Autoprotolysis Constant of Water
Your textbook is correct, the $K_\mathrm{w}$ expression is as follows:
$$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}]=1.0\times10^{-14}$$

Since for every $\ce{OH-}$ there must be one $\ce{H+}$ ($\ce{H+}$ in water is synonomous with $\ce{H3O+}$) the mol ratio is 1:1, if moles are equal the molarities are equal. I'm going to substitute $x$ in for both
$$K_\mathrm{w}=[x][x]=[x]^2=1.0\times10^{-14}; x= 1.0\times10^{-7}.$$

So the molarity of $[\ce{OH-}]$ and $[\ce{H3O+}]$ are both $1.0\times10^{-7}$ That's a really really low amount, small enough to ignore.

However, a lot of reactions (especially in Organic Chemistry) you'll see a negative ion, for example, $\ce{Cl-}$ pulling the $\ce{H+}$ off of a water. So when $[\ce{OH-}]$ or $[\ce{H3O+}]$ are needed, water can be torn apart to make it.