Definition 15.15.1. A ring $R$ is said to be *auto-associated* if $R$ is local and its maximal ideal $\mathfrak m$ is weakly associated to $R$.

## 15.15 Auto-associated rings

Some of this material is in [Autour].

Lemma 15.15.2. An auto-associated ring $R$ has the following property: (P) Every proper finitely generated ideal $I \subset R$ has a nonzero annihilator.

**Proof.**
By assumption there exists a nonzero element $x \in R$ such that for every $f \in \mathfrak m$ we have $f^ n x = 0$. Say $I = (f_1, \ldots , f_ r)$. Then $x$ is in the kernel of $R \to \bigoplus R_{f_ i}$. Hence we see that there exists a nonzero $y \in R$ such that $f_ i y = 0$ for all $i$, see Algebra, Lemma 10.23.4. As $y \in \text{Ann}_ R(I)$ we win.
$\square$

Lemma 15.15.3. Let $R$ be a ring having property (P) of Lemma 15.15.2. Let $u : N \to M$ be a homomorphism of projective $R$-modules. Then $u$ is universally injective if and only if $u$ is injective.

**Proof.**
Assume $u$ is injective. Our goal is to show $u$ is universally injective. First we choose a module $Q$ such that $N \oplus Q$ is free. On considering the map $N \oplus Q \to M \oplus Q$ we see that it suffices to prove the lemma in case $N$ is free. In this case $N$ is a directed colimit of finite free $R$-modules. Thus we reduce to the case that $N$ is a finite free $R$-module, say $N = R^{\oplus n}$. We prove the lemma by induction on $n$. The case $n = 0$ is trivial.

Let $u : R^{\oplus n} \to M$ be an injective module map with $M$ projective. Choose an $R$-module $Q$ such that $M \oplus Q$ is free. After replacing $u$ by the composition $R^{\oplus n} \to M \to M \oplus Q$ we see that we may assume that $M$ is free. Then we can find a direct summand $R^{\oplus m} \subset M$ such that $u(R^{\oplus n}) \subset R^{\oplus m}$. Hence we may assume that $M = R^{\oplus m}$. In this case $u$ is given by a matrix $A = (a_{ij})$ so that $u(x_1, \ldots , x_ n) = (\sum x_ i a_{i1}, \ldots , \sum x_ i a_{im})$. As $u$ is injective, in particular $u(x, 0, \ldots , 0) = (xa_{11}, xa_{12}, \ldots , xa_{1m}) \not= 0$ if $x \not= 0$, and as $R$ has property (P) we see that $a_{11}R + a_{12}R + \ldots + a_{1m}R = R$. Hence see that $R(a_{11}, \ldots , a_{1m}) \subset R^{\oplus m}$ is a direct summand of $R^{\oplus m}$, in particular $R^{\oplus m}/R(a_{11}, \ldots , a_{1m})$ is a projective $R$-module. We get a commutative diagram

with split exact rows. Thus the right vertical arrow is injective and we may apply the induction hypothesis to conclude that the right vertical arrow is universally injective. It follows that the middle vertical arrow is universally injective. $\square$

Lemma 15.15.4. Let $R$ be a ring. The following are equivalent

$R$ has property (P) of Lemma 15.15.2,

any injective map of projective $R$-modules is universally injective,

if $u : N \to M$ is injective and $N$, $M$ are finite projective $R$-modules then $\mathop{\mathrm{Coker}}(u)$ is a finite projective $R$-module,

if $N \subset M$ and $N$, $M$ are finite projective as $R$-modules, then $N$ is a direct summand of $M$, and

any injective map $R \to R^{\oplus n}$ is a split injection.

**Proof.**
The implication (1) $\Rightarrow $ (2) is Lemma 15.15.3. It is clear that (3) and (4) are equivalent. We have (2) $\Rightarrow $ (3), (4) by Algebra, Lemma 10.81.4. Part (5) is a special case of (4). Assume (5). Let $I = (a_1, \ldots , a_ n)$ be a proper finitely generated ideal of $R$. As $I \not= R$ we see that $R \to R^{\oplus n}$, $x \mapsto (xa_1, \ldots , xa_ n)$ is not a split injection. Hence it has a nonzero kernel and we conclude that $\text{Ann}_ R(I) \not= 0$. Thus (1) holds.
$\square$

Example 15.15.5. If the equivalent conditions of Lemma 15.15.4 hold, then it is not always the case that every injective map of free $R$-modules is a split injection. For example suppose that $R = k[x_1, x_2, x_3, \ldots ]/(x_ i^2)$. This is an auto-associated ring. Consider the map of free $R$-modules

For any integer $n$ the restriction of $u$ to $\bigoplus _{i = 1, \ldots , n} Re_ i$ is injective as the images $u(e_1), \ldots , u(e_ n)$ are $R$-linearly independent. Hence $u$ is injective and hence universally injective by the lemma. Since $u \otimes \text{id}_ k$ is bijective we see that if $u$ were a split injection then $u$ would be surjective. But $u$ is not surjective because the inverse image of $f_1$ would be the element

which is not an element of the direct sum. A side remark is that $\mathop{\mathrm{Coker}}(u)$ is a flat (because $u$ is universally injective), countably generated $R$-module which is not projective (as $u$ is not split), hence not Mittag-Leffler (see Algebra, Lemma 10.92.1).

