Is $\mathbb{Z}_{18}$ a ring?
A Ring R is a set with 2 binary operations, often denoted as addition and multiplication, that satisfy the following conditions:
1. $a+b=b+a$ for $a,b \in R$ (commutative)
2. $(a+b)+c = a+(b+c)$ for $a,b,c \in R$ (associative)
3. There is an additive identity element 0. $0\in R$ s.t. $a+0=a, \forall a\in R$
4. Every element has an additive inverse. $\forall a \in R, \exists (-a) \in R$ such that $a+(-a)=0$
5. $(ab)c = a(bc)$ for $a,b,c \in R$ (associative)
6. For $a,b,c \in R$ $a(b+c) = ab+ac$ and $(a+b)c = ac+bc$ (distributive property)
My initial answer was that this wasn't a ring for the following reason:
$\mathbb{Z}_{18}$ = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17}.
This is not a ring since $\not\exists(-a)\in R, \forall a\in R$ such that $a+(-a)=0$
Best Answer
$\mathbb{Z}_n$ is always a ring for $n \geq 1$. Given $a \in \mathbb{Z}$, we call $\overline{a}$ the equivalence class of $a$ modulo $n$. It's the set of all integers a multiple of $n$ away from $a$, or the set of all integers equivalent to $a$ mod $n$. So $\overline{3}$ when $n=18$ would be $\{... -33, -15, 3, 21, 39, ...\}$.
$\mathbb{Z}_n$ itself is defined as the set of equivalence classes mod $n$ that partition $\mathbb{Z}$. A common misconception is that the integers from $0$ to $n-1$ are the elements of $\mathbb{Z}_n$. It is actually their equivalence classes: $\mathbb{Z}_n = \{\overline{0}, \overline{1}, ..., \overline{n-1}\}$. This really just pushes the problem back a level, because I haven't defined how we add and multiply these equivalence classes. Luckily, there is a very natural way to do so, and it allows us to quickly show that the elements of $\mathbb{Z}_n$ form a ring.
We define addition and multiplication of equivalence classes by looking at addition and multiplication of representative integers. Given two equivalence classes $A, B \in \mathbb{Z}_n$, choose some $a \in A, b \in B$, and define $A+B := \overline{a+b}$. There are two important observations: first, the LHS of this equation is dealing with equivalence classes; the RHS deals with integers. We use the + symbol in both cases, but they technically refer to different operations. Second, this addition is well-defined because it doesn't matter which representatives we choose. If instead we had chosen different representatives $a+kn$ and $b+ln$ from $A$ and $B$ respectively, we would still have that $A+B = \overline{(a+kn)+(b+ln)}=\overline{(a+b)+(k+l)n}=\overline{a+b}$, where the last equality uses the fact that two numbers are part of the same equivalence class if they are a multiple of $n$ apart.
Similarly, we can define $AB=\overline{ab}$ for any representatives $a \in A, b \in B$, and you can check this definition is well-defined despite arbitrary choice of representatives by a similar argument as above.
It's worth verifying at this point, especially if the idea of equivalence classes are unfamiliar, that these operations match your regular understanding of addition and multiplication mod $n$. For instance, saying $3 + 15 \equiv 0 \pmod{18}$ is equivalent to saying $\overline{3}+\overline{15}=\overline{0}$.
The beautiful result of thinking in these terms is that we immediately recognize that $\mathbb{Z}_n$ is a ring because $\mathbb{Z}$ itself is a ring. For each of the properties defining a ring, as you list in your question, think of operations among equivalence classes in terms of operations in terms of representative integers, then verify that the property is satisfied.
For example, to show that multiplication of equivalence classes is commutative, we see that $AB=BA$ for any $A,B \in \mathbb{Z}_n$ exactly because however we choose representatives $a \in A, b\in B$, commutativity of multiplication in integers guarantees that $ab = ba$. Another example: additive inverses. Given an equivalence class $A$, choose $a \in A$ and recognize that $\overline{-a}$ is the additive inverse of $A$ exactly because $\overline{a}+\overline{-a}=\overline{a + (-a)} = \overline{0}$. Of course, $\overline{0}$ is the additive identity among equivalence classes since its elements are multiples of $n$, which do not change the equivalence classes of any representative integers they are added to.
This may feel like cheating on some level, but what this argument is ultimately saying is that addition and multiplication between equivalence classes is defined in terms of the equivalent operations among integers, so we shouldn't be surprised that they obey similar properties.