How do we define $\frac{\mathbb{Z}}{p\mathbb{Z}}$ for any $p\in \mathbb{N}$?

Question

Can someone explain in simple terms, how $\mathbb{Z}/p\mathbb{Z}$ is defined as? Is it the same as $\mathbb{Z}_p$ or 'integers mod p'? If not what is the difference? Please keep it simple.

Answer 1

You probably mean the group "$\mathbb{Z}/p \mathbb{Z}, +$", which is indeed the group of integers modulo $p$ (for a $p \in \mathbb{N}_0$). This notation comes from the fact that it is the quotient group of the group of integers $\mathbb{Z}$ (i.e. $\{0,1,2,3...\}, +$) by the normal subgroup $n \mathbb{Z}$ of integer multiples of $n$ (i.e. $\{0,n,2n,3n,...\}, +$).

Since $p$ is not necessarily a prime number it is best to use $n$ instead, to avoid confusion.

I would not use $\mathbb{Z}_p$, as this notation is often used to designate the ring of p-adic integers (for a prime number $p$).

Answer 2

$\newcommand{\SES}[1]{\mathbb Z /#1 \mathbb Z }$ Well, there are many equivalent definitions, one classical definition is to take the integers over the congruence relation $a\sim b \iff \exists n\in \mathbb Z, a-b=pn$
Or, simply, $a\sim b$ if and only if their difference $a-b$ is divisible by p.
this congruence relation divides $\mathbb Z$ into $p$ equivalence classes, where an equivalence class is a set all whose elements are equivalent under $\sim$.
The equivalence classes are $\{...-3p,-2p,-p,0,p,2p,3p...\},\{...-3p+1,-2p+1,-p+1,1,p+1,2p+1,3p+1...\}...\{...-3p+p-1,-2p+p-1,-p+p-1,p-1,2p+p-1,39+p-1...\}$
Now I will use a concrete example, $p=3$: our equivalence relation, $a\sim b$ is just if the difference of $a$ and $b$ is divisible by $3$ or simply if they are equivalent modulo $3$. the equivalence classes formed are $\{...-6,-3,0,3,6\},\{...-5,-2,1,4,7...\},\{...-4,-1,2,5,8...\}$
The construct created is $\SES p$ where each element is an equivalence class.
It is provable that these classes respect multiplication and addition, i.e. we can interchange elements of classes and still have the result.
For example, $4+2=6$ if we think of this in terms of $\SES 2$ we get $0+0=0$, in terms of $\SES 3$, we get $1+2=0$, which seems not to make sense, but then we remember that in $\SES 3$, $3$ and $0$ are interchangeable. Note: We denote it by $\SES p$ because $p\mathbb Z$ is the set of integers dividing $p$ and a normal subgroup(or subring) of $\mathbb Z$. and we use the symbol of division as we divide $\mathbb Z$ using(or "by") $p\mathbb Z$.