If $M_1$ and $M_2$ are R-modules, $R$ commutative. Show that $M_1 \oplus M_2$ is flat $\iff M_1$ and $M_2$ are both flat [duplicate]

Question

If $M_1$ and $M_2$ are R-modules, $R$ commutative. Show that $M_1 \oplus M_2$ is flat $\iff M_1$ and $M_2$ are both flat

My try:

By definition an $R$-module $P$ is defined to be flat if the functor $-\otimes_R P$ is exact

$(\implies)$

Suppose $M_1 \oplus M_2$ is flat , ie $-\otimes_R (M_1 \oplus M_2)$ is exact, ie it preserve short exact sequences.

Let $0 \to M' \to M \to M'' \to 0$ be a s.e.s

then $0 \to M'\otimes_R (M_1 \oplus M_2) \to M\otimes_R (M_1 \oplus M_2) \to M''\otimes_R (M_1 \oplus M_2) \to 0$ is short exact

$\iff$

$0 \to (M'\otimes_R M_1)\oplus (M' \otimes_R M_2) \to (M\otimes_R M_1)\oplus (M \oplus M_2) \to (M''\otimes_R M_1)\oplus (M'' \otimes_R M_2) \to 0 \tag {*}$

is short exact

To prove that $M_1$ is short exact I would like to get

$0 \to (M'\otimes_R M_1) \to (M\otimes_R M_1) \to (M''\otimes_R M_1) \to 0$

and similarly for $M_2$.

However I don't know how to get rid of the second summands in (*) to get what I want.

I am stuck here

$(\impliedby)$:

Let $M_1, M_2 $ be short exact, ie $-\otimes_R M_1 , -\otimes_R M_2$ are exact, ie they preserve short exact sequences.

Let $0 \to M' \to M \to M'' \to 0$ be a s.e.s

then $0 \to M'\otimes_R M_1 \to M\otimes_R M_1 \to M''\otimes_R M_1 \to 0$

and $0 \to M'\otimes_R M_2 \to M\otimes_R M_2 \to M''\otimes_R M_2 \to 0$

are short exact

Then I read in many places that the sum of two short exact sequences is exact (although I would like a proof of this)So using this

$0 \to (M'\otimes_R M_1)\oplus (M' \oplus M_2) \to (M\otimes_R M_1)\oplus (M \oplus M_2) \to (M''\otimes_R M_1)\oplus (M'' \oplus M_2) \to 0 $

is short exact

which is the same as

$0 \to M'\otimes_R (M_1 \oplus M_2) \to M\otimes_R (M_1 \oplus M_2) \to M''\otimes_R (M_1 \oplus M_2) \to 0$ is short exact

is short exact. Therefore we have the thesis

1) How do I complete my argument in $(\implies)$ ? Or how would you prove it?

2) Is $(\impliedby)$ correct?

Answer 1

Your argument in 2) is correct, and in fact you can finish the proof in 1) if you use the converse to the fact you use in 2). Let us show it by checking it on elements.

Lemma. Suppose given for $i\in\{0,1\}$ a sequence $0\to M'_i\xrightarrow{f_i} M_i\xrightarrow{g_i} M''_i\to 0$ of $R$-modules such that $g_if_i=0$. Then these two sequences are short exact if and only if the sequence $$0\to M'_1\oplus M'_2\xrightarrow{f_1\oplus f_2} M_1\oplus M_2\xrightarrow{g_1\oplus g_2} M''_1\oplus M''_2\to 0$$ is short exact.

Proof. It suffices to check the following: for any two maps $h_1\colon A_1\to B_1$ and $h_2\colon A_2\to B_2$ of $R$-modules, we have equalities

• $\mathrm{ker}(h_1\oplus h_2)=\mathrm{ker}(h_1)\oplus\mathrm{ker}(h_2)$ as submodules of $A_1\oplus A_2$;
• $\mathrm{im}(h_1\oplus h_2)=\mathrm{im}(h_1)\oplus\mathrm{im}(h_2)$ as submodules of $B_1\oplus B_2$.

Namely, if you have these isomorphisms, then it is clear that the forward direction "$\implies$" of the lemma holds. But conversely, in general, if you have inclusions $j\colon A\hookrightarrow A'$ and $k\colon B\hookrightarrow B'$ such that $j\oplus k\colon A\oplus B\to A'\oplus B'$ is an isomorphism, then $j$ and $k$ must have been identity maps and $A=A'$ and $B=B'$. Namely, $j\oplus k$ is apparently surjective, so given $a'\in A'$ there exists $(a,b)\in A\oplus B$ such that $(j(a),k(b))=(a',0)\in A'\oplus B'$. Therefore, $j(a)=a'$ and $j\colon A\hookrightarrow A'$ is surjective. Likewise, $k$ is surjective.

Applying this observation to the inclusions $\mathrm{im}(f_i)\hookrightarrow\mathrm{ker}(g_i)$, we see that if $$0\to M'_1\oplus M'_2\xrightarrow{f_1\oplus f_2} M_1\oplus M_2\xrightarrow{g_1\oplus g_2} M''_1\oplus M''_2\to 0$$ is short exact, then the fact that $$\mathrm{im}(f_1)\oplus \mathrm{im}(f_2)\cong \mathrm{im}(f_1\oplus f_2)\to \mathrm{ker}(g_1\oplus g_2)\cong\mathrm{ker}(g_1)\oplus \mathrm{ker}(g_2)$$ is an isomorphism shows that $\mathrm{im}(f_i)=\mathrm{ker}(g_i)$ for $i=0,1$, thereby showing exactness of the sequences $0\to M'_i\xrightarrow{f_i} M_i\xrightarrow{g_i} M''_i\to 0$ the middle place. You show exactness at the other place in a similar manner, so you have to show the two bullets above.

But this is straightforward. An element of $\mathrm{ker}(h_1\oplus h_2)$ is an element $(a_1,a_2)\in A_1\oplus A_2$ for which $(h_1(a_1),h_2(a_2))=(0,0)\in B_1\oplus B_2$. Therefore, this is precisely saying that $a_1\in\mathrm{ker}(h_1)$ and $a_2\in\mathrm{ker}(a_2)$. This shows the first bullet. For the second, note that an element $(b_1,b_2)\in B_1\oplus B_2$ lies in the image of $h_1\oplus h_2$ precisely when there exists $(a_1,a_2)\in A_1\oplus A_2$ such that $(h_1(a_1),h_2(a_2))=(b_1,b_2)$, and this is the case iff there exists $a_1\in A_1$ and $a_2\in A_2$ such that $h_1(a_1)=b_1$ and $h_2(a_2)=b_2$, and this holds iff $b_1\in\mathrm{im}(h_1)$ and $b_2\in\mathrm{im}(h_2)$. With this, we have shown both bullets hold, and we win.

Another proof of the lemma goes like this, if you know enough categorical input. The image of a map is the kernel of its cokernel. Since (finite) direct sums are both products and coproducts, they commute with limits (among which kernels) and colimits (among which cokernels). Therefore, the two bullets follow immediately, and via the same arguments as before this finishes the proof of the lemma.