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Let n be a positive integer and consider $[a]$ in $Zn$. Prove that $[a]$ is a zero divisor in $Zn$ if and only if it does not have an inverse in $Zn$.

Bad proof.


An element $[a]$ of $Zn$ has a multiplicative inverse in $Zn$ if and only if $(a,n)=1$ in $Zn$ where n is prime this will always be true and the theorem holds.

i start with for all $[a] $s.t $(a,n)$$> 1$ and $[a]$ is not $[0]$ i state that there exists a $[b]$ s.t $[a][b] = [0]$ and $[b] \neq [0]$

from this i attempt to draw the conclusion that for all $[a]$ that satisfied $(a,n)>1$ are not invert-able as $[0]$ is not invert-able. then using the theorem when $(a,n)=1$ then for all of these class's $[a]$ they are invert-able

i feel like i am missing half of my proof because i haven't proved if its not invert-able then it is a $[0]$ divisor. i also haven't proved that $[0]$ is not invert-able.

Perhaps i can say [0][x]=[0] for all [x] thus $($($[0][0]^-1$)$) \neq [1]$ therefor $[0]$ is not invert-able?

Please help :)

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The statement of the theorem is not quite right. For usually a zero-divisor is defined as a non-zero $x$ for which there is $\dots$. Thus $[kn]=[0]$ is non-invertible but is not a zero-divisor. (That's the only exception.) – André Nicolas Jan 31 '13 at 17:38
Are you sure? its the congruence class of [0] i had that as anything of the form pk where p is the prime? – Faust7 Jan 31 '13 at 17:41
Yes, I am sure. Definitions of zero-divisor differ, the one I quoted is the common one. But this is a minor technical point. The main thing to do is to show that if $d\gt 1$ is a common divisor of $a$ and $n$, then $[a]$ is a zero-divisor. That has already been done in an answer. – André Nicolas Jan 31 '13 at 17:47
up vote 2 down vote accepted

$$\,[a]\,\text{ non-invertible in}\,\,\Bbb Z_n\Longleftrightarrow g.c.d.(n,a)=d>1\Longleftrightarrow\,\,\exists\,x,y\in\Bbb Z\,\,s.t.\,\,n=xd\,,\,a=yd\Longleftrightarrow$$

$$ax=xyd=yn=0\pmod n\Longleftrightarrow \,\,[a]\,\,\text{is a zero divisor}$$

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Oh wow that makes it much more obvious i am a little disappointed in myself i just proved something that looked exactly like that :( – Faust7 Jan 31 '13 at 17:37
your proving both directions at once that way aren't you? – Faust7 Jan 31 '13 at 17:44
Yup, I am...but can you see where exactly you must say 2-3 words about $\,d>1\,$ ? – DonAntonio Jan 31 '13 at 17:47
seriously wow though i could barely read that its so concise and i get it now i need to use the definition of the gcd being the set {ax +ny} when d>1 there will exist integers dq1=n and dq2=a then like you did pick q1 to be x and pick q2 to be y. nicely done! – Faust7 Jan 31 '13 at 17:54

Hint $\ $ The map $\rm\:x \to ax\:$ is $1$-$1\!\iff\!$ onto, since $\rm\,\Bbb Z/n\,$ is finite.

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