The Stacks project

Proof. Let $0 \to A \to B \to C \to 0$ be an exact sequence as in (2). To prove that (1) implies (2) it suffices to prove that $\mathop{\mathrm{Hom}}\nolimits _ R(R/fR, B) \to \mathop{\mathrm{Hom}}\nolimits _ R(R/fR, C)$ is surjective for every $f \in R$. Let $\psi : R/fR \to C$ be a map. Say $\psi (1)$ is the image of $b \in B$. Then $fb \in A$. Hence there exists an $a \in A$ such that $fa = fb$. Then $f(b - a) = 0$ hence we get a morphism $\varphi : R/fR \to B$ mapping $1$ to $b - a$ which lifts $\psi$.

Conversely, assume that (2) holds. Let $I$ be the set of pairs $(f, \varphi )$ where $f \in R$ and $\varphi : R/fR \to P$. For $i \in I$ denote $(f_ i, \varphi _ i)$ the corresponding pair. Consider the map

which sends the element $r$ in the summand $R/f_ iR$ to $\varphi _ i(r)$ in $P$. Let $A = \mathop{\mathrm{Ker}}(B \to P)$. Then we see that (1) is true if the sequence

is an exact sequence as in (2). To see this suppose $f \in R$ and $a \in A$ maps to $f b$ in $B$. Write $b = (r_ i)_{i \in I}$ with almost all $r_ i = 0$. Then we see that

in $P$. Hence there is an $i_0 \in I$ such that $f_{i_0} = f$ and $\varphi _{i_0}(1) = \sum \varphi _ i(r_ i)$. Let $x_{i_0} \in R/f_{i_0}R$ be the class of $1$. Then we see that

is an element of $A$ and $fa' = a$ as desired. $\square$

Proof. Assume (2) and let $a, b \in R$. Then $(a, b) = (c)$. If $c = 0$, then $a = b = 0$ and $a$ divides $b$. Assume $c \not= 0$. Write $c = ua + vb$ and $a = wc$ and $b = zc$. Then $c(1 - uw - vz) = 0$. Since $R$ is local, this implies that $1 - uw - vz \in \mathfrak m$. Hence either $w$ or $z$ is a unit, so either $a$ divides $b$ or $b$ divides $a$. Thus (2) implies (1).

Assume (1). If $R$ has two maximal ideals $\mathfrak m_ i$ we can choose $a \in \mathfrak m_1$ with $a \not\in \mathfrak m_2$ and $b \in \mathfrak m_2$ with $b \not\in \mathfrak m_1$. Then $a$ does not divide $b$ and $b$ does not divide $a$. Hence $R$ has a unique maximal ideal and is local. It follows easily from condition (1) and induction that every finitely generated ideal is principal. Thus (1) implies (2).

It is straightforward to prove that (1) and (3) are equivalent. The final statement is Algebra, Lemma 10.50.4. $\square$

Proof. Let $M$ be a finitely presented $R$-module. We will use all the equivalent properties of $R$ from Lemma 15.124.2 without further mention. Denote $\mathfrak m \subset R$ the maximal ideal and $\kappa = R/\mathfrak m$ the residue field. Let $I \subset R$ be the annihilator of $M$. Choose a basis $y_1, \ldots , y_ n$ of the finite dimensional $\kappa$-vector space $M/\mathfrak m M$. We will argue by induction on $n$.

By Nakayama's lemma any collection of elements $x_1, \ldots , x_ n \in M$ lifting the elements $y_1, \ldots , y_ n$ in $M/\mathfrak m M$ generate $M$, see Algebra, Lemma 10.20.1. This immediately proves the base case $n = 0$ of the induction.

