Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.4
Find all ring homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}/(30)$. In each case describe the kernel and the image.
Solution: We begin with some lemmas.
Lemma 1: Every ring homomorphism $\varphi : \mathbb{Z} \rightarrow R$ is uniquely determined by $\varphi(1)$.
Proof: This follows because $\varphi$ is an additive group homomorphism, and $\mathbb{Z}$ is free on $\{1\}$. $\square$
Lemma 2: Let $R$ be a ring and let $\varphi : \mathbb{Z} \rightarrow R$ be the additive group homomorphism defined by $\varphi(1) = z$. Then $\varphi$ is a ring homomorphism if and only if $z^2 = z$.
Proof: Suppose $\varphi$ is a ring homomorphism. Then $$z = \varphi(1) = \varphi(1 \cdot 1) = \varphi(1) \varphi(1) = z^2.$$ Suppose now that $z^2 = z$. Then \begin{align*}\varphi(ab) =&\ \varphi\left( \sum_{i=1}^{ab} 1 \right) = \sum_{i=1}^{ab} z = \sum_{i=1}^a \sum_{j=1}^b z \\=&\ \sum_{i=1}^a \sum_{j=1}^b z \cdot z = \sum_{i=1}^a \left( z \sum_{j=1}^b z \right)\\ =&\ \left( \sum_{i=1}^a z \right) \left( \sum_{j=1}^b z \right)\\ =&\ (az)(bz) = \varphi(a)\varphi(b).\end{align*} Thus \varphi is a ring homomorphism. $\square$
By the lemma, the group homomorphism $\varphi_{\overline{a}} : \mathbb{Z} \rightarrow \mathbb{Z}/(30)$ defined by extending $\varphi(1) = \overline{a}$ is a ring homomorphism if and only if $\overline{a}^2 = \overline{a}$ in $\mathbb{Z}/(30)$, and moreover every ring homomorphism $\mathbb{Z} \rightarrow \mathbb{Z}/(30)$ is obtained in this way. Evidently, $$\overline{a} \in \{ \overline{1}, \overline{6}, \overline{10}, \overline{15}, \overline{21}, \overline{25} \}.$$ Next we prove another lemma.
Lemma 3: Let $\varphi : \mathbb{Z} \rightarrow \mathbb{Z}/(m)$ be a ring homomorphism with $\varphi(1) = \overline{a}$. Then $\mathsf{ker}\ \varphi = \langle m/\mathsf{gcd}(a,m) \rangle$ and $\mathsf{im}\ \varphi = \langle \overline{\mathsf{gcd}(a,m)} \rangle$.
Proof: (1) ($\subseteq$) Let $b \in \mathsf{ker}\ \varphi$. Then $ab = 0 \pmod m$, and we have $ab = mk$ for some $k$. Now write $\mathsf{gcd}(a,m) = d$ and $a = da_2$; note that $\mathsf{gcd}(a_2,m) = 1$ by definition. Now $b = mk/da_2$ is an integer. Since $a_2$ is relatively prime to $m$, it must divide $k$. Let $k_2 = k/a_2$; then $b = mk_2/d \in \langle m/\mathsf{gcd}(a,m) \rangle$.
($\supseteq$) Note that $$\varphi(m/\mathsf{gcd}(a,m)) = am/\mathsf{gcd}(a,m) = \mathsf{lcd}(a,m) = mk \equiv 0 \pmod m,$$ so that $\langle m/\mathsf{gcd}(a,m) \rangle \subseteq \mathsf{ker}\ \varphi$.
(2) ($\subseteq$) Suppose $\overline{b} \in \mathsf{im}\ \varphi$. Then $\overline{b} = \varphi(k) = \overline{ka}$ for some $k \in \mathbb{Z}$, and we have $b = m\ell + ak$ for some $\ell \in \mathbb{Z}$. Thus $\mathsf{gcd}(a,m)$ divides $b$, and we have $\overline{b} \in \langle \overline{\mathsf{gcd}(a,m)} \rangle$.
($\supseteq$) Recall from Bezout’s Identity that $\mathsf{gcd}(a,m) = mx + ay$ for some integers $x$ and $y$. Mod $m$, we have $$\overline{\mathsf{gcd}(a,m)} = \overline{ay} = \varphi(y),$$ so that $\langle \overline{\mathsf{gcd}(a,m)} \rangle \subseteq \mathsf{im}\ \varphi$. $\square$
(1) $\mathsf{ker}\ \varphi_{\overline{1}} = \langle 30 \rangle$ and $\mathsf{im}\ \varphi_{\overline{1}} = \langle \overline{1} \rangle$
(2) $\mathsf{ker}\ \varphi_{\overline{6}} = \langle 5 \rangle$ and $\mathsf{im}\ \varphi_{\overline{6}} = \langle \overline{6} \rangle$
(3) $\mathsf{ker}\ \varphi_{\overline{10}} = \langle 3 \rangle$ and $\mathsf{im}\ \varphi_{\overline{10}} = \langle \overline{10} \rangle$
(4) $\mathsf{ker}\ \varphi_{\overline{15}} = \langle 2 \rangle$ and $\mathsf{im}\ \varphi_{\overline{15}} = \langle \overline{15} \rangle$
(5) $\mathsf{ker}\ \varphi_{\overline{21}} = \langle 10 \rangle$ and $\mathsf{im}\ \varphi_{\overline{21}} = \langle \overline{3} \rangle$
(6) $\mathsf{ker}\ \varphi_{\overline{25}} = \langle 6 \rangle$ and $\mathsf{im}\ \varphi_{\overline{25}} = \langle \overline{5} \rangle$
