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  "body": "\n## Proof\n\n1. Let $f \\colon R \\to S$ be a [[ring homomorphism|0GGE0Z0JQ1]] with [[kernel|0AAHE7GEW6]] $\\ker f = \\set{a \\in R : f(a) = 0}$.\n2. We have $f(0) = f(0 + 0) = f(0) + f(0)$. Subtracting $f(0)$ gives $f(0) = 0$. So $0 \\in \\ker f$.\n3. For $a \\in R$, $f(a) + f(-a) = f(a + (-a)) = f(0) = 0$. So $f(-a) = -f(a)$.\n4. Hence $f(a - b) = f(a) + f(-b) = f(a) - f(b)$ for all $a, b \\in R$.\n5. Suppose that $f$ is [[injective|0PVG1CRAJN]], and let $a \\in \\ker f$. Then $f(a) = 0 = f(0)$, so $a = 0$. With step 2, $\\ker f = \\set{0}$.\n6. Conversely, suppose that $\\ker f = \\set{0}$. Let $a, b \\in R$ with $f(a) = f(b)$.\n7. By step 4, $f(a - b) = f(a) - f(b) = 0$. So $a - b \\in \\ker f$, and $a - b = 0$.\n8. Thus $a = b$, and $f$ is injective.\n\n## Idea\n\nTwo elements have the same image exactly when their difference lies in the kernel. If only $0$ lies in the kernel, the difference is $0$ and the elements are equal. The argument uses only that $f$ preserves addition.\n",
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