MATHEMATICS

V. E. GOVOROV

RINGS OVER WHICH FLAT MODULES ARE FREE

(Presented by Academician A. I. Mal'cev, February 1, 1962)

In Bass’s paper (²) it was established that the direct limit of a direct spectrum of projective \(\Lambda\)-modules is a projective \(\Lambda\)-module if and only if \(\Lambda\) is perfect. In this case flat modules are projective. In the present paper it is shown that flat modules over an arbitrary ring are the direct limit of a direct spectrum of free modules. Hence it follows that the problem: over which rings are flat modules free (projective), is equivalent to the problem: over which rings is the direct limit of a direct spectrum of free (projective) modules a free (projective) module. It is further proved that, in order that all left flat \(\Lambda\)-modules be free, it is necessary and sufficient that \(\Lambda\) be a local ring with left \(T\)-nilpotent maximal ideal.

Let \(\Lambda\) be an arbitrary associative ring with identity. All modules considered will be assumed to be unitary.

Lemma 1. A left \(\Lambda\)-module \(A\) is flat if and only if it is the direct limit of a direct spectrum of projectives.

That the direct limit of a direct spectrum of projective modules is flat follows from the commutativity of the functor \(\operatorname{Tor}^{\Lambda}_{1}(A,B)\) with the operation of taking the direct limit.

Now let \(A\) be a flat left \(\Lambda\)-module; then there exists a free module \(F\) and an exact sequence

\[ 0 \to A' \to F \to A \to 0. \tag{1} \]

The module \(A'\) is pure (see (¹)) in \(F\). Indeed, sequence (1) induces, for an arbitrary right \(\Lambda\)-module \(B\), the exact sequence

\[ 0 \to \operatorname{Tor}^{\Lambda}_{1}(B,A) \to B \otimes_{\Lambda} A' \to B \otimes_{\Lambda} F \to B \otimes_{\Lambda} A \to 0; \]

but \(\operatorname{Tor}^{\Lambda}_{1}(B,A)=0\), since \(A\) is flat; consequently, \(A'\) is pure in \(F\).

Take an arbitrary finite set of elements \(a_{1}, a_{2}, \ldots, a_{n} \in A'\), and let \(\{u_{\alpha}\}\) be a system of free generators of the module \(F\). Then

\[ a_j=\sum_{i=1}^{s_j}\lambda_{ji}u_i \quad (j=1,2,\ldots,n). \]

By Theorem 2.4 of (¹) there exist elements

\[ u'_1,u'_2,\ldots,u'_s \in A' \]

such that

\[ a_j=\sum_{i=1}^{s_j}\lambda_{ji}u'_i, \]

and, consequently, there exists an epimorphism \(\varphi\) of the module \(F\) onto the submodule \(C'\) of the module \(A'\), generated by the elements \(u'_1,u'_2,\ldots,u'_s\), defined by the equalities: \(\varphi(u_i)=u'_i\) \((i=1,2,\ldots,s)\), \(\varphi(u_k)=0\) \((k\ne 1,2,\ldots,s)\), i.e. \(C'\) is a direct summand in \(F\). Thus, for every finite subset of elements of \(A'\) one can find a projective submodule \(C'\) of the module \(A'\) containing this subset; that is, the module \(A'\) is the direct limit of a direct spectrum of projective modules.

\(\{C_\alpha\}\); but then the factor modules \(F/C_\alpha\) are projective and form a direct spectrum with respect to the ordering \(F/C_\alpha < F/C_\beta\), if \(C_\alpha \subset C_\beta\), and the natural homomorphisms. The limit of the spectrum \(\{F/C_\alpha\}\) is the module \(F/A' \simeq A\) (see \((^4)\), p. 284).

Lemma 2. A \(\Lambda\)-module \(A\) is flat if and only if it is the limit of a direct spectrum of free \(\Lambda\)-modules.

