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Is there a Set of integer modulo m, Zm, where m is composite and the set is a field, (all of its elements having a multiplicative inverse) ?

I have heard that the set of integer modulo 4 cannot be a field. Why is that?

Asaf Karagila
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user3467433
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    if $ab=n$ with $0<a,b<n$, then suppose $ac\equiv 1$. What happens when we multiply both sides by $b$? – David P Apr 03 '14 at 01:09

1 Answers1

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Hint $\ $ If $\ m = ab\ $ is composite then mod $\,ab\!:\ ab\equiv 0\,$ but $\,a,b\not\equiv 0.\,$ If $\,a^{-1}\,$ existed then multiplying through by it we obtain that $\,0\equiv a^{-1}(ab) \equiv (a^{-1}a)b \equiv b,\,$ contradiction.

Alternatively, $ $ recall, by Bezout, that: $\ a\,$ is invertible mod $\,m\iff a\,$ is coprime to $\,m.\,$ Hence, if $\,m\,$ is composite then there exist noninvertibles besides $0,\,$ e.g. any proper divisor of $\,m.$

Bill Dubuque
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