Is multiplication always repeated addition?

Question

I understand that a field is defined as a set and two operations and some properties for those operations.

As a consequence of those properties is it always the case that the operation conventionally called "multiplication" can be implemented as a repeated application of the operation called "addition", or is that a special property restricted only to some sets (e.g., $\mathbb{C}$)?

Answer 1

Yes, multiplication by an integer* is always repeated addition due to the distributivity of multiplication over addition:

$$a \cdot (b + c) = (a \cdot b) + (a \cdot c)$$

*I define an integer as repeated addition of the multiplicative identity ($1$) to the additive identity ($0$).

Then it's easy to see that

$$a \cdot 0 = 0$$

$$a \cdot (1 + 0) = (a \cdot 1) + (a \cdot 0) = a$$ $$a \cdot (1 + 1 + 0) = (a \cdot 1) + (a \cdot 1) + (a \cdot 0) = a + a$$

Etc...

However a field may contain many elements that aren't integers by this definition, for example in the rationals $\frac{1}{2}$ can not be written as $0 + 1 + 1 + \cdots$. As long as at least one side of the multiplication (thanks to commutativity) is an integer, you can write it as repeated multiplication.

Answer 2

No, this is not true in general, not even in the finite field case.

The problem is that the word "repeated" makes explicit reference to counting, which may not map in any obvious way to the field that we're talking about.

So for example in GF(4) we have four elements, $0, 1, x, y$ such that for all elements $A$ we have $A + A = 0$, but we have a nontrivial multiplication structure $xy = yx = 1.$There is no way to view this as a repeated addition of either $x$ or $y$ to itself, as the condition $x + x = y + y = 0$ means that repeated additions of $x$ to itself or $y$ to itself just live in the sets $\{0, x\}$ and $\{0, y\}$ respectively, never hitting the $1$ element.