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., $\mathbb{C}$)?

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    $\begingroup$ This is in fact only rarely the case, even in the most common examples. Take for instance multiplication by $\frac12$. $\endgroup$ – Tobias Kildetoft Jul 26 '17 at 18:34

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

$$a \cdot 0 = 0$$

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


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.

  • $\begingroup$ So distributivity and the existence of a multiplicative and additive identity are the crucial properties? $\endgroup$ – orome Jul 26 '17 at 18:43
  • $\begingroup$ @raxacoricofallapatorius Correct. Be careful though, what can be considered an "integer" inside a field can be counter-intuitive. $\endgroup$ – orlp Jul 26 '17 at 18:44
  • $\begingroup$ (I'd be really convinced if I could see this illustrated graphically with $\mathbb{C}$.) $\endgroup$ – orome Jul 26 '17 at 18:45

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.

  • $\begingroup$ Can the additional properties of a field that make multiplication always repeated addition be easily stated? $\endgroup$ – orome Jul 26 '17 at 18:47
  • $\begingroup$ @raxacoricofallapatorius Yes, that is if every field element is an "integer" per the definition in my answer. That is, every field element must be able to be written as $0 + 1 + 1 + \ldots$. $\endgroup$ – orlp Jul 26 '17 at 18:48
  • $\begingroup$ @orlp: So if every element of the set can be expressed as repeated addition of the multiplicative identity to the additive identity, but otherwise not? $\endgroup$ – orome Jul 26 '17 at 18:49
  • $\begingroup$ @raxacoricofallapatorius Yes. And this is the case if and only if the number of elements in the field is a prime. $\endgroup$ – Tobias Kildetoft Jul 26 '17 at 18:50
  • $\begingroup$ @TobiasKildetoft: Or infinite? (It's not clear to me why.) $\endgroup$ – orome Jul 26 '17 at 18:51

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