Fields and Modular Arithmetic

I only recently became aware of the fact that a set of elements like {0, 1, 2, 3, 4, 5, 6} forms a field under modular arithmetic with modulus 7 (the number of elements in the set). The condition is that there are a prime number of elements in the set. Let's firstly define this idea of modular arithmetic:

Figure 1: extract from
https://crypto.stanford.edu/pbc/notes/numbertheory/arith.html

Now what is a field? Here's a definition from http://www.applet-magic.com/field.htm:

A field $F$ consists of a set $S$ of elements and two binary functions $f$ and $g$ defined on $S$ and whose values are in $S.$ Both binary functions are associative, $f(a,f(b,c)$ is equal to $f(f(a,b),c)$ for all $a$, $b$ and $c$ in $S$ and likewise for $g$. The binary function $f$, called addition is necessarily commutative; i.e., $f(a,b)$ is equal to $f(b,a)$ for all $a$ and $b$ in $S$. The other function $g$, called multiplication, is not necessarily commutative. There is an identity for each functions; i.e. there exist $e$ such that $f(e,a)=a$ for all $a$ in $S$ and an $l$ such that $g(l,a)=a$ for all $a$ in $S$. There exist additive inverses for all elements of $S$ and multiplicative inverses for all elements of $S$ except the additive identity. This means that for any element $a$ in $S$ there exist $a$ and $b$ such that $f(a,b)=e$. Such $a$ $b$ is denoted as $-a$. For any element $a$ in $S$ that is not $e$ there exists $a$ and $b$ such that $g(a,b)$ is equal to $l.$ The element $b$ is usually denoted as $a^{-1}$. Furthermore there is distributivity of multiplication with respect to addition; i.e. $g(a,f(b,c))$ is equal to $f((g(a,b),g(a,c))$ for all $a,$, $b$ and $c$ in $S$.

The set of $n$ x $n$ matrices is an example of a field in which multiplication is not commutative. Putting the whole business less formally, we can say that a set of elements form a field if the set is closed under the operations of addition, subtraction, multiplication and division. In other words, these operations when applied to two elements in the set always produce a third element that is also in the set. An example is {0, 1, 2, 3, 4, 5, 6} and it is easy to set up tables for all four operations verifying the set is indeed closed, as it is under all sets with a prime number of elements. Figure 2 shows what happens to the elements of the set under the operation of multiplication by 2:

Figure 2: mapping of elements of mod 7 under multiplication by 2

Figure 3 shows the full multiplication table:

Figure 3: multiplication table for mod 7

Of course, as stated above there is no multiplicative inverse for the additive identity element. In the case of the mod 7, this element is 0 and it means we can't divide by this element just as in regular arithmetic. Figure 4 shows the full division table:

Figure 4: division table for mod 7 with case of 5 divided by 3 highlighted

Figure 5 explains the condition for division to be possible in a modular system:

Figure 4: extract from
https://crypto.stanford.edu/pbc/notes/numbertheory/arith.html