element in field, expressed as multiple of elements? I'm new to the formal concept of a field, although I'm aware of the defining axioms. I've been struggling to think through the following idea:
Let F be a field. If I take an element a in F , is it always the case that I can find b and c in F such that  b•c = a?
Any thoughts would be helpful, thanks :)
Edit:
[Forgot to add this, but I mean b,c different from a]
 A: Yes, that is true$^{(*)}$: simply take $b=a$ and $c=1$, for example. Then $bc = a\cdot1 = a$.
You can make a much stronger claim. For any $a \in F$ and any nonzero $b \in F$, you can find $c \in F$ such that $bc = a$. Namely, let $c = ab^{-1}$, so that $bc = b(ab^{-1}) = a$.
The idea is that fields support division. To give an example in the rationals, say, let $a = 5$ and $b = 7$. Is there a (rational) multiple of $b$ that gives $a$? Yes: let $c = 5\cdot\frac17 = \frac57$. Then you can check that $bc = a$. On the other hand, the integers do not behave this way because they do not form a field. Keeping $a=5$ and $b=7$, you can see that there is no (integer) multiple of $b$ that gives $a$.

(*) You edited your question to require that $b$ and $c$ be different from $a$. The answer is now no: take $a=0$. Then since $b$ and $c$ are different from $a$, they are both nonzero. But then $bc \neq 0$, because of the field axioms.
Suppose your further require that $a$ be nonzero. Then your field would need to have at least three nonzero elements to make $a,b,c$ distinct, so that clearly eliminates $GF(2)$ and $GF(3)$ (the finite fields with $2$ and $3$ elements). Any other field will be big enough.
A: The answer to your question is "yes" but it isn't a particularly informative question.
Just think about the rational numbers for a moment. Suppose $a=2$. Then you have infinitely many solutions to your problem:
$$
2 =   1 \times 2  = 6 \times (1/3)  = (12/5) \times (5/6) = \cdots
$$
You should see how to generalize this for any $a$ in any field.
