Endomorphisms with small image is in the jacobson radical of endomorphims.

Question

Let $P$ be a projective $R$-module ($R$ commutative and with unity), $S = End_R(P)$ and $J(S)$ the Jacobson radical of $S$.

Let $\Delta S = \{f \in End_R(P) \mid Im(f)$ small in $P\}$. Recall that $N$ is a small submodule of $P$ if $N + K = P \Rightarrow K = P$.

I want to proof the following theorem :

PROPOSITION 1.1. Let $P$ be a projective $R$-module, $S=Hom_R(P, P)$ the ring of endomorphisms of $P$, and $\Delta S= \{f \in S \mid Im(f) \ \text{is small in}\ P\}$. Then
(1) $J(S)=\Delta S$.
(2) $J(S) \subset Hom_R(P, J(P))$.
(3) There is a ring epimorphism $\Phi: S \rightarrow Hom_R(P/J(P),P/J(P))$ with $Ker (\Phi) = Hom_R(P, J(P))$.
(4) If $J(P)$ is small in $P$ then $J(S) = Hom_R(P, J(P))$ and hence $$ S/J(S) \cong Hom_R(P/J(P), P/J(P)).$$ Proof. (1) The proof that $\Delta S \subset J(S)$ is contained in the proof of Theorem 2.4 in [8] and will be omitted here.
(...)
src : Endomorphism Rings of Projective Modules by Roger Ware.

I understand the proof of (2),(3) and (4) but as you can see, the proof of $\Delta S \subset J(S)$ is not here. I found the reference :

Theorem 2.4) If $P$ is semi-perfect, then $S/J(S)$ is regular and $J(S)= \Delta S$.
Proof. We remark first that for this theorem we need only the assumption that $P$ is self-projective and that for every submodule $Q$ there exists a sub-module $U$ with the property that $P=Q+U$ and $Q \cap U$ is small in $P$.
i) $\Delta S \subset J(S)$: For $s \in \Delta S$ we show that $1-s$ is an isomorphism, which means that $s$ is quasi-regular; hence $s \in J(S)$. Since $Im(s) + Im(1— s) = P$ and $Im (s)$ is small, we have $Im (1—s) = P$. It remains to prove that $1 — s$ is a monomorphism. Since $1 — s$ is an epimorphism and $M$ is (self-)projective, then $P$ = $Ker(1-s)\oplus M$. Then $(1-s) (M) = P$, and hence $M + s(M) = M + Im (s) = P$. Since $Im (s)$ is small, this implies $M= P$; hence $Ker (1- s) =0$.
(...)
src : Semi-perfect modules by Erika A. Mares, Math. Zeitschr. 82, 347-360 (1963)

But I'm having some issues with it... My first remark is $s$ quasi-regular not imply $s \in J(S)$, according to the Wikipedia page on Jacobson radical :

J(R) is the unique right ideal of R maximal with the property that every element is right quasiregular (or equivalently left quasiregular).(...). Although every element of the J(R) is necessarily quasiregular, not every quasiregular element is necessarily a member of J(R).

However, I believe this is not a problem because we have shown that all $s \in \Delta S$ are quasi-regular, and $\Delta S$ is an ideal, so we have $\Delta S \subseteq J(S)$ by maximality of $J(S)$, right?

My second problem is about the proof of "$1-s$ is a monomorphism". We know that $1-s$ is epimorphisms, $1-s : P \twoheadrightarrow P$ and because $P$ is projective, we obtain $i : P \rightarrow P$ with $(1-s)i = id_P$. So we have the following (split) short exact sequence : $$ 0 \longrightarrow ker(1-s) \longrightarrow P \stackrel{1-s}{\longrightarrow} P \longrightarrow 0$$ We obtain $P \cong ker(1-s) \oplus P$ by the splitting lemma. I don't know why the author did not use this fact... anyway we obtain the same result. But now why this is imply $ker(1-s) = 0$ ? I know that we don't always have $N \oplus P \cong P \Rightarrow N = 0$. But is the result true in this proof ??

Example : For a field $k$, $P = \bigoplus_{i \in \mathbb{N}} k$ we have $P \oplus ker(1-s) \cong P$, but $ker(1-s) \cong k \neq 0$, where $s : P \rightarrow P$ with $s(a_1,a_2,\dots) = (a_1,a_2-a_1,a_3-a_2,\dots)$. However, $Im(s) = P$ is not small, so this is not a counter example of $\Delta S \subset J(S)$.

Thanks for any help, and apologies for the bad English. - M.T.

Answer 1

Okay, I have found a proof that $\Delta S \subset J(S)$, but I am unsure if in our case $\ker(1-s) \oplus P \cong P$ implies $\ker(1-s) = 0$. Actually, we only need each $s \in S$ to be right quasi-regular (i.e., $1-s$ has a right inverse), as shown in the following theorem:

Theorem: Let $R$ be any ring.
a) Every element of $J(R)$ is quasi-regular.
b) If $I$ is a right ideal and every element of $I$ is right quasi-regular, then $I \subset J(R)$.
Source: Algebra, a graduate course by I. Martin Isaacs.

Finally, I can state and prove the following lemma, which is a direct consequence of the previous theorem:

Lemma: $\Delta S \subset J(S)$
Proof: For $s \in \Delta S$, we have $P = \operatorname{Im}(1-s) + \operatorname{Im}(s)$. Since $\operatorname{Im}(s)$ is small, we obtain $\operatorname{Im}(1-s) = P$. Since $P$ is a projective module, there exists $i : P \rightarrow P$ such that $(1-s) i = \operatorname{id}_P$. Then, $s$ is a right quasi-regular element. By the previous theorem, $\Delta S \subset J(S)$. (We repeat the argument for convenience: Let $M$ be a maximal ideal of $S$. If $\Delta S$ is not included in $M$, we have $\Delta S + M = S$ by the maximality of $M$. In particular, $1 = m + s$ with $m \in M$ and $s \in S$, so we obtain: $$ S = (1-s)S = mS \subset M$$ ( $(1-s)S = S$ because $\forall f \in S,\ f = (1-s) i f$). This contradiction shows that $\Delta S \subset M$ for any maximal ideal of $S$, i.e., $\Delta S \subset J(S)$.)