Difference between $\operatorname{Spec }\mathbb Z$ and $\operatorname{Spec }\mathbb Z\left[\frac1n\right]$ 
What is the difference between the affine schemes, $\operatorname{Spec }\mathbb Z$ and $\operatorname{Spec }\mathbb Z\left[\frac1n\right]$ ?

I know the spectrum of a ring is the set of its prime ideals, but when the ring is enlarged does its spectrum shrink down (always)?
Can we say if $X$ is a $\operatorname{Spec }\mathbb Z[\frac1n]$ scheme then it is also a $Spec\mathbb Z$ scheme ?
 A: Expanding a ring doesn't always mean shrinking the associated some schemes. However, expanding a ring by specifically adding inverses of elements that didn't already have inverses will shrink the Spec space. The inclusion $\Bbb Z\to\Bbb Z[1/n]$ will induce an inclusion $\operatorname{Spec}\Bbb Z[1/n]\subseteq \operatorname{Spec}\Bbb Z$.
In contrast, $\operatorname{Spec}\Bbb Z$ is well-known and easy to describe, while $\operatorname{Spec}\Bbb Z[X]$ is a complicated beast, and I don't think we are anywhere near a description of $\operatorname{Spec}\Bbb Z[X,Y]$.
If you have two rings $R, S$ and a ring homomorphism $R\to S$, then that induces a morphism of affine schemes $\operatorname{Spec}S\to\operatorname{Spec}R$. And if $X$ is a $\operatorname{Spec}S$ scheme, that just means there is a morphism $X\to\operatorname{Spec}S$. Composing that with the morphism $\operatorname{Spec}S\to\operatorname{Spec}R$ gives you a morphism $X\to \operatorname{Spec}R$, which is to say that $X$ is indeed a scheme over  $\operatorname{Spec}R$. This is true for $R = \Bbb Z$ and $S = \Bbb Z[1/n]$ in particular.
There is a morphism from any scheme $X$ to $\operatorname{Spec}\Bbb Z$. It takes each point in $X$ to the (prime ideal generated by the) characteristic of the residue field over that point. So any scheme is a $\Bbb Z$-scheme.
