\mathfrak q)$.

Proof. Let $\mathfrak p$ be a prime of $R$. The fibre of the map is the spectrum of the fibre ring $S \otimes _ R \kappa (\mathfrak p)$. By Lemma 10.107.3 the map $\kappa (\mathfrak p) \to S \otimes _ R \kappa (\mathfrak p)$ is an epimorphism, and hence by Lemma 10.107.8 we have either $S \otimes _ R \kappa (\mathfrak p) = 0$ or $S \otimes _ R \kappa (\mathfrak p) = \kappa (\mathfrak p)$ which proves (1) and (2). $\square$

Lemma 10.107.10. Let $R$ be a ring. Let $M$, $N$ be $R$-modules. Let $\{ x_ i\} _{i \in I}$ be a set of generators of $M$. Let $\{ y_ j\} _{j \in J}$ be a set of generators of $N$. Let $\{ m_ j\} _{j \in J}$ be a family of elements of $M$ with $m_ j = 0$ for all but finitely many $j$. Then

\[ \sum \nolimits _{j \in J} m_ j \otimes y_ j = 0 \text{ in } M \otimes _ R N \]

is equivalent to the following: There exist $a_{i, j} \in R$ with $a_{i, j} = 0$ for all but finitely many pairs $(i, j)$ such that

\begin{align*} m_ j & = \sum \nolimits _{i \in I} a_{i, j} x_ i \quad \text{for all } j \in J, \\ 0 & = \sum \nolimits _{j \in J} a_{i, j} y_ j \quad \text{for all } i \in I. \end{align*}

Proof. The sufficiency is immediate. Suppose that $\sum _{j \in J} m_ j \otimes y_ j = 0$. Consider the short exact sequence

\[ 0 \to K \to \bigoplus \nolimits _{j \in J} R \to N \to 0 \]

where the $j$th basis vector of $\bigoplus \nolimits _{j \in J} R$ maps to $y_ j$. Tensor this with $M$ to get the exact sequence

\[ K \otimes _ R M \to \bigoplus \nolimits _{j \in J} M \to N \otimes _ R M \to 0. \]

The assumption implies that there exist elements $k_ i \in K$ such that $\sum k_ i \otimes x_ i$ maps to the element $(m_ j)_{j \in J}$ of the middle. Writing $k_ i = (a_{i, j})_{j \in J}$ and we obtain what we want. $\square$

Lemma 10.107.11. Let $\varphi : R \to S$ be a ring map. Let $g \in S$. The following are equivalent:

  1. $g \otimes 1 = 1 \otimes g$ in $S \otimes _ R S$, and

  2. there exist $n \geq 0$ and elements $y_ i, z_ j \in S$ and $x_{i, j} \in R$ for $1 \leq i, j \leq n$ such that

    1. $g = \sum _{i, j \leq n} x_{i, j} y_ i z_ j$,

    2. for each $j$ we have $\sum x_{i, j}y_ i \in \varphi (R)$, and

    3. for each $i$ we have $\sum x_{i, j}z_ j \in \varphi (R)$.

Proof. It is clear that (2) implies (1). Conversely, suppose that $g \otimes 1 = 1 \otimes g$. Choose generators $\{ s_ i\} _{i \in I}$ of $S$ as an $R$-module with $0, 1 \in I$ and $s_0 = 1$ and $s_1 = g$. Apply Lemma 10.107.10 to the relation $g \otimes s_0 + (-1) \otimes s_1 = 0$. We see that there exist $a_{i, j} \in R$ such that $g = \sum _ i a_{i, 0} s_ i$, $-1 = \sum _ i a_{i, 1} s_ i$, and for $j \not= 0, 1$ we have $0 = \sum _ i a_{i, j} s_ i$, and moreover for all $i$ we have $\sum _ j a_{i, j}s_ j = 0$. Then we have

\[ \sum \nolimits _{i, j \not= 0} a_{i, j} s_ i s_ j = -g + a_{0, 0} \]

and for each $j \not= 0$ we have $\sum _{i \not= 0} a_{i, j}s_ i \in R$. This proves that $-g + a_{0, 0}$ can be written as in (2). It follows that $g$ can be written as in (2). Details omitted. Hint: Show that the set of elements of $S$ which have an expression as in (2) form an $R$-subalgebra of $S$. $\square$