Every flat \(\Lambda\)-module \(A\) can, by Lemma 1, be represented as the limit of a direct spectrum \(\{\pi_\alpha^\beta, P_\alpha\}\) of projective \(\Lambda\)-modules \(P_\alpha\). If for some free module \(F\) there is a decomposition \(F=P_\alpha \oplus Q_\alpha\), then the module \(P_\alpha\) can be represented as the limit of a direct spectrum of free \(\Lambda\)-modules
\[ F \xrightarrow{\varphi_1} F \xrightarrow{\varphi_2} \cdots, \]
where \(\operatorname{Ker}\varphi_i=Q_\alpha\). Let now \(\{\pi_\alpha^\beta, P_\alpha\}\) be a direct spectrum of projective \(\Lambda\)-modules and \(\{q_{\alpha k}^{k+1}, F_{\alpha,k}\}\) a direct spectrum of free \(\Lambda\)-modules constructed as indicated above. The set of all \(F_{\alpha k}\) is naturally ordered. We define the homomorphisms \(\varphi_{\alpha k}^{\beta l}: F_{\alpha k}\to F_{\beta l}\) as follows:
\[ \varphi_{\alpha k}^{\beta l}=i_{\beta l}\pi_\alpha^\beta\varphi_{\alpha k}, \]
where \(i_{\beta l}\) is a monomorphism of \(P_\beta\) into \(F_{\beta l}\) such that \(\varphi_{\beta l}i_{\beta l}\) is the identity mapping. With respect to the homomorphisms \(\varphi_{\alpha k}^{\beta l}\), the set \(F_{\alpha k}\) forms a direct spectrum; indeed:
\[ \varphi_{\beta l}^{\gamma s}\varphi_{\alpha k}^{\beta l} =i_{\gamma s}\pi_\beta^\gamma\varphi_{\beta l}i_{\beta l}\pi_\alpha^\beta\varphi_{\alpha k} =i_{\gamma s}\pi_\beta^\gamma\pi_\alpha^\beta\varphi_{\alpha k} =i_{\gamma s}\pi_\alpha^\gamma\varphi_{\alpha k} =\varphi_{\alpha k}^{\gamma s}. \]
It is obvious that the limit of \(\{\varphi_{\alpha k}^{\beta l}, F_{\alpha k}\}\) is the same as that of the spectrum \(\{\pi_\alpha^\beta, P_\alpha\}\).

Theorem 1. In order that all flat left \(\Lambda\)-modules be free, it is necessary and sufficient that \(\Lambda\) be a local ring with a \(T\)-nilpotent \((^2)\) maximal ideal on the left.

Lemma 3. If all left flat \(\Lambda\)-modules are free, then every element \(\lambda\in\Lambda\) is either invertible or is a zero divisor.

Proof. Suppose \(\lambda\in\Lambda\) is neither invertible nor a zero divisor; then \(\Lambda\), as a left \(\Lambda\)-module, has a proper submodule \(\Lambda\lambda\), isomorphic to the \(\Lambda\)-module \(\Lambda\). This permits one to construct the following direct spectrum of modules isomorphic to \(\Lambda\):
\[ \Lambda_1 \xrightarrow{\varphi_1} \Lambda_2 \xrightarrow{\varphi_2} \cdots \xrightarrow{\varphi_{n-1}} \Lambda_n \xrightarrow{\varphi_n} \cdots, \]
where \(\varphi_n\) is a monomorphism of \(\Lambda\) onto \(\Lambda\lambda\).

By assumption, the limit of this spectrum is a free \(\Lambda\)-module \(F\) with an infinite basis, since if the basis were finite, there would be a number \(n\) such that all elements of the basis would be contained in \(\Lambda_n\), but \(\Lambda_n\) is a proper submodule of the module \(\Lambda_{n+1}\).

Let \(w_1,w_2,\ldots,w_n,\ldots\) be a basis of \(F\), and let \(a_1,a_2,\ldots,a_n,\ldots\) be free generators of the modules \(\Lambda_1,\Lambda_2,\ldots,\Lambda_n,\ldots\), respectively. Take \(w_1\) and \(w_2\); for them there is a number \(n\) and elements \(\mu,\nu\in\Lambda\) such that \(\mu a_n=w_1,\ \nu a_n=w_2\). On the other hand, \(a_n\) lies in \(F\) and, therefore, \(a_n=\sum \lambda_i w_i\); hence we obtain
\[ \mu\lambda_1=1,\qquad \mu\lambda_2=0,\qquad \nu\lambda_1=0,\qquad \nu\lambda_2=1. \]
But then \(\lambda_1\mu\) and \(\lambda_2\mu\) are orthogonal idempotents not equal to zero; if a ring has an idempotent different from zero and one, then the ring decomposes into a direct sum of left ideals, i.e. there exists a projective but not free \(\Lambda\)-module. The lemma is proved.

