# Proving that free modules are flat (without appealing projective modules)

Suppose $R\neq 0$ is a commutative ring with $1$. Let $M$ be a free $R$-module. I would like to prove that $M$ is a flat $R$-module. Everywhere I have looked (mostly online) this is proved by first proving that every free module is projective, and then proving that every projective module is flat. Unfortunately, Atiyah & Macdonald's "Introduction to Commutative Algebra" (Chapter 2) does not discuss projective modules. But the result that every free module is flat comes very handy in the exercises.

So my question is,

Is it possible to prove that every free module is flat just by definitions and without appealing to projective modules?

Thanks!

• The way people usually prove that projective modules are flat is to reduce to the free case by using that every projective module is a direct summand of a free module, so I'm not sure I understand the question. In fact, it is an somewhat challenging exercise to show that a projective module is flat without appealing to free modules. May 31, 2013 at 9:59
• @Martin: You are right. I just wanted to see a proof that doesn't involve discussion of projective modules... But I was ignorant enough not to look at the proof of "projective implies flat" and to see that it does indeed use the result of seemingly weaker statement "free implies flat". Thanks for the heads-up! May 31, 2013 at 10:06

You can just show that the functor $M\otimes_R(-)$ is left-exact. Write $M=\bigoplus_{i\in I} R$ and let $$0\to N'\xrightarrow{\quad\iota\quad} N\to N''\to 0$$ be an exact sequence of $R$-modules. Note that $M\otimes N=\bigoplus_{i\in I} R\otimes_R N = \bigoplus_{i\in I} N$, so the sequence

$$0 \to M\otimes N' \xrightarrow{\quad\mathrm{id}\otimes\iota\quad} M\otimes N \to M\otimes N'' \to 0$$

is the same as

$$0 \to \bigoplus_{i\in I} N' \xrightarrow{\textstyle\quad\bigoplus_{i\in I} \iota\quad} \bigoplus_{i\in I}N \to \bigoplus_{i\in I}N'' \to 0$$

and the morphism $\bigoplus_{i\in I} \iota$ is clearly injective if $\iota$ is injective.

• If flatness is defined by asserting that $M \otimes (\_)$ is exact, should you not also show right-exactness?
– gen
Dec 24, 2018 at 22:49
• Dear @gen, you are correct. However, it is a rather well-known lore result that taking the tensor product with a fixed module is right-exact in general, so I will weasel out of this by arguing that you may define flatness over general rings by requiring that the tensor product preserves monomorphisms. Dec 24, 2018 at 22:56
• Understood, thanks for clarifying. It's just that not all of us did phd's in algebra, as you can imagine
– gen
Dec 24, 2018 at 22:58
• @gen I am honestly very sorry if my comment came across as patronizing. I firmly believe in asking every question and clarifying every detail in Mathematics. Shouting "well-known" is a last resort; Please forgive me for using it here. I answered this question 5 years ago and it seemed to me at the time that the OP would know the result about right-exactness, so I only went for left-exactness. I have little time these days, but if I can clear my schedule a bit, I will try to include more detail in my answer. Dec 25, 2018 at 22:36
• @JeskoHüttenhain The world needs more humble Mathematicians like you :) May 11, 2020 at 17:22
1. The identity functor is exact.

2. If $\{F_i\}$ is a family of exact functors, then $\oplus_i F_i$ is also exact.

Now apply this to $M \otimes -$ for a free module $M$. 1. deals with the case $M=R$, and 2. generalizes this to $M = \oplus_i R$.

• For 2. is the reverse implication also true?
– gen
Dec 24, 2018 at 23:23
• @gen Yes it is. Dec 22, 2019 at 14:57
• Thank you for this
– gen
Dec 22, 2019 at 15:34

Assume our free module $M$ is of the form $\oplus_{i\in S}R_{i}$, then the statement just follows. The details may be best to leave out to you.

For your question on the projective module, you can consider the projective module as a direct summand of a free module and "piece" up the maps together.