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  "body": "\n## Proof\n\n1. Let $R$ be an [[integral domain|09S3PFBEPM]] and let $n$ be its [[characteristic|0P7CH1CND5]]. For a positive integer $k$, write $k \\cdot 1$ for the sum of $k$ copies of $1$ in $R$.\n2. If $n = 0$, there is nothing to prove. So assume $n \u003e 0$. Then $n \\cdot 1 = 0$, and $k \\cdot 1 \\neq 0$ for $0 \u003c k \u003c n$.\n3. Since $1 \\cdot 1 = 1 \\neq 0$ in an integral domain, $n \\neq 1$. So $n \\ge 2$.\n4. We argue by contradiction. Suppose that $n$ is not a [[prime number|0SN0MAEKYX]]. Then $n$ is [[composite|0N3ADXHNTQ]], so it has a positive divisor $a$ with $a \\neq 1$ and $a \\neq n$.\n5. Put $b = n / a$. Then $1 \u003c a \u003c n$ and $1 \u003c b \u003c n$.\n6. For positive integers $a$ and $b$, $(a \\cdot 1)(b \\cdot 1) = (a b) \\cdot 1$. Indeed, by the distributive laws of the [[ring|0894J980RK]] $R$, the product is a sum of $a b$ terms, each equal to $1 \\cdot 1 = 1$.\n7. So $(a \\cdot 1)(b \\cdot 1) = n \\cdot 1 = 0$.\n8. Since $R$ is an integral domain, $a \\cdot 1 = 0$ or $b \\cdot 1 = 0$.\n9. Both $a$ and $b$ are positive and less than $n$. So step 8 contradicts the minimality of $n$ in step 2.\n10. Hence $n$ is a prime number. So the characteristic of $R$ is $0$ or a prime number.\n\n## Idea\n\nIf the characteristic $n$ factors as $a b$ with smaller factors, then $(a \\cdot 1)(b \\cdot 1) = 0$. In an integral domain, one of the two factors must be $0$. That contradicts the choice of $n$ as the least positive integer with $n \\cdot 1 = 0$.\n",
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    "name": "Proof that the characteristic of an integral domain is zero or prime",
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    "tag": "G5TFN4G1RM",
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  "msc": [
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  "name": "Proof that the characteristic of an integral domain is zero or prime",
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