direct product of modules as categorical product

1. Proposition

Let \(R\) be a ring and \(M_i\) a family of \(R\)-modules. Then the direct product of modules

\begin{align*} \prod_{i \in I} M_i \end{align*}

with the projection

\begin{align*} \pi_i: \prod_{i \in I} M_i \rightarrow M_i \\ (m_i) \mapsto& m_i \end{align*}

is the categorical product

2. Proof

Let \(N\) be a module with module-homomorphisms

\begin{align*} f_i: N \rightarrow M_i \end{align*}

Then

\begin{align*} f: N \rightarrow \prod_{i \in I} M_i n \mapsto& (f_i(n)) \end{align*}

is the unique morphism

2.1. module homomorphism

For \(n_1,n_2 \in N\) and \(r \in R\), we get by construction of the direct product

\begin{align*} f(n_1 + r \cdot n_2) =& (f_i(n_1) + f_i(r \cdot n_2)) \\ =& (f_i(n_1)) + (f_i(r \cdot n_2)) \\ =& (f_i(n_1)) + (r \cdot f_i(n_2)) \\ =& (f_i(n_1)) + r \cdot (f_i(n_2)) \\ \end{align*}

2.2. commuting

Follows from

\begin{align*} \pi_i \circ f(n) =& \pi_i((f_i(n))) \\ =& f_i(n) \end{align*}

2.3. uniqueness

Suppose \(g \neq f\), then there exists an element \((n)\) such that

\begin{align*} (m_i) = g(n) \neq f(n) \end{align*}

and hence for some component \(m_i \neq f(n)_i\) Therefore, we get

\begin{align*} \pi_i(g(n)) = m_i \neq f_i(m_i) \end{align*}

hence \(f\) is unique