We claim there exists an index $i$ such that for any choice of $x_ i \in M$ mapping to $y_ i$ the annihilator of $x_ i$ is $I$. Namely, if not, then we can choose $x_1, \ldots , x_ n$ such that $I_ i = \text{Ann}(x_ i) \not= I$ for all $i$. But as $I \subset I_ i$ for all $i$, ideals being totally ordered implies $I_ i$ is strictly bigger than $I$ for $i = 1, \ldots , n$, and by total ordering once more we would see that $\text{Ann}(M) = I_1 \cap \ldots \cap I_ n$ is bigger than $I$ which is a contradiction. After renumbering we may assume that $y_1$ has the property: for any $x_1 \in M$ lifting $y_1$ the annihilator of $x_1$ is $I$.

We set $A = Rx_1 \subset M$. Consider the exact sequence $0 \to A \to M \to M/A \to 0$. Since $A$ is finite, we see that $M/A$ is a finitely presented $R$-module (Algebra, Lemma 10.5.3) with fewer generators. Hence $M/A \cong \bigoplus _{j = 1, \ldots , m} R/f_ jR$ by induction. On the other hand, we claim that $A \to M$ satisfies the property: if $f \in R$, then $fA = A \cap fM$. The inclusion $fA \subset A \cap fM$ is trivial. Conversely, if $x \in A \cap fM$, then $x = gx_1 = f y$ for some $g \in R$ and $y \in M$. If $f$ divides $g$, then $x \in fA$ as desired. If not, then we can write $f = hg$ for some $h \in \mathfrak m$. The element $x'_1 = x_1 - hy$ has annihilator $I$ by the previous paragraph. Thus $g \in I$ and we see that $x = 0$ as desired. The claim and Lemma 15.124.1 imply the sequence $0 \to A \to M \to M/A \to 0$ is split and we find $M \cong A \oplus \bigoplus _{j = 1, \ldots , m} R/f_ jR$. Then $A = R/I$ is finitely presented (as a summand of $M$) and hence $I$ is finitely generated, hence principal. This finishes the proof. $\square$

Proof. Let $M$ be a finitely presented $R$-module. We first show that $M$ is a summand of a direct sum of modules of the form $R/fR$ and at the end we argue the direct sum can be taken to be finite. Let

be a short exact sequence of $R$-modules such that $fA = A \cap fB$ for all $f \in R$. By Lemma 15.124.1 we have to show that $\mathop{\mathrm{Hom}}\nolimits _ R(M, B) \to \mathop{\mathrm{Hom}}\nolimits _ R(M, C)$ is surjective. It suffices to prove this after localization at maximal ideals $\mathfrak m$, see Algebra, Lemma 10.23.1. Note that the localized sequences $0 \to A_\mathfrak m \to B_\mathfrak m \to C_\mathfrak m \to 0$ satisfy the condition that $fA_\mathfrak m = A_\mathfrak m \cap fB_\mathfrak m$ for all $f \in R_\mathfrak m$ (because we can write $f = uf'$ with $u \in R_\mathfrak m$ a unit and $f' \in R$ and because localization is exact). Since $M$ is finitely presented, we see that

by Algebra, Lemma 10.10.2. The module $M_\mathfrak m$ is a finitely presented $R_\mathfrak m$-module. By Lemma 15.124.3 we see that $M_\mathfrak m$ is a direct sum of modules of the form $R_\mathfrak m/fR_\mathfrak m$. Thus we conclude by Lemma 15.124.1 that the map on localizations is surjective.

At this point we know that $M$ is a summand of $\bigoplus _{i \in I} R/f_ i R$. Consider the map $M \to \bigoplus _{i \in I} R/f_ i R$. Since $M$ is a finite $R$-module, the image is contained in $\bigoplus _{i \in I'} R/f_ i R$ for some finite subset $I' \subset I$. This finishes the proof. $\square$

