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Let $\mathbb Z_{n}=\{0,1,...,n-1\}$ and $+_{n}$ and $\times_{n}$ be the modulo addition and multiplication. For $n=3$, the set $\mathbb Z_{n}$ is not group wrt $\times_{3}$. I have the following questions.

(1) What should be the value of n (e.g., $n\geq ?$ ) so that $Z_{n}$ form ring with the above operations.

(2) I have saw that $Z_{p}$ form field for $p$ being a prime number. Now $p=3$ is prime but $Z_{3}$ doesn't form field.

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    $\begingroup$ Why do you believe that $\Bbb Z_3$ is not a field? $\endgroup$
    – Ben Grossmann
    Mar 23 at 12:53
  • $\begingroup$ Oh Thanks dear. I have forgot that the F-{0} should be abelian $\endgroup$
    – Junaid
    Mar 23 at 12:55
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    $\begingroup$ Better consider it as the quotient ring $\Bbb Z/n\Bbb Z$ by the ideal $(n)=n\Bbb Z$. If $n$ is not prime then we have zero divisors. For $n=p$ prime this is not the case and we have a field. $\endgroup$
    – Dietrich Burde
    Mar 23 at 14:56

1 Answer 1

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A commutative ring $R$ forms a field if the set $R \setminus \{0\}$ is a group under multiplication. The field $\Bbb Z_3$ satisfies this property: $\Bbb Z_3 \setminus \{0\} = \{\pm 1\}$ forms a group under multiplication.

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