What does it mean to be integral over something I don't understand what it means to say $\mathbb Q[\sqrt d]$ to be integral over $\mathbb Z$?
 A: This terminology comes from commutative algebra. (Really it originates in number theory, but that's beside the point.)

Definition: Let $R$ be a commutative ring with unity and $S \subseteq R$ a subring of $R$. (Here "subring" means $S$ is a commutative ring with the same unity element as $R$.) An element $r$ of $R$ is said to be integral over $S$ if $r$ is a root of a monic polynomial in $S[x]$, i.e., if there exist $n \in \mathbb{N}$ and $s_0,\ldots,s_{n-1} \in S$ such that
  $$
    s_0 + s_1 r + \ldots + s_{n-1}r^{n-1} + r^n = 0.
$$
  The ring $R$ is called integral over $S$ if every element of $R$ is integral over $S$.

In algebraic number theory, one often studies finite extensions $K$ of the rational numbers, such as $K = \mathbb