So I managed to find all of them units being {1,2,4,7,8,11,13,14} and zero-divisors {3,5,6,9,10,12}. I found these values using a table of $\bmod15$ but I wanted to know is there a quicker way or a faster formula I could use for example $$\Bbb Z_{30}$$. Thanks, all suggestions welcome
*corrected from earlier mistakenly gave the zero-divisors and units for $\Bbb Z_{14}$ rather than $\Bbb Z_{15}$
Best Answer
Look at the map $f : \mathbb{Z}_{15} \rightarrow \mathbb{Z}_3 \times \mathbb{Z}_5$. You see that if $x=y \mod 15$, you also have $x=y \mod 3$ and $x=y \mod 5$ so this is a well-defined map, which as you can check is a group morphism.
When is $f(x \mod 15)$ zero ? $x$ has to be both a multiple of $3$ and $5$. $3$ and $5$ being coprime, this means $x$ has to be a multiple of $15$ so $(x \mod 15)=0$. In other words, f is an injective map. Since $15=3\times 5$, the orders of the groups are equal (and finite) so $f$ is in fact an isomorphism.
So $\mathbb{Z}_{15} = \mathbb{Z}_3 \times \mathbb{Z}_5$. This is a more general fact, known as the Chinese Remainder Theorem. You can look it up if you want, but the proof is as above.
Know, it is easier to answer your question :
$x \mod 15$ is a unit in $\mathbb{Z}_{15}$ iff $f(x \mod 15)$ is a unit in $\mathbb{Z}_{3} \times \mathbb{Z}_{5}$, which means that both $x \mod 3$ and $x \mod 5$ are units in $\mathbb{Z}_{3}$ and $\mathbb{Z}_{5}$ respectively. As you may know, or may prove with a simple use of Bézout's identity, $x \mod p$ is a unit in $\mathbb{Z}_p$ (when $p$ is prime) iff it is nonzero, i.e. $p$ does not divide $x$.
$3$ and $5$ being prime, this means that units in $\mathbb{Z}_{15}$ are precisely elements which aren't multiples of either $3$ or $5$, that is elements of $\{1,2,4,7,8,11,13,14\} \mod 15$. You can do exactly the same thing for $\mathbb{Z}_{30}$ by showing $\mathbb{Z}_{30}=\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5$. I'll let you figure it out, and also think about what to do when you have prime factors with exponent $>1$.
For zero-divisors, this is pretty similar : zero-divisors in $\mathbb{Z}_{15} = \mathbb{Z}_3 \times \mathbb{Z}_5$ are elements which are zero divisors in either $\mathbb{Z}_3$ or $\mathbb{Z}_5$ (because if $xy=0$ in $\mathbb{Z}_3$, you have $(x,0)(y,0)=0$, and similarly if $(x,x')(y,y')=0$ then $xy=0$ and the same holds for $\mathbb{Z}_5$). So they are multiples of either $3$ or $5$. This means they are exactly the elements which aren't units : $\{0,3,5,9,10,12\}$.
So zero divisors are exactly non-units. Why is that ? Take an element $x$ in a finite ring $R$. The ring morphism $g : y \mapsto xy$ is either injective, in which case it is an isomorphism (because $R$ and $R$ are "both" finite and have same cardinality...) and thus $x$ is a unit (because by surjectivity there exists $y$ such that $xy=1$), or it isn't injective which means $x$ is a zero-divisor (because there exists nonzero $y$ such that $xy=0$). So elements of a finite ring are either units or zero-divisors. The same holds for rings which are finite-dimensional algebras on their center.