Step in Bourbaki Commutative algebra exercise

Question

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 $R$.) — darij grinberg, Aug 16 '19 at 09:40
With the equivalence mentioned by @darijgrinberg, this is given as Lemma 2.10(c) of Milne's "Etale Cohomology". — Alex Wertheim, Aug 16 '19 at 09:43
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. — Fabio Lucchini, Aug 16 '19 at 09:46
@AlexWertheim: Nice! And Milne's argument does not even use the freeness of $F$. (Link to the relevant part of the book.) — darij grinberg, Aug 16 '19 at 09:50
@AlexWertheim: your reference is exactly what I'm looking for; I could accept it as an answer. — Fabio Lucchini, Aug 16 '19 at 14:03
@FabioLucchini: done, thanks! And thank you for the link, Darij. — Alex Wertheim, Aug 17 '19 at 03:09

score 4 · Answer 1 · answered Aug 17 '19 at 03:49

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 $i\geq 1$.

score 3 · Accepted Answer · answered Aug 17 '19 at 03:08

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.

Step in Bourbaki Commutative algebra exercise

2 Answers2