Remark 10.107.12. Let $R \to S$ be a ring map. Sometimes the set of elements $g \in S$ such that $g \otimes 1 = 1 \otimes g$ is called the epicenter of $S$. It is an $R$-algebra. By the construction of Lemma 10.107.11 we get for each $g$ in the epicenter a matrix factorization

\[ (g) = Y X Z \]

with $X \in \text{Mat}(n \times n, R)$, $Y \in \text{Mat}(1 \times n, S)$, and $Z \in \text{Mat}(n \times 1, S)$. Namely, let $x_{i, j}, y_ i, z_ j$ be as in part (2) of the lemma. Set $X = (x_{i, j})$, let $y$ be the row vector whose entries are the $y_ i$ and let $z$ be the column vector whose entries are the $z_ j$. With this notation conditions (b) and (c) of Lemma 10.107.11 mean exactly that $Y X \in \text{Mat}(1 \times n, R)$, $X Z \in \text{Mat}(n \times 1, R)$. It turns out to be very convenient to consider the triple of matrices $(X, YX, XZ)$. Given $n \in \mathbf{N}$ and a triple $(P, U, V)$ we say that $(P, U, V)$ is a $n$-triple associated to $g$ if there exists a matrix factorization as above such that $P = X$, $U = YX$ and $V = XZ$.

Lemma 10.107.13. Let $R \to S$ be an epimorphism of rings. Then the cardinality of $S$ is at most the cardinality of $R$. In a formula: $|S| \leq |R|$.

Proof. The condition that $R \to S$ is an epimorphism means that each $g \in S$ satisfies $g \otimes 1 = 1 \otimes g$, see Lemma 10.107.1. We are going to use the notation introduced in Remark 10.107.12. Suppose that $g, g' \in S$ and suppose that $(P, U, V)$ is an $n$-triple which is associated to both $g$ and $g'$. Then we claim that $g = g'$. Namely, write $(P, U, V) = (X, YX, XZ)$ for a matrix factorization $(g) = YXZ$ of $g$ and write $(P, U, V) = (X', Y'X', X'Z')$ for a matrix factorization $(g') = Y'X'Z'$ of $g'$. Then we see that

\[ (g) = YXZ = UZ = Y'X'Z = Y'PZ = Y'XZ = Y'V = Y'X'Z' = (g') \]

and hence $g = g'$. This implies that the cardinality of $S$ is bounded by the number of possible triples, which has cardinality at most $\sup _{n \in \mathbf{N}} |R|^ n$. If $R$ is infinite then this is at most $|R|$, see [Ch. I, 10.13, Kunen].

If $R$ is a finite ring then the argument above only proves that $S$ is at worst countable. In fact in this case $R$ is Artinian and the map $R \to S$ is surjective. We omit the proof of this case. $\square$

Lemma 10.107.14. Let $R \to S$ be an epimorphism of rings. Let $N_1, N_2$ be $S$-modules. Then $\mathop{\mathrm{Hom}}\nolimits _ S(N_1, N_2) = \mathop{\mathrm{Hom}}\nolimits _ R(N_1, N_2)$. In other words, the restriction functor $\text{Mod}_ S \to \text{Mod}_ R$ is fully faithful.

Proof. Let $\varphi : N_1 \to N_2$ be an $R$-linear map. For any $x \in N_1$ consider the map $S \otimes _ R S \to N_2$ defined by the rule $g \otimes g' \mapsto g\varphi (g'x)$. Since both maps $S \to S \otimes _ R S$ are isomorphisms (Lemma 10.107.1), we conclude that $g \varphi (g'x) = gg'\varphi (x) = \varphi (gg' x)$. Thus $\varphi $ is $S$-linear. $\square$

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