Question about the definition of a field... Just out of curiosity - when we define a field, why bother mention multiplication, when its nothing more then repeating the same addition operation?
Here's the definition we were taught in calculus for physicists class:

A field F is a non-empty set on which two binary operations are
  defined: an operation which we call addition, and denote by +, and an
  operation which we call multiplication and denote by $\cdot$ (or by
  nothing, as in a  b = ab). The operations on elements of a field
  satisfy nine defining properties, which we list now...

And then of course you have the axioms of field.
 A: For an abstract field, $+$ and $\times$ are just symbols for two binary operations which need not be related in any way except by the distributive requirement(i.e. $a\times(b+c)=(a\times b)+(a \times c) $). We use $+$ and $\times$ because they represent operations in the fields we know and love best, the rational numbers, the real numbers and the complex numbers. You could use $\heartsuit $ and $\clubsuit $ , if you like them better.But, as Arturo pointed out, to think of multiplication as repeated addition in even these fields is dangerous. So, if your fields had elements which were say,sequences, it becomes worse, how do I add something like $(0,1,0,\cdots)$ to itself $(1,1,1\cdots)$ times?
But, this idea of "adding" elements $n$(for a natural integer) times has been thought about before and you might consider reading this to see how different things are in abstract fields.
http://en.wikipedia.org/wiki/Characteristic_(algebra)
A: We need to mention multiplication in the definitions of a field in order to work with more general fields than just the real numbers.  In many cases multiplication has no relationship with repeated addition.
For example you can define a field where the elements are certain kinds of functions.  Where the product of two functions $f$ and $g$ is the pointwise product $f \cdot g (x)=f(x)g(x)$.  This product is in no way related to a sum of multiple copies of $f$. 
