Ma · Mathematics

How to Prove That Free Modules Are Flat Without Projective Modules

Published By the 9to5Science editorial team

At a glance

  1. 1. Fix the conventions

    Treat F as a left R-module and tensor it with right R-modules.

  2. 2. Write the free module

    Express F as a direct sum of copies of R, with no finite-rank assumption.

  3. 3. Tensor the sum

    Use the natural isomorphism between tensoring a direct sum and taking a direct sum of tensor products.

  4. 4. Check injectivity

    The tensor map acts coordinatewise, so an injection remains injective.

A free module is flat for a direct reason: tensoring with it amounts to taking a direct sum of copies of the module you started with. Direct sums preserve injections, so the tensor operation cannot turn an injective map into a non-injective one.

The proof works for free modules of any rank and does not use projective modules. To keep the tensor products well-defined even if the ring is noncommutative, we state the conventions explicitly.

The definition and module conventions

Let RR be a ring with identity, and let FF be a left RR-module. A left module FF is flat if, for every injective map u:N→N′u:N\to N' of right RR-modules, the induced map

u\otimes_R \operatorname{id}_F:N\otimes_R F\longrightarrow N'\otimes_R F$$ is injective. If your ring is commutative, you can regard all modules as left modules and use the same proof. It is enough to check injections in this definition: tensor product is always right exact, so preserving injections makes the tensor functor exact. Equivalently, flatness means that tensoring preserves short exact sequences. ## Proof using a basis Since $F$ is free, it has a basis indexed by some set $I$, and as a left module it is isomorphic to $$F\cong \bigoplus_{i\in I}R.$$ The direct sum here consists of families with only finitely many nonzero entries. The set $I$ may be infinite. Let $N$ be any right $R$-module. Tensor products commute with direct sums, and $N\otimes_R R\cong N$. Therefore

\begin{aligned} N\otimes_R F &\cong N\otimes_R\left(\bigoplus_{i\in I}R\right)\ &\cong \bigoplus_{i\in I}(N\otimes_R R)\ &\cong \bigoplus_{i\in I}N. \end{aligned}

Now take an injection $u:N\to N'$. Under these identifications, the map $u\otimes_R\operatorname{id}_F$ is the direct sum of copies of $u$: $$\bigoplus_{i\in I}N\longrightarrow\bigoplus_{i\in I}N',\qquad (n_i)_{i\in I}\longmapsto (u(n_i))_{i\in I}.$$ This map is injective. If its output is zero, then $u(n_i)=0$ for every $i$. Because $u$ is injective, every $n_i=0$, so the input is zero. Thus tensoring with $F$ preserves injections, and $F$ is flat. The key point is the naturality of the identifications: the map induced by $u$ really does act separately in every summand. The fact that elements of a direct sum have finite support ensures these coordinatewise maps are well-defined. ## A concrete example Take $R=\mathbb{Z}$ and the free module $F=\mathbb{Z}^{(I)}$, the group of finitely supported integer-valued families indexed by $I$. Consider the injective homomorphism $u:\mathbb{Z}\to\mathbb{Z}$ defined by $u(n)=2n$. After tensoring with $F$, the resulting map identifies with $$\mathbb{Z}^{(I)}\longrightarrow\mathbb{Z}^{(I)},\qquad (a_i)\longmapsto(2a_i).$$ If every output coordinate is zero, then $2a_i=0$ in $\mathbb{Z}$ for each $i$, so every $a_i=0$. The map is injective, including when $I$ is infinite. The general proof is the same argument with an arbitrary injection in place of multiplication by $2$. ## Common mistakes - **Confusing a direct sum with a direct product.** A free module with basis $I$ is $\bigoplus_{i\in I}R$, not generally $\prod_{i\in I}R$. Its elements have finite support. - **Assuming the basis must be finite.** No finite-generation assumption is needed; the proof works for an arbitrary basis set. - **Using the wrong module sides.** Over a possibly noncommutative ring, the tensor product $N\otimes_R F$ requires $N$ to be a right module and $F$ to be a left module. - **Invoking projectivity unnecessarily.** Free modules are projective, but the direct-sum argument proves flatness without that fact. ## Why the direct-sum step is valid For modules $F_i$, the natural map from $\bigoplus_i(N\otimes_R F_i)$ to $N\otimes_R(\bigoplus_i F_i)$ sends a finite family of tensors to their sum. Its inverse sends $n\otimes(f_i)_i$ to the finite sum of the tensors $n\otimes f_i$. This describes why tensoring commutes with direct sums and why the proof applies even when the free module has infinite rank.

References

  1. An Introduction to Homological Algebra — Charles A. Weibel
  2. An Introduction to Homological Algebra — Joseph J. Rotman

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