Do the field axioms imply that $a+a+ ... + a$ ($n$ times) $= na$? In trying to understand the notion of field characteristics, I've come across:
(1) $\Sigma_{i=1}^n a = a\cdot n$
Herstein's Abstract Algebra and other stackexchange posts I've found relating to field characteristics seem to take (1) for granted.
Obviously (1) is clear for the standard definitions of addition and multiplication (...not that I could prove it), but for other definitions of the operations, I have a lot of trouble seeing why this connection between addition and multiplication is true and implied by the Field axioms.
Any points in the right direction would be appreciated!
 A: You can define the multiples of a field (or ring) element $a$ inductively as follows:
$0a=e$ (zero element) and $(n+1)a = na + a$ for each $n\geq 0$. Then you have $na = a+a+\ldots+a$ ($n$ times). I guess this is what is meant.
A: Let me be a bit more formal than perhaps necessary.
Let $R$ be any commutative ring with identity. Then, we can define ring homomorphism $\varphi\colon \mathbb Z\to R$ with $\varphi(1) = 1_R$. We extend it by additivity, $$\varphi(n) = \underbrace{1_R+1_R+\ldots+1_R}_n = n\cdot1_R.$$ When we write $nr$, for some integer $n$ and $r\in R$, what we mean is $\varphi(n)r=r+r+\ldots +r$ (by distributivity). Thus, your identity becomes trivial under this convention.
But, we can even do this for any abelian group! There is no need for ring structure at all. Basically, abelian groups are same as $\mathbb Z$-modules.
Take any abelian group $A$ and define $n\cdot a=a+a+\ldots +a$. You can easily check that this action of $\mathbb Z$ gives structure of module. And conversely, any $\mathbb Z$-module is automatically abelian group.
A: Let's put it this way.  Either it is true or it is meaningless.
Bear in mind that $n$ is a natural number and not an element of the field $F$ and that "$\cdot$" is not the multiplicative binary operation on the field.  
So before we can prove $n\cdot a = \underbrace{a\oplus a\oplus\cdots\oplus a}_{n\text{ times}}$ (where "$\oplus$" is the additive binary operation) , we have to define what $n\cdot a$ is for $n \in \mathbb N$ and $a\in F$ means.
.... and it means $n\cdot a := \underbrace{a\oplus a\oplus\cdots\oplus a}_{n\text{ times}}$.  That is simply the definition.
There are consequences of this definition that can be verified but basically this behaves "nicely".  $n\cdot a + m \cdot a = (a+m)\cdot a$ because $\oplus$ is associative.  And $n\cdot a = a\odot (n\cdot 1_F)$ (where $1_F$ is the multiplicative identity of the field, and "$\odot$" is the multiplicative operation of the field) follows as by the distributive law.
That last result implies that it could be very useful to define $n_F \in F$ as $n_F := n\cdot 1_F = \underbrace{1_F\oplus 1_F\oplus\cdots\oplus 1_F}_{n\text{ times}}$.  With that definition it is true that $a\odot n_F = a\odot ( \underbrace{1_F\oplus 1_F\oplus\cdots\oplus 1_F}_{n\text{ times}})=\underbrace{a\oplus a\oplus\cdots\oplus a}_{n\text{ times}}$ (by distribution).
But there are two very important caveats.  1) For $b \in F$ it doesn't follow that there is any $n \in \mathbb N$ so that $\underbrace{1_F\oplus 1_F\oplus\cdots\oplus 1_F}_{n\text{ times}}$ actually exists.  And 2) it will be possible (especially if $F$ is finite) that $1_n = 1_m$ even if $n \ne m$.  (This will happen if $\underbrace{1_F\oplus 1_F\oplus\cdots\oplus 1_F}_{m-n\text{ times}}= 0_F$ where $0_F$ is the additive identity of the field).
....
Technically speaking:  $\cdot: \mathbb N \times F \rightarrow F$ is a function that maps an ordered pair $(n, a)$ to an element of the field whereas $\odot: F \times F \rightarrow F$ is a function that maps an ordered pair $(g,a)$ of elemments of the filed to an element of the field.
Notice we can extend $\cdot: \mathbb Z \times F \rightarrow F$ by defining $0\cdot a := 0_F$ and for $n < 0$ defining $n\cdot a = |n|\cdot (-a)$.
We can't extend $\cdot: \mathbb Q \times F \rightarrow F$ unless for each element $a \in F$ there would exits a $n \in \mathbb N$ so that $\underbrace{1_F\oplus 1_F\oplus\cdots\oplus 1_F}_{n\text{ times}}=a$.  We have no reason to believe that is true, so we can not define $\frac nm\cdot a = n_F\odot \frac 1m_F \odot a$ where $\underbrace{\frac 1m_F\oplus \frac 1m_F\oplus\cdots\oplus \frac 1m_F}_{m\text{ times}}= 1_F$ as there may not be any such $\frac 1m_F$ element in $F$.