From the absence of nontrivial idempotents it follows that a left inverse coincides with a right inverse; indeed, if \(ab=1\), then \(ba\) is an idempotent not equal to zero and, consequently, \(ba=1\).

Lemma 4. The noninvertible elements in the ring \(\Lambda\) form a \(T\)-nilpotent ideal.

First of all, let us prove the \(T\)-nilpotence of the set of all noninvertible elements, i.e., if \(a_1, a_2,\ldots, a_n,\ldots\) are noninvertible, then there exists a natural number \(s\) such that \(a_1a_2\ldots a_s=0\). Take a free left \(\Lambda\)-module \(F\) with basis \(x_1,x_2,\ldots,x_n,\ldots\), and in it the submodule \(G\) generated by the elements \(z_i=x_i-a_ix_{i+1}\) \((i=1,2,\ldots)\). The module \(G\) is free. By \(G_n\) we shall denote the submodule of the module \(G\) generated by \(z_1,z_2,\ldots,z_n\). Then \(F/G_n\) is a free \(\Lambda\)-module. The module \(F/G\) is the limit of a direct spectrum of free \(\Lambda\)-modules and, consequently, is free by hypothesis. Denote by \(t_1,t_2,\ldots\) the images of the elements \(x_1,x_2,\ldots\) under the mapping of \(F\) onto \(F/G\). Suppose that \(F/G\) is a free \(\Lambda\)-module with free generators \(v_1,v_2,\ldots\). Then

\[ v_1=\sum \lambda_i t_i, \]

and since \(t_i=a_i t_{i+1}\), there exists a \(k\) such that

\[ v_1=\sum_i \lambda_i a_i\ldots a_k t_{k+1}, \]

where

\[ t_{k+1}=\sum_i \mu_j v_j-\mu_1v_1,\quad v_1=\left(\sum_i \lambda_i a_i\ldots a_k\right)\left(\sum \mu_j v_j+\mu_1v_1\right), \quad 1=\sum_i \lambda_i a_i\ldots a_k\mu_1, \]

and, since left inverses are right inverses, it follows that

\[ 1=\mu_1\sum \lambda_i a_i\ldots a_k, \]

\[ \mu_1v_1=t_{k+1}=a_{k+1}t_{k+2}, \quad\text{but}\quad t_{k+2}=\sum \nu_j v_j, \quad \mu_1v_1=\sum a_{k+1}\nu_j v_j, \quad \mu_1=a_{k+1}\nu_1, \]

i.e., we have obtained that \(a_{k+1}\) is invertible, contrary to the assumption. Thus it has been proved that all noninvertible elements in the ring \(\Lambda\) form a \(T\)-nilpotent ideal. In particular, every noninvertible element \(a\) of the ring \(\Lambda\) is nilpotent and, consequently, \(1-a\) is invertible, and all elements that have no inverses form an ideal which is obviously maximal and \(T\)-nilpotent. Lemma 4 is proved.

Suppose now that \(\Lambda\) is a local ring with a maximal \(T\)-nilpotent ideal. Then \(\Lambda\) is perfect (see \((^2)\)) and, consequently, all flat \(\Lambda\)-modules are projective, and since \(\Lambda\) is local, all projective modules are free (see \((^3)\)). Theorem 1 is proved.

The author expresses deep gratitude to Prof. L. A. Skornyakov, under whose supervision this work was written.

Moscow State University
named after M. V. Lomonosov

Received
25 I 1962

REFERENCES

1. P. M. Cohn, Math. Zs., 71, No. 4, 380 (1959).
2. R. W. Bass, Trans. Am. Math. Soc., 95, 466 (1960).
3. I. Kaplansky, Sbornik: Mathematics, 4, No. 1, 3 (1960).
4. N. Steenrod, S. Eilenberg, Foundations of Algebraic Topology, Moscow, 1958.