and this is the solution given
Why do we need the characteristic to be 3? Why wouldn't this work if over $\mathbb{Z}/\mathbb{9Z}$?
2026-04-11 23:25:47.1775949947
Why does the characteristic need to be 3?
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We need $9$ to be the same as $0$ in the base field. As a field contains no zero divisors and $3\times 3 = 9$ which is $0$, we must also have $3$ equal to $0$. This is only true in a field of characteristic $3$. An example of a field of characteristic $3$ is $\mathbb{Z}/3\mathbb{Z}$. Any other field of characteristic $3$ would also be a suitable answer, but $\mathbb{Z}/9\mathbb{Z}$ is not a field (for example, $3\times 3 = 0$ in $\mathbb{Z}/9\mathbb{Z}$, but $3 \neq 0$ in $\mathbb{Z}/9\mathbb{Z}$).