# All non-zero elements in a field have same additive order?

I'm trying to understand the proof of this theorem in my course notes, which states:

Let $K, +, \cdot$ be a field. Then all elements $x \in K \setminus \left\{0\right\}$ have the same additive order.

The proof goes as follows:

If all elements $x \in K \setminus \left\{0\right\}$ have infinite order, then there is nothing to prove. Suppose now that an element $x \in K \setminus \left\{0\right\}$ has finite order $n \in \mathbb{N} \setminus \left\{0\right\}$. We now show that some other arbitrary element $y \in K \setminus \left\{0\right\}$ has also additive order $n$. From the commutative property it follows that $$x \cdot (ny) = (nx) \cdot y = 0 \cdot y = 0$$ so that $ny = x^{-1} (x \cdot (ny)) = 0.$ The additive order of $y$ is thus a divisor of $n$ and hence is finite. By analogy, we can prove that $n$ is a divisor of the order of $y$. This shows that all elements in $K \setminus \left\{0\right\}$ have the same additive order $n$.

My questions are: how exactly does it follow from the equation $ny = x^{-1} (x \cdot (ny)) = 0$ that the additive order of $y$ is a divisor of $n$? Also, how does one explicitly show that $n$ is a divisor of the order of $y$? I'm not seeing the 'analogy' that is being mentioned.

Thank you in advance for any clarification.

The order of $y$ is the smaller $p$ such that $py=0$, divide $n$ by $p$, $n=qp+r$, $ny=(qp+r)y=ry=0$ since $p$ is minimal, $r=0$.
Say for some $s\in F$, the additive order of $s$ is $m$. Then if $k$ is some natural number which has remainder $r$ when divided by $m$, it's easy to show that $ks = rs$. Therefore, $ks = 0$ if and only if $k$ is a multiple of $m$.
Going backwards, if we first observe that $ks=0$, then we can assume $k$ is a multiple of the order of $s$, i.e. the order of $s$ divides $k$.
One can show that $n$ is a divisor of the order of $y$ be repeating the first argument, swapping $x$ and $y$.
More simply: the proof shows $\,nx=0\,\Rightarrow\,ny=0\,$ (and conversely by symmetry). Therefore $\ \{ n>0:\ nx = 0\} = \{n>0:\ ny=0\},\,$ so both sets have the same least element (= order), $\$ i.e. $\$ the order or $x =\,$ the order of $y$