0
$\begingroup$

I'm trying to solve this exercise:

6.48 EXERCISE. Let $M$, $M_1$, $\ldots$, $M_n$ (where $n\in\mathbb{N}$ with $n\ge 2$) be modules over the commutative ring $R$.

(i) Show that there is an exact sequence $$0 \rightarrow M_1 \xrightarrow{q_1} \bigoplus_{i=1}^n M_i \xrightarrow{p_1'} \bigoplus_{i=2}^n M_i \to 0$$ (of $R$-modules and $R$-homomorphisms) in which $q_1$ is the canonical injection and $$p_1'((m_1,\ldots,m_n)) = (m_2,\ldots,m_n)$$ for all $(m_1,\ldots,m_n) \in \bigoplus_{i=1}^n M_i$.

(ii) Suppose that there exist, for each $i=1,\ldots,n$, homomorphisms $\tilde{p}_i:M\to M_i$ and $\tilde{q}_i:M_i \to M$ such that, for all $i,j$ with $1\le i, j\le n$, $$\tilde{p}_i \circ \tilde{q}_i = \DeclareMathOperator{\Id}{Id} \textstyle \Id_{M_i} \qquad \text{and} \qquad \tilde{p}_i \circ \tilde{q}_j = 0 \text{ for } i\neq j,$$ and $\sum_{i=1}^n \tilde{q}_i \circ \tilde{p}_i = \Id_M$. Show that the mapping $\,f:M \to \oplus_{i=1}^n M_i$ defined by $$f(m) = (\tilde{p}_1(m), \ldots, \tilde{p}_n(m)) \qquad \text{for all } m\in M$$ is an isomorphism.

I almost solved this question, I'm having troubles to prove the surjectivity in part (ii).

$\endgroup$
2
  • $\begingroup$ @YACP ohhhh, sorry I've already done this part, wait $\endgroup$
    – user42912
    Jun 29, 2013 at 9:14
  • $\begingroup$ @YACP my problem is with the surjectivity part. I was mistaken. $\endgroup$
    – user42912
    Jun 29, 2013 at 9:15

2 Answers 2

1
$\begingroup$

$g : \oplus_i M_i \to M, (m_i)_i \mapsto \sum_i \tilde{q_i}(m_i)$ is easily seen to be inverse to $f$.

$\endgroup$
2
  • $\begingroup$ this is a two-sided inverse? $\endgroup$
    – user42912
    Jun 29, 2013 at 9:19
  • $\begingroup$ Yes, sure. And this is the fastest way to show that $f$ is an isomorphism. $\endgroup$ Jun 29, 2013 at 14:53
1
$\begingroup$

$\DeclareMathOperator{\Id}{Id}$I assume you’re trying to prove that $\,f$ is an isomorphism by showing that it’s injective and surjective, and hence bijective?

You can actually jump straight from homomorphism to isomorphism, by trying to find a map $g:\bigoplus_{i=1}^n M_i \to M$ with $g \circ f = \Id_M$ and $f\circ g=\Id_{\bigoplus_{i=1}^n M_i}$ (then $\,f$ is invertible and so bijective).

But the second statement is sufficient for surjectivity, which is what you asked. You’re given that $\tilde{p}_i \circ \tilde{q}_i = \Id_{M_i}$ for all $i$, and that $\tilde{p}_i \circ \tilde{q}_j=0$ for $i\ne j$. So define $g$ by $$g((m_1,\ldots,m_n)) = \sum_{i=1}^n \tilde{q}_i(m_i)$$ and then you have \begin{align*} (\,f\circ g)((m_1,\ldots,m_n)) &= f\,\left(\sum_{i=1}^n \tilde{q}_i(m_i)\right) \\ &= \left(\sum_{i=1}^n (\,\tilde{p_1} \circ \tilde{q_i})(m_i), \ldots, \sum_{i=1}^n (\,\tilde{p_n} \circ \tilde{q_i})(m_i)\right) \\ \end{align*} The condition on the $\tilde{p}_i,\tilde{q}_i$ then tells you that in each sum, every cross-term vanishes, and you’re left with \begin{align*} &= \left((\,\tilde{p}_1 \circ \tilde{q}_1)(m_1), \ldots, (\,\tilde{p}_1 \circ \tilde{q}_1)(m_n)\right) \\ &= (m_1,\ldots,m_n), \end{align*} and this is true for all $(m_1,\ldots,m_n) \in \bigoplus_{i=1}^n M_i$. So the preimage of $(m_1,\ldots,m_n)$ is $g((m_1,\ldots,m_n))$, whence $\,f$ is surjective.

You can prove that this is an inverse in the opposite direction (which you might have used for your proof of injectivity), and then you get the bijection immediately.

$\endgroup$
1
  • $\begingroup$ @YACP: yes, although I started writing it before I saw Martin’s answer. $\endgroup$
    – alexwlchan
    Jun 29, 2013 at 14:09

You must log in to answer this question.

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