-1 * -1 = 1 in any ring. I have been confusing myself over this one.
How do I prove that $$(-1) \cdot  (-1) = 1$$ in any ring? Where $\cdot$ is the multiplication operation.
That is, the additive inverse of the multiplicative identity multiplied with the additive inverse of the multiplicative identity is the multiplicative identity.
 A: Law of Signs proof: $\rm\,\ (-x)(-y) = (-x)(-y) + x(\overbrace{-y + y}^{\large =\,0}) = (\overbrace{-x+x}^{\large =\,0})(-y) + xy = xy$
Equivalently, evaluate $\rm\:\overline{(-x)(-y) +} \overline{ \underline {x(-y)}} \underline{ +xy_{\phantom{.}}}\ $ in two ways, noting each over/under term $ = 0$.
Said more conceptually, $\rm\:(-x)(-y)\ $ and $\rm\:xy\:$ are both inverses of $\rm\ x(-y)\ $ so they are equal by uniqueness of inverses.
Note that the proof uses only ring laws (most notably the distributive law), so the law of signs holds true in every ring. In fact every nontrivial ring theorem (i.e. one that does not degenerate to a theorem about the underlying additive group or multiplicative monoid), must employ the distributive law, since that is the only law that connects the additive and multiplicative structures that combine to form the ring structure. Without the distributive law a ring would simply be a set with two completely unrelated additive and multiplicative structures. So, in a certain sense, the distributive law is the keystone of the ring structure.
A: Simplify $(-1+1)\cdot(-1+1)$ in two different ways.
A: Hint: $$
0 = (1-1)*(1-1) = 1*1 + 1*-1 + -1*1 + -1 * -1
$$
A: Every ring $R$ has a ring homomorphism $i:\mathbb{Z}\to R$ that comes from $R$ being a $\mathbb{Z}$-module --- that is, an abelian group.  It is determined by $i(1)=1$.  (I've sort of swept what you are trying to prove into these statements.)
Since $(-1)(-1)=1$ in $\mathbb{Z}$, we have $i((-1)(-1))=i(1)$, or expanded $i(-1)i(-1)=1$.  This is $(-1)(-1)=1$ in $R$.