The following lemma is a special case of Algebra, Proposition 10.101.9 in case the local ring is Noetherian.

Lemma 15.15.6. Let $R$ be a ring. Suppose that $\varphi : R^ m \to R^ n$ is a map of finite free modules. The following are equivalent

$\varphi $ is injective,

the rank of $\varphi $ is $m$ and the annihilator of $I(\varphi )$ in $R$ is zero.

If $R$ is Noetherian these are also equivalent to

the rank of $\varphi $ is $m$ and either $I(\varphi ) = R$ or it contains a nonzerodivisor.

Here the rank of $\varphi $ and $I(\varphi )$ are defined as in Algebra, Definition 10.101.5.

**Proof.**
If any matrix coefficient of $\varphi $ is not in $\mathfrak m$, then we apply Algebra, Lemma 10.101.2 to write $\varphi $ as the sum of $1 : R \to R$ and a map $\varphi ' : R^{m-1} \to R^{n-1}$. It is easy to see that the lemma for $\varphi '$ implies the lemma for $\varphi $. Thus we may assume from the outset that all the matrix coefficients of $\varphi $ are in $\mathfrak m$.

Suppose $\varphi $ is injective. We may assume $m > 0$. Let $\mathfrak q \in \text{WeakAss}(R)$ so that $R_\mathfrak q$ is an auto-associated ring. Then $\varphi $ induces a injective map $R_\mathfrak q^ m \to R_\mathfrak q^ n$ which is universally injective by Lemmas 15.15.2 and 15.15.3. Thus $\varphi : \kappa (\mathfrak q)^ m \to \kappa (\mathfrak q)^ n$ is injective. Hence the rank of $\varphi \bmod \mathfrak q$ is $m$ and $I(\varphi \otimes \kappa (\mathfrak q))$ is not the zero ideal. Since $m$ is the maximum rank $\varphi $ can have, we conclude that $\varphi $ has rank $m$ as well (ranks of matrices can only drop after base change). Hence $I(\varphi ) \cdot \kappa (\mathfrak q) = I(\varphi \otimes \kappa (\mathfrak q))$ is not zero. Thus $I(\varphi )$ is not contained in $\mathfrak q$. Thus none of the weakly associated primes of $R$ are weakly associated primes of the $R$-module $\text{Ann}_ R I(\varphi )$. Thus $\text{Ann}_ R I(\varphi )$ has no weakly associated primes, see Algebra, Lemma 10.65.4. It follows from Algebra, Lemma 10.65.5 that $\text{Ann}_ R I(\varphi )$ is zero.

Conversely, assume (2). The rank being $m$ implies $n \geq m$. Write $I(\varphi ) = (f_1, \ldots , f_ r)$ which is possible as $I(\varphi )$ is finitely generated. By Algebra, Lemma 10.14.5 we can find maps $\psi _ i : R^ n \to R^ m$ such that $\psi \circ \varphi = f_ i \text{id}_{R^ m}$. Thus $\varphi (x) = 0$ implies $f_ i x = 0$ for $i = 1, \ldots , r$. This implies $x = 0$ and hence $\varphi $ is injective.

For the equivalence of (1) and (3) in the Noetherian local case we refer to Algebra, Proposition 10.101.9. If the ring $R$ is Noetherian but not local, then the reader can deduce it from the local case; details omitted. Another option is to redo the argument above using associated primes, using that there are finitely many of these, using prime avoidance, and using the characterization of nonzerodivisors as elements of a Noetherian ring not contained in any associated prime. $\square$

Lemma 15.15.7. Let $R$ be a ring. Suppose that $\varphi : R^ n \to R^ n$ be an injective map of finite free modules of the same rank. Then $\mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Coker}}(\varphi ), R) = 0$.

**Proof.**
Let $\varphi ^ t : R^ n \to R^ n$ be the transpose of $\varphi $. The lemma claims that $\varphi ^ t$ is injective. With notation as in Lemma 15.15.6 we see that the rank of $\varphi ^ t$ is $n$ and that $I(\varphi ) = I(\varphi ^ t)$. Thus we conclude by the equivalence of (1) and (2) of the lemma.
$\square$

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