Proof. Let $a, b \in R$ be nonzero. Consider the $1 \times 2$ matrix $A = (a\ b)$. Then we see that $u(a\ b)V = (f\ 0)$ with $u \in R$ invertible and $V = (g_{ij})$ an invertible $2 \times 2$ matrix. Then $f = u a g_{11} + u b g_{2 1}$ and $(g_{11}, g_{2 1}) = R$. It follows that $(a, b) = (f)$. An induction argument (omitted) then shows any finitely generated ideal in $R$ is generated by one element. $\square$

Proof. We omit the proof of the statement on localizations. The final statement is Algebra, Lemma 10.50.15. The second statement follows from the other two. $\square$

Proof. Proof of (1). Let $M \subset F$ be a finite submodule of a free module $F$. Since $M$ is finite, we may assume $F$ is a finite free module (details omitted). Say $F = R^{\oplus n}$. We argue by induction on $n$. If $n = 1$, then $M$ is a finitely generated ideal, hence principal by our assumption that $R$ is Bézout. If $n > 1$, then we consider the image $I$ of $M$ under the projection $R^{\oplus n} \to R$ onto the last summand. If $I = (0)$, then $M \subset R^{\oplus n - 1}$ and we are done by induction. If $I \not= 0$, then $I = (f) \cong R$. Hence $M \cong R \oplus \mathop{\mathrm{Ker}}(M \to I)$ and we are done by induction as well.

Let $M$ be a finitely presented $R$-module. Since the localizations of $R$ are maximal ideals are valuation rings (Lemma 15.124.7) we may apply Lemma 15.124.4. Thus $M$ is a summand of a module of the form $R^{\oplus r} \oplus \bigoplus _{i = 1, \ldots , n} R/f_ iR$ with $f_ i \not= 0$. Since taking the torsion submodule is a functor we see that $M_{tors}$ is a summand of the module $\bigoplus _{i = 1, \ldots , n} R/f_ iR$ and $M/M_{tors}$ is a summand of $R^{\oplus r}$. By the first part of the proof we see that $M/M_{tors}$ is finite free. Hence $M \cong M_{tors} \oplus M/M_{tors}$ as desired. $\square$

Proof. A PID is a Noetherian Bézout ring. By Lemma 15.124.8 it suffices to prove the result if $M$ is torsion. Since $M$ is finite, this means that the annihilator of $M$ is nonzero. Say $fM = 0$ for some $f \in R$ nonzero. Then we can think of $M$ as a module over $R/fR$. Since $R/fR$ is Noetherian of dimension $0$ (small detail omitted) we see that $R/fR = \prod R_ j$ is a finite product of Artinian local rings $R_ i$ (Algebra, Proposition 10.60.7). Each $R_ i$, being a local ring and a quotient of a PID, is a generalized valuation ring in the sense of Lemma 15.124.2 (small detail omitted). Write $M = \prod M_ j$ with $M_ j = e_ j M$ where $e_ j \in R/fR$ is the idempotent corresponding to the factor $R_ j$. By Lemma 15.124.3 we see that $M_ j = \bigoplus _{i = 1, \ldots , n_ j} R_ j/\overline{f}_{ji}R_ j$ for some $\overline{f}_{ji} \in R_ j$. Choose lifts $f_{ji} \in R$ and choose $g_{ji} \in R$ with $(g_{ji}) = (f_ j, f_{ji})$. Then we conclude that

as an $R$-module which finishes the proof. $\square$

Proof. This follows from Lemma 15.124.8 but we can also prove it directly as follows. By induction on $n$. The result holds for $n = 1$. Assume $n > 1$. We may assume $f_1 \not= 0$ after renumbering. Choose $f \in R$ such that $(f) = (f_1, \ldots , f_{n - 1})$. Let $A$ be an $(n - 1) \times (n - 1)$ matrix whose first row is $f_1/f, \ldots , f_{n - 1}/f$. Choose $a, b \in R$ such that $af - bf_ n = 1$ which is possible because $1 \in (f_1, \ldots , f_ n) = (f, f_ n)$. Then a solution is the matrix

Observe that the left matrix is invertible because it has determinant $1$. $\square$