Does $x\cdot 0 = 0$ follow from the field axioms alone?

Question

From the field axioms alone, does it follow that $x \cdot 0 = 0$ for all $x$?

All I would like is a statement that it can or cannot be done (hints not necessary). I would like to do it myself; I just don't want to waste my time if it's not true. Then I will post my own answer when I get it.

I have tried to prove it by taking some $y\neq 0$, and then using multiplication's distributivity over addition to argue that: $x\cdot 0 = x \cdot (y + -y) = x\cdot y + x\cdot -y$. However I cannot find a way to show that $x\cdot -y = -(x\cdot y)$ without using what I want to prove in the first place.

Answer 1

Hint: In any ring we have $0=0+0$ and distributivity.

Answer 2

This two equations can be taken from field axioms: $$x.(0+ 0)= x. 0$$ $$x.(0+ 0)= x. 0 + x. 0$$ thus, we have $$x. 0 = x. 0 + x. 0$$ and with this, we have proved that $x. 0 = 0$.