First Isomorphism Theorem (Modules)

Theorem

Given modules \(M\) and \(N\), with a homomorphism \(\phi : M \to N\)

\[ M/\ker(\phi) \cong \mathrm{im}(\phi)\]

with explicit isomorphism \(m + \mathrm{ker}(\phi) \mapsto \phi(m)\).

Proof

Let \(\phi : M \to N\) be a module homomorphism. This means it must also be a group homomorphism of the additive abelian groups \(M\) and \(N\). Hence, applying the first isomorphism theorem for groups we have that the map \(\psi: m + \mathrm{ker}(\phi) \mapsto \phi(m)\) is a group isomorphism from \(M/\ker(\phi) \to \mathrm{im}(\phi)\).

Thus we just need to prove that this map respects the scalar multiplication, that is

\[\begin{align*} \psi(s.(g + \mathrm{ker}(\phi))) &= \psi(s.g + \mathrm{ker}(\phi)) \\ &= \phi(s.g) \\ &= s.\phi(g) \\ &= s.\psi(g + \mathrm{ker}(\phi)). \\ \end{align*}\]