# Step in Bourbaki Commutative algebra exercise

I'm trying to solve this step from a Bourbaki's commutative algebra exercise (Chapter I, Flat Modules, exercise 23 (a) page 47, implication $$(\gamma)\Rightarrow(\delta)$$):

Let $$R$$ be a ring, $$F$$ be a free $$R$$-module and $$K\subseteq F$$ be an $$R$$-submodule. Assume that for every $$x\in K$$ there exists an $$R$$-module homomorphism $$\upsilon:F\to K$$ such that $$\upsilon(x)=x$$. Then for every finite sequence $$x_1,\ldots,x_n$$ of elements of $$K$$ there exists an $$R$$-module homomorphism $$\upsilon:F\to K$$ such that $$\upsilon(x_i)=x_i$$ for every $$1\leq i\leq n$$.

My try. I'm starting with the case $$n=2$$. By assumption there exists $$\upsilon_1,\upsilon_2:F\to K$$ such that $$\upsilon_1(x_1)=x_1$$ and $$\upsilon_2(x_2)=x_2$$. I'm failed to build $$\upsilon$$ as a linear combination of $$\upsilon_1$$ and $$\upsilon_2$$. Moreover, I cannot find a useful way to use assumption about $$F$$ to be a free $$R$$-module. Any hint will be appreciated.

• Note that another equivalent condition, according to Bourbaki, is that the quotient $R$-module $F / K$ is flat. (Caution: What you call $R$ is not what Bourbaki calls $R$.) Aug 16, 2019 at 9:40
• Also, $R$ is supposed to be commutative, right? Aug 16, 2019 at 9:40
• With the equivalence mentioned by @darijgrinberg, this is given as Lemma 2.10(c) of Milne's "Etale Cohomology". Aug 16, 2019 at 9:43
• I'm already proved $(\alpha)\Rightarrow(\beta)\Rightarrow(\gamma)$ and $(\delta)\Rightarrow(\alpha)$; thus only remains $(\gamma)\Rightarrow(\delta)$. I assume the ring not commutative, since Bourbaki distinguish between right and left modules. Aug 16, 2019 at 9:46
• @AlexWertheim: Nice! And Milne's argument does not even use the freeness of $F$. (Link to the relevant part of the book.) Aug 16, 2019 at 9:50

Here is a self-contained proof, adapted from the proof in Milne's Étale Cohomology referenced in the other answer. We use induction on $$n$$, the base case $$n=1$$ being given. Now suppose $$n>1$$ and the result is known for $$n-1$$. We can choose $$u:F\to K$$ such that $$u(x_1)=x_1$$ and then using the induction hypothesis we can choose $$v:F\to K$$ such that $$v(x_i-u(x_i))=x_i-u(x_i)$$ for $$i>1$$. Defining $$w(x)=u(x)+v(x-u(x))$$ we then have $$w(x_1)=x_1+v(x_1-x_1)=x_1$$ and $$w(x_i)=u(x_i)+v(x_i-u(x_i))=x_i$$ for $$i>1$$, so $$w(x_i)=x_i$$ for all $$i\geq 1$$.
As requested above, I am posting my comment as an answer. The implication $$(\delta) \implies (\alpha)$$ is addressed (modulo the equivalence of the given hypotheses with the condition that $$F/K$$ be a flat $$R$$-module) in Lemma 2.10(c) of Milne's "Etale Cohomology". A link to the relevant portion (kindly provided by Darij Grinberg) can be found here.