# Exact sequence with $F$ free splits

Prove that every exact sequence of $$R$$ modules $$0 \rightarrow M \rightarrow N \xrightarrow{\phi} F \rightarrow 0$$ splits. This is an exercise from Algebra chapter 0, which does not go over projective modules before giving this exercise. It says to use the following prior exercise

Let $$R$$ be a ring, $$F$$ a nonzero free $$R$$-module, and let $$\varphi : M \to N$$ be a homomorphism of $$R$$-modules. Prove that $$\varphi$$ is onto if and only if for all $$R$$-module homomorphisms $$\alpha : F \to N$$ there exists an $$R$$-module homomorphism $$\beta : F \to M$$ such that $$\alpha=\varphi \circ \beta$$.

To solve it. So I will try to use this.

Attempt: Since $$\phi$$ is onto by definition of short exact sequence, for $$\text{id}:F \rightarrow F$$, there exists $$\beta:F \rightarrow N$$ such that $$\text{id}=\phi \circ \beta$$. Thus, $$\beta$$ is a splitting homomorphism for the sequence and it splits. Is this ok? I tried to come up with an explicit isomorphism $$M \oplus F \cong N$$ but could not figure this out. Can help me construct an isomorphism?

## 3 Answers

You can obtain an explicit isomorphism considering a basis $$(e_i)_{i\in I}$$ of $$F$$ and an element $$n_i\in \phi^{-1}\bigl(\{e_i\}\bigr)$$ for each $$i\in I$$ (of course, if $$I$$ is not finite, this requires the axiom of choice).

Then set $$\beta(e_i)=n_i,\: i\in I$$. It is easy to check that $$N=\langle\, n_i\,\rangle_{i\in I} \oplus \ker\phi\simeq F\oplus M.$$

• What is $x_i$ here? Dec 6, 2022 at 16:14
• Could you clarify how to show your last step? I suppose we want to associate $n \in N$ to an element of the direct sum you have written (call it $D$). My idea was to consider cases, of (1) $n \in ker(\varphi)$, so we associate $n$ to $e_F + n$ in D, and (2) $n \notin ker(\varphi)$, so that $\varphi(n) = \sum r_i e_i$, for $r_i \in R$, say - and so we associate $n$ to $\sum r_i n_i + e_N$ in $D$. Dec 6, 2022 at 16:36
• @Bernand What is $x_i$ here?
– user1167379
Oct 17, 2023 at 15:56
• @DTPW $x_i$ was clearly a typo for $n_i.$ Oct 17, 2023 at 16:04
• @algebroo Your (2) is wrong. The canonical (unique) decomposition of any $n\in N$ as the sum of an element of $\langle\, n_i\,\rangle_{i\in I}$ and an element of $\ker\varphi$ is: let $(r_i)_{i\in I}\in R^{(I)}$ be defined by $\varphi(n)=\sum r_ie_i,$ then $n=\sum r_in_i+(n-\sum r_in_i).$ Oct 17, 2023 at 16:10

For "Aluffi Algebra Chapter 0" context only:

Using the hint "6.9 th Exercise"

Let $$R$$ be a ring, $$F$$ a nonzero free $$R$$-module, and let $$\varphi : M \to N$$ be a homomorphism of $$R$$-modules. Prove that $$\varphi$$ is onto if and only if for all $$R$$-module homomorphisms $$\alpha : F \to N$$ there exists an $$R$$-module homomorphism $$\beta : F \to M$$ such that $$\alpha=\varphi \circ \beta$$.

Since $$0 \rightarrow M \rightarrow N \xrightarrow{\phi} F \rightarrow 0$$ exact, $$\phi : N\to F$$ is onto.

$$\require{AMScd}\begin{CD} F @>\exists ! \psi >> N \\ @V Id VV \swarrow_{\phi} \\ F \end{CD}$$

Such that the diagram commutative, so it means $$Id_F=\phi\psi$$. Means $$\phi$$ has right inverse, and using the splitting lemma we are done.

Hint. A splitting homomorphism $$\beta$$ yields a equality $$N=M\oplus\beta(F)$$ (internal direct sum) . This is a good exercise and a general fact about split sequences. Note now that $$\beta$$ is necessarily injective (why?).

It follows from that you have almost one only natural choice for your isomorphism $$M\times F\to N$$, in view of the equality $$N=M\oplus\beta(F)$$. Which one?