Additive inverse of multiplicative identity, must it be it's own multiplicative inverse?

Ok another probably very basic algebra question.

With real numbers and integers and complex numbers, one is used to $(-1) \cdot (-1) = 1$, i.e. the additive inverse of the multiplicative identity is it's own multiplicative inverse. Does this have to be the case for fields or does it just happen to be for $\mathbb{Z,R,C}$?

Distributivity gives: $(0-1)(0-1) = 0^2 +(-1)0+ 0(-1) + (-1)(-1)$ but I cant get from there to "$1$" in any way.

• $(-1)\cdot x = (-x)$ for all $x$. – Daniel Fischer Oct 9 '15 at 20:06

Yes, this happens in any ring. Here's one way to prove it. First, prove that $0\cdot x=0$ for all $x$: $$0\cdot x=(0+0)\cdot x=0\cdot x+0\cdot x,$$ so subtracting $0\cdot x$ from both sides we find $0=0\cdot x$. Now apply this with $x=0$ as follows: $$0=0\cdot 0=(1+(-1))(1+(-1))=1\cdot 1+1\cdot (-1)+(-1)\cdot 1+(-1)\cdot(-1).$$
The right-hand side simplifies to $$0=1+(-1)+(-1)+(-1)^2=(-1)+(-1)^2.$$
Adding $1$ to both sides, we get $$1=1+(-1)+(-1)^2=(-1)^2.$$