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Let $n\in\mathbb{Z}_{\geq2}$. Considering the ring $\mathbb{Z}/n\mathbb{Z}$, we denote $(\mathbb{Z}/n\mathbb{Z})^*$ to be its ring of units. Furthermore, the map $\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z};\,a\mapsto\overline{a}$ is meant to be the canonical projection modulo $n$.

claim 1: $\overline{a}\in\mathbb{Z}/n\mathbb{Z}$ zero-divisor $\iff$ $1<\gcd(a,n)<n$.

I am wondering if claim 1 is true or not. I was able to prove $''\implies''$ using Bézout as follows:

$$\gcd(a,n)=1\stackrel{\text{Bézout}}\iff\exists x,y\in\mathbb{Z}: ax+ny=1\iff ax\equiv 1\mod n \iff \overline{a}\in(\mathbb{Z}/n\mathbb{Z})^*.$$ So in particular $\overline{a}$ is not a zero-divisor.

If claim 1 is true, would it then imply the following proposition?

claim 2: $\overline{a}\in\mathbb{Z}/n\mathbb{Z}$ zero-divisor $\iff \overline{a}\notin(\mathbb{Z}/n\mathbb{Z})^*$

Bernard
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mahlovic
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    Hint: is a proper divisor of $n$ that is not equal to $1$ a zero divisor in $\mathbb{Z}/n\mathbb{Z}$? As to claim 2, do you consider $0$ to be a zero divisor? Some people do, some people do not. If you do not, then claim 2 needs to be tweaked a bit. – Arturo Magidin Mar 12 '21 at 18:40
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    In general you have that for finite rings all elements are units or zero divisors (if you include 0 as a zero divisor). But as to if claim 1 implies claim 2 it does except for the case the $gcd(a,n) = n$ in that case $a=0$ which is not a unit. If you consider $0$ to be a zero divisor you then you should include $n$ as a possible gcd in claim 1 otherwise for claim 2 have a is a zero divisor or a=0 on the left side of the biconditional – MIO Mar 12 '21 at 18:51
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    A brief proof that in any finite ring, a non-zero-divisor is a unit: if $a$ is not a zero divisor, that implies that the function $R \to R, x \mapsto a \cdot x$ is injective. $R$ being finite, that implies that this function is also surjective, so in particular 1 is in the image. And similarly for the function $x \mapsto x \cdot a$. – Daniel Schepler Mar 12 '21 at 18:51

1 Answers1

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Let $R$ be a finite ring. Then every non-zero element of $R$ is either a zero-divisor or a unit, but not both.

Proof: suppose that $a$ is a zero-divisor. Then clearly, $a$ cannot be a unit. For if $ab = 1$, and if we have $c \neq 0$ such that $ca = 0$, then we would have $cab = c1 = c = 0$. This is a contradiction.

On the other hand, suppose $a$ is not a zero-divisor. Then the map $x \mapsto a \cdot x$ is injective. For suppose we have $ax = ay$; then $a(x - y) = 0$; then $x - y = 0$; then $x = y$. Since this map is injective and its domain and codomain are both $R$, which is a finite set, the map must be surjective. Then there must exist $x$ such that $a \cdot x = 1$.

Thus, for $\overline{a} \neq 0$, we have $\overline{a}$ zero-divisor iff $\overline{a} \notin (\mathbb{Z} / n \mathbb{Z})^*$.

Now note that $\overline{a} \in (\mathbb{Z} / n \mathbb{Z})^*$ iff $gcd(a, n) = 1$. Clearly, if we have $\overline{a} \overline{x} = 1$, then $ax = 1 + bn$ for some $b \in \mathbb{Z}$, and thus $ax - bn = 1$, and thus $gcd(a, n) = 1$. And conversely, if $gcd(a, n) = 1$, then we have some $x, b$ such that $ax + bn = 1$. Then $ax = 1 - bn$. Then $\overline{a} \overline{x} = 1$.

Then $\overline{a}$ is a zero-divisor iff ($\overline{a}$ is not a unit and $\overline{a}$ is not zero) iff ($gcd(a, n) \neq 1$ and $gcd(a, n) \neq n$) iff $1 < gcd(a, n) < n$.

Mark Saving
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