Show that $\mathbb Z/p\mathbb Z$ contains no zero-divisors except $0$

Question

Let $p$ ∈ $\mathbb N$ be a prime number. Show that $\mathbb Z/p\mathbb Z$ contains no zero-divisors except $0$, hence ($\mathbb Z/p\mathbb Z$)$^∗$ = {$1, 2, . . . , p − 1$}, and $\mathbb Z/p\mathbb Z$ is field. Where ($\mathbb Z/p\mathbb Z$)$^∗$ is the set of units.

My attempt:

I showed this by contradiction:

Let $a$ be a zero-divisor in $\mathbb Z/p\mathbb Z$, then there exists $b$ $\not=$ $0$ in $\mathbb Z/p\mathbb Z$ such that $a.b = 0$ mod $p$.

We know that $\mathbb Z/p\mathbb Z$ = {$0,1,2, ... , p-1$}.

Without loss of generality, let $a = 1$, and $b = p-1$. Then $a.b = 1.(p-1) = p-1 = 0$ and hence $p=1$, which contradicts the fact that $p$ is a prime number since $1$ is not prime. And hence $\mathbb Z/p\mathbb Z$ contains no zero-divisors except $0$.

Is my proof correct? Do you have any other suggestions or alternatives? And how do I show the other parts of the problem?

For instance, I know that an element $a$ is said to be a unit if there is $b$ such that $a.b = 1$ and so $b$ is the inverse of $a$.

And a field ($F,0,1,.+$) is a commutative ring where for all $x$ in $F$ {$0$}, there exists $x^{-1}$ such that $x.x^{-1} = 1 = x^{-1}.x$

Answer 1

Take $a \neq 0 \in \mathbb{Z} / p\mathbb{Z}$. Then $\gcd (a,p) = 1$, since $p$ is prime and $0 < a < p$. By Bezout's lemma, there exist $x,y \in \mathbb{Z} / p\mathbb{Z}$ such that $$ ax + py = 1 $$ Since $p = 0$ in $\mathbb{Z} / p\mathbb{Z}$, this can be simplified to $$ ax = 1 $$ Now suppose that $ab = 0$. Then $$ 0 = (0)x = (ab)x = b(ax) = b(1) = b $$ Thus, we have shown that if $a \neq 0$, then $ab = 0$ implies $b = 0$. This is just the statement that $\mathbb{Z} / p\mathbb{Z}$ has no $0$ divisors.

Answer 2

''Let a be a zero-divisor in Z/pZ, then there exists b ≠ 0 in Z/pZ such that a.b=0 mod p.

We know that Z/pZ = {0,1,2,...,p−1}.

Without loss of generality, let a=1, and b=p−1''

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Technically, zero is not considered as a zero divisor in a ring.

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How can you assume that $a=1$? It's the unit element of the ring with $b\cdot 1 = b$ for all $b$ and so cannot be a zero-divisor.

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Proof:

Suppose $ab=0$ for $b\ne 0$. Then $ab\equiv 0\mod p$ and so $p$ divides the product $ab$. Since $p$ is a prime, it follows that $p$ divides a factor. So $p\mid a$ or $p\mid b$. But $b\ne 0$ and so $p\mid a$, i.e., $a\equiv 0\mod p$, i.e., $a=0$ in ${\Bbb Z}_p$.

Answer 3

If $n,m$ are integers and $p$ is a prime number then: $$p\mid nm\implies p\mid n\text{ or } p\mid m$$

In terms of modularity: $$[nm\equiv 0\mod p]\implies [n\equiv0\mod p\text{ or } m\equiv\mod p]$$

In terms of ring $\mathbb Z/p\mathbb Z$:$$\overline n\overline m=0\implies \overline n=0\text{ or }\overline m=0$$where $\overline n:=n+p\mathbb Z\in\mathbb Z/\mathbb p\mathbb Z$.

This makes evident that the ring has no zero-divisors.

Answer 4

I suggest you to read the definition of the ring $\mathbb{Z}/p\mathbb{Z}$. A ring comes with a set of elements and two binary operations satisfying some conditions. I was asking you what does the set $\{0,1,2,\cdots,p-1\}$ mean how you are defining multiplication of two elements in $\mathbb{Z}/p\mathbb{Z}$. One way to define $\mathbb{Z}/p\mathbb{Z}$ is, as a set, the collection $\{0,1,2,\cdots,p-1\}$ where $0,1,2,\cdots, p-1$ have their standard meaning as natural numbers but multiplication here is defined as $a\circ b=ab\mod p$.

An element $a\in R$ is called a zero divisor if $a\circ b=0$ for some $b\in R $ implies $b=0$. An element $a\in R$ is called a unit if there exists an element $b\in R$ such that $a\circ b=b\circ a=1$. Even if you can prove that no non zero element of $\mathbb{Z}/p\mathbb{Z}$ is a zero divisor, that does not prove that every non zero element of $\mathbb{Z}/p\mathbb{Z}$ is a unit.

Let $a\in \mathbb{Z}/p\mathbb{Z}$ be a non zero element such that $a\circ b=0$ for some non zero element $b\in \mathbb{Z}/p\mathbb{Z}$. By definintion, $a\circ b=ab\mod p$. Thus, we have $0=ab\mod p$; that is $p\mid ab$. As $p$ is a prime number, the condition $p\mid ab$ implies $p\mid a $ or $p\mid b$ (the notation $a\mid b$ means $a$ divides $b$). So, we see that $p\mid a$ or $p\mid b$ which means $a=0\in \mathbb{Z}/p\mathbb{Z}$ or $b=0\in \mathbb{Z}/p\mathbb{Z}$; a contradiction to the assumption that $a\neq 0\in \mathbb{Z}/p\mathbb{Z}$ and $b\neq 0\in \mathbb{Z}/p\mathbb{Z}$. Thus, if $a\neq 0$, there can be no non zero element $b\in \mathbb{Z}/p\mathbb{Z}$ such that $a\circ b$. Thus, no non zero element of $\mathbb{Z}/p\mathbb{Z}$ is a zero divisors.