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Whilst reading I came across the strange claim that multiplicative inverses exist for only prime values of $n$ in $Z_n$. I am a little puzzled as contrary to that, I know that additive inverses exist for all values of $n$. What would be an approachable proof of the above result?

Thanks in advance.

JohnK
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    Elements of $Z_n$ will have multiplicative inverses for composite $n$, it just won't be all of the elements. If $n$ is composite then do its divisor's residues have inverses? Why or why not? The best way to approach these problems is to actually experiment with numbers, which is within your power to do if you know what $Z_n$ is. – anon Dec 14 '13 at 20:25

2 Answers2

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By Bezout's theorem, we have $$\gcd(k,n)=1\iff\exists u,v\;|\; uk+vn=1\iff\overline u\cdot\overline k=\overline 1\iff \overline k\;\text{is invertible}. $$

Matthew Conroy
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If $\rm\,n = p\,$ is prime, every $\rm\:0\neq a\in\Bbb Z_p\:$ is invertible, since, by little Fermat $\rm\:1 = a^{p-1} = a (a^{p-2}).$

If $\rm\:n = ab\:$ is composite then $\rm\:a\:$ is a zero-divisor in $\rm\,\Bbb Z_n.\,$ But zero-divisors cannot be invertible: $$\rm\begin{eqnarray} 1&=&\rm\ \color{#0A0}{\bar a\,a},\ \ \ 0\ =\,&\rm\color{#C00}{ab},&\,\rm a,b\ne 0 \\[.2em] \rm \Rightarrow\ \ b &=&\rm (\color{#0A0}{\bar a\,a})b = \bar a&\!\!\!\rm (\color{#C00}{ab})&\rm = 0\ \ \Rightarrow\!\Leftarrow\end{eqnarray}\qquad$$

That zero-divisors are not invertible is a ring-theoretic generalization of the well-known fact that zero cannot be invertible, i.e. we cannot divide by zero in a nontrivial ring (i.e. the quotient does not uniquely exist)

Bill Dubuque
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