The question
Obviously, a field has a $1$ and a $0$ element, the former being neutral regarding addition, the latter being neutral regarding the multiplication operator. Additionally, we know that in the best-known numerical fields, e.g. $\mathbb{F} = \mathbb{R}$, holds $\forall x \in \mathbb{F}: x + x = 2x$.
Does this and similar behavior ($x + x + x = 3x$, $\frac{3}{2} \cdot x = x + \frac{1}{2} \cdot x$, $\dots$) hold for all fields $\mathbb{F}$? Apart from the fact that the distributive property holds, I can't see any connection between the $+$ and $\cdot$ operation in a given field $\mathbb{F}$.
The reason I'm asking
I'm asking this question because I originally thought that this would not hold in every field. For one thing, how would you know which element represents e.g. the $2 \in \mathbb{R}$ in $x + x = 2x$? What would guarantee that such an element even exists?
But looking at this question and its answer, it seems there must be a way to relate the well-known $\frac{1}{2}$ in e.g. $\mathbb{R}$ to some element in any body $\mathbb{F}$. The exact way they've come to the solutions there isn't quite clear, but I'd do it by solving the system of equations $$ X_{ij} = A_{ij} + B_{ij}\\ X_{ji} = A_{ij} - B_{ij}\\ \Updownarrow\\ X_{ij} = A_{ij} + B_{ij}\\ X_{ij} + X_{ji} = A_{ij} + A_{ij} \overset{?}{=} 2 A_{ij} $$
Well, consider a commutative ring $R$ with unity $1$.
For $n\geq 1$, the $n$-fold of $x\in R$ is defined inductively as $$1\cdot x = x,\quad (n+1)\cdot x = n\cdot x + x.$$ Write $nx$ instead of $n\cdot x$.
Moreover, $0x = 0$, since $0x = (x+(-x))x = x^2-x^2=0$.
For $n<0$, the $n$-fold of $x\in R$ is defined as $$n\cdot x = (-n)(-x).$$
Example: $2x = x+x$ and $(-2)x = (-x) + (-x)$.