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I need help with this:

If a is an element of $\mathbb Z_n$ and $\gcd(a,n) > 1$ , then a is not invertible.

First you show that if $a$ is an element of $\mathbb Z_n$ and $\gcd(a,n)>1$, then there is an element $b$ of $\mathbb Z_n$ and ($b$ is not equal to the zero element) from which $ab = 0$.

The second part I cant get. i.e. show if $b$ is not the zero element and $ab=0$ then $a$ is not invertible.

Bill Dubuque
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Alex77
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    Suppose to the contrary that $ab\equiv 1\pmod{n}$. Then $ab=1+kn$ for some integer $k$. If $d$ divides $a$ and $n$, then $d$ divides $1$. – André Nicolas Mar 24 '15 at 15:55

3 Answers3

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It is the special case $\,\color{0a0}{b = 1}\,$ of the following solvability criterion.

$\qquad \exists\, x\!:\ ax\equiv b\pmod{\! n}\iff \exists\, x,y\!:\ ax+ny = b\color{#c00}{\overset{\rm Bezout\!\!\!\!}\Longleftarrow}\!\color{}{\Longrightarrow}\, \gcd(a,n)\mid b $

Remark $\ $ The OP only requires the simpler black direction $\,(\Longrightarrow),\,$ not the deeper reverse arrow $(\color{#c00}\Longleftarrow)$ using $\rm\color{#c00}{Bezout's}\,$ identity for the gcd.

A more general ring-theoretic understanding comes from viewing it as a special case of the fact that zero-divisors are never invertible (in nonzero-rings).

Bill Dubuque
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Assume $ab = 0$ and at the same time that $a$ is invertible. That means that there exists an $a^{-1}$ such that $a^{-1}a = 1$ But then we have $$ b = 1\cdot b = a^{-1}ab = a^{-1}\cdot 0 = 0 $$ so there cannot be a $b \neq 0$ such that $ab = 0$. That means that if there is such a $b$, then $a$ cannot be invertible.

Arthur
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    i.e. a zero-divisor is not invertible (in a nonzero ring). This is a generalization of the more widely known fact that zero is not invertible in a field or domain. – Bill Dubuque Mar 24 '15 at 15:42
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Suppose gcd(a,n)=1 We can write this as: ax+ny=1 And class of ax+ny is equal to class of 1 cl(ax+ny)=cl(1) cl(ax)+cl(ny)=cl(1) We know about class of zero i.e (cl(0)={0+ny\y belongs to Z}) From this we can write: cl(ax)+cl(0)=cl(1) cl(ax)=cl(1) cl(a).cl(x)=cl(1) cl(a)^-1=cl(x) So cl(a) is Invertible

Conversely, Suppose that cl(a) is Invertible Then there exist an element cl(x) of Zn other than cl(0) such that cl(a).cl(x)=cl(1) cl(ax)=cl(1) ax=1 mod(n) n|(ax-1) ax-1=ny (y belongs to Integers) ax-ny=1 ax+n(-y)=1 gcd(a,n)=1

Ravians Fun
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