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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.

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    $\begingroup$ $(-1)\cdot x = (-x)$ for all $x$. $\endgroup$ – Daniel Fischer Oct 9 '15 at 20:06
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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 $$1=1+(-1)+(-1)^2=(-1)^2.$$

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