1. **Problem Statement:** Prove that the map $\rho : \mathbb{Z}[i] \to \mathbb{Z}/10\mathbb{Z}$ defined by $\rho(a + bi) = [a + 7b]$ is a homomorphism and find its kernel. 2. **Recall the definition of a homomorphism:** A function $\rho$ between two rings is a homomorphism if for all $x,y$ in the domain: $$\rho(x + y) = \rho(x) + \rho(y) \quad \text{and} \quad \rho(xy) = \rho(x) \cdot \rho(y)$$ 3. **Check addition:** Let $x = a + bi$, $y = c + di$ with $a,b,c,d \in \mathbb{Z}$. $$\rho(x + y) = \rho((a + c) + (b + d)i) = [(a + c) + 7(b + d)] = [a + 7b + c + 7d]$$ On the other hand, $$\rho(x) + \rho(y) = [a + 7b] + [c + 7d] = [a + 7b + c + 7d]$$ Thus, $$\rho(x + y) = \rho(x) + \rho(y)$$ 4. **Check multiplication:** $$xy = (a + bi)(c + di) = (ac - bd) + (ad + bc)i$$ Apply $\rho$: $$\rho(xy) = [(ac - bd) + 7(ad + bc)] = [ac - bd + 7ad + 7bc]$$ Calculate $\rho(x) \cdot \rho(y)$: $$[a + 7b] \cdot [c + 7d] = [ (a)(c) + (a)(7d) + (7b)(c) + (7b)(7d) ] = [ac + 7ad + 7bc + 49bd]$$ Since we are in $\mathbb{Z}/10\mathbb{Z}$, reduce $49bd$ modulo 10: $$49bd \equiv 9bd \pmod{10}$$ So, $$\rho(x) \cdot \rho(y) = [ac + 7ad + 7bc + 9bd]$$ Compare with $\rho(xy)$: $$\rho(xy) = [ac - bd + 7ad + 7bc] = [ac + 7ad + 7bc - bd]$$ For $\rho$ to be a homomorphism, these must be equal: $$[ac + 7ad + 7bc - bd] = [ac + 7ad + 7bc + 9bd]$$ Subtracting common terms, $$[-bd] = [9bd] \implies [-bd - 9bd] = [0] \implies [-10bd] = [0]$$ Since $-10bd$ is divisible by 10, this holds for all $b,d \in \mathbb{Z}$. Therefore, $$\rho(xy) = \rho(x) \cdot \rho(y)$$ 5. **Conclusion:** $\rho$ preserves addition and multiplication, so it is a ring homomorphism. 6. **Find the kernel:** The kernel is all $x = a + bi$ such that $$\rho(a + bi) = [a + 7b] = [0] \in \mathbb{Z}/10\mathbb{Z}$$ This means $$a + 7b \equiv 0 \pmod{10}$$ or equivalently, $$a \equiv -7b \equiv 3b \pmod{10}$$ So the kernel is $$\ker(\rho) = \{a + bi \in \mathbb{Z}[i] : a \equiv 3b \pmod{10}\}$$ **Final answer:** $\rho$ is a ring homomorphism and $$\ker(\rho) = \{a + bi \mid a + 7b \equiv 0 \pmod{10}\}$$