Let $R$ be a ring with unit. A submodule $N$ of an $R$-module $M$ is called superfluous if the only sumbodule $T$ of $M$ for which $N+T = M$ is $M$ itself.

It is shown, for example, in

[1] F. W._Anderson, K. R. Fuller "Rings and Categories of Modules" (1974)

that if every submodule of $M$ is contained in a maximal submodule, then the radical of $M$ is superfluous (Proposition 9.18). This, in particular, implies that for every finitely generated module $M$ its radical is superfluous. In exercise 9.2. it is explained that divisible abelian groups coincide with their radicals, and therefore their radicals are not superfluous. Divisible abelian groups are not projective objects.

I was curious if it is possible to construct a projective module with non-superfluous radical.

Question: is there an example of a ring $R$ and a projective $R$-module $P$ such that the radical $JP$ of $P$ is not superfluous?

The existence of such module (or, at least, that its non-existence is non-obvious) is somehow hinted by the formulation of Corollary 17.12 in [1]:

Let $J = J(R)$. If $P$ is a projective left $R$-module such that $JP$ is superfluous in $P$ (e.g., if ${}_RP$ is finitely generated), then $J(End({}_RP)) = Hom_R(P,JP)$ and $End({}_RP)/J(End_RP) \cong End({}_RP/JP)$.


1 Answer 1


According to Proposition 17.10 in the Anderson-Fuller book (I am using the 1992 second edition; don't know if the first 1974 edition is any different), for any projective module $P$ over any (unital associative) ring $R$, the radical of $P$ is computable as $Rad\,P=JP$, where $J$ is the Jacobson radical of the ring $R$ (just as you say).

Let $p$ be a prime number. Consider the commutative ring $R=\mathbb Z_{(p)}$, that is, the localization of the ring of integers $\mathbb Z$ at the prime ideal $(p)\subset\mathbb Z$. Alternatively, one can consider the ring of $p$-adic integers $R=\mathbb Z_p$, that is the completion of the local ring $\mathbb Z_{(p)}$ at its maximal ideal. In both cases, $J(R)=pR$ is the unique maximal ideal of $R$.

Consider the free $R$-module $F$ with a countable set of generators, $F=R^{(\omega)}$. Let us show that $JF$ is not superfluous in $F$. For this purpose, we will construct a proper submodule $T\subset F$ such that $JF+T=F$.

Consider the $R$-module $Q=R[p^{-1}]$. In other words, $Q$ is just the ring of fractions of the local domain $R$. The $R$-module $Q$ is generated by the sequence of elements $1$, $p^{-1}$, $p^{-2}$, $\dots$; so $Q$ is a countably generated $R$-module. Hence $Q$ is a quotient $R$-module of the $R$-module $F$.

Denote by $T\subset F$ a submodule such that $F/T\cong Q$. So we have a short exact sequence of $R$-modules $0\to T\to F\to Q\to 0$. We want to check that $T+JF=F$.

Indeed, we have $JF=pF$, since $J=pR$. The desired equation $T+pF=F$ is equivalent to $p(F/T)=F/T$. Now $F/T\cong Q$ and we have $pQ=Q$ by construction.

In fact, as it is clear now, any discrete valuation ring can be used in the role of $R$ in this construction (with a prime number $p$ replaced by any uniformizing element).

  • $\begingroup$ As far as I can see, it is used only that $p$ lies in $JR$ and that it is not a zero-divisor. It is easier for me to think in terms of basis and generators. If we write $e_0$, $e_1$, ... for the basis elements of $F$, then $T$ is generated by the elements $e_i - p e_{i+1}$. As a result $T + JF$ contains all the elements $e_i$ and thus coincides with $F$. On the other hand every element in $T$ is a linear combination $a_{i_1} (e_{i_1} - pe_{i_1 + 1}) + ... + a_{i_k} (e_{i_k} - pe_{i_k + 1})$ with $i_1 < ... < i_k$ and all $a_{i_s}$ non-zero. $\endgroup$
    – Ivan Yudin
    Jun 21, 2020 at 7:27
  • $\begingroup$ Now since $p$ is not a zero divisor the coefficients of $e_{i_1}$ and of $e_{i_k + 1} $ are not zero in the above linear combinations. This shows that every element of $T$ has support of size at least $2$ and thus $T$ does not contain the basis elements $e_i$. $\endgroup$
    – Ivan Yudin
    Jun 21, 2020 at 7:33
  • $\begingroup$ After thinking a little bit more, I realized that the construction can be generalized further, if there is an infinite sequence of elements $p_1$, $p_2$, .... in $JR$ such that each initial product $p_1 p_2 .... p_k$ is non-zero in $JR$. Namely, if such sequence exists, we can take as $T$ the submodule of $F$ generated by the elements $e_i - p_{i+1}e_{i+1}$. $\endgroup$
    – Ivan Yudin
    Jun 21, 2020 at 16:44
  • 1
    $\begingroup$ Now, existence of such sequence means that $JR$ is not left T-nilpotent. Knowing this I immediately landed with Lemma 28.3 of Anderon-Fuller, which says, among other things, that $JR$ is left T-nilpotent if and only if $JF$ is superfluous in $F$. $\endgroup$
    – Ivan Yudin
    Jun 21, 2020 at 16:46

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