7) For the \( \mathbb{Z} \) modules \( M = \mathbb{Z} / 7 \mathbb{Z} \) and \( N = \mathbb{Z} / 6 \mathbb{Z} \), find all
the elements in \( \text{Hom}_{\mathbb{Z}} (M, N). \)
Proof. Ex. §10.2,#4 – #6 in Dummit & Foote bear on this. Ex. #6 Says there are \( (7, 6) = 1 \) such homomorphisms, namely
just the \( 0 \) homomorphism.
But playing this out, put \( \mathbb{Z} / n \mathbb{Z} = \mathbb{Z}_{n} \), and consider Ex. #4, which says that \( \text{Hom}_{\mathbb{Z}}
(\mathbb{Z}_{7}, \mathbb{Z}_{6}) \cong \{a \in \mathbb{Z}_{6} \; | \; 7a \equiv 0 \; (\text{mod} \; 6) \} \), where the homomorphism \( \varphi_a \) associated with
\( a \) is \( \varphi_a(\overline{k}) = ka \). So \( 7a \equiv a \equiv 0 \; (\text{mod} 6) \), and \( \varphi_0 \) is the only homomorpism,
which multiplies by \( 0 \) and is therefore the 0 homomorphism.
To do this from scratch, let \( \psi \in \text{Hom}_{\mathbb{Z}} (\mathbb{Z}_{7}, \mathbb{Z}_{6}) \). Then \( \psi(7) = \psi(0) = 0 \) and
\( \psi(6) = \psi(6 \cdot 1) = 6 \cdot \psi(1) = 0 \cdot \psi(1) = 0 \). So \( \psi(1) = \psi(7 - 6) = \) \( \psi(7) - \psi(6) = 0 - 0 = 0 \).
Therefore \( \psi(k) = \psi(k \cdot 1) = k \cdot \psi(1) = k \cdot 0 = 0, \) all \( k \in \mathbb{Z}_{7} \). That is, \( \psi \) is
the \( 0 \) homomorphism, and \( \text{Hom}_{\mathbb{Z}} (\mathbb{Z}_{7}, \mathbb{Z}_{6}) = 0. \) QED.