Invertible or zero divisor in $\mathbb Z_n$

Question

Is it true (for $n \le 10$) that every nonzero element in $\mathbb Z_n$ is either invertible or a zero divisor?

Can anyone please help me. Thank you

Answer 1

This is true in $\mathbb Z_n$ for every $n$.

Consider $a\neq 0$ and look at the set of products $$ P = \left\{\ ab \ \big|\ b\in\mathbb Z_n, b\neq 0\ \right\}. $$ If $0\in P$, then $a$ is a zero divisor. If $1\in P$ then $a$ is invertible.

If $0,1\notin P$ we have $|P|\le n-2$, since there are $(n-1)$ different $b\neq 0$, there must be $b\neq b'$ with $ab=ab'$, thus $$ a(b-b') = ab-ab' = 0, $$ so $a$ is again a zero divisor.

Answer 2

Let $a \in \mathbb Z_n$. Look at $f: \mathbb Z_n \to \mathbb Z_n$ $f(x)=ax$.

If $f$ is one to one, then, it must be onto and hence $f(b)=1$ for some $b$.

If $f$ is not one to one, you can find $b \neq c$ so that $f(b)=f(c)$. Thus $$a(b-c) =0 \,.$$

Answer 3

Let $1 \le a < n$ and let $\mathrm{gcd}(a,n) = d$. By Bézout's identity there exist integers $x,y$ such that $ax+ny=d$.

• If $d=1$ then you have $ax \equiv 1 \pmod{n}$, so $a$ is invertible with inverse $x$.

• If $d>1$ then there exists $0<k<n$ such that $dk=n$, and then $$a(xk) \equiv (ax+ny)k \equiv dk \equiv n \equiv 0 \pmod{n}$$ so $a$ is a zero divisor.