3
$\begingroup$

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?

$\endgroup$
0

3 Answers 3

2
$\begingroup$

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 $$N=\langle\, n_i\,\rangle_{i\in I} \oplus \ker\phi\simeq F\oplus M.$$

$\endgroup$
9
  • $\begingroup$ What is $x_i$ here? $\endgroup$
    – algebroo
    Dec 6, 2022 at 16:14
  • 1
    $\begingroup$ 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 $D$. $\endgroup$
    – algebroo
    Dec 6, 2022 at 16:36
  • $\begingroup$ @Bernand What is $x_i$ here? $\endgroup$
    – user1167379
    Oct 17, 2023 at 15:56
  • $\begingroup$ @DTPW $x_i$ was clearly a typo for $n_i.$ $\endgroup$ Oct 17, 2023 at 16:04
  • $\begingroup$ @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 $n=\sum r_in_i+(n-\sum r_in_i).$ $\endgroup$ Oct 17, 2023 at 16:10
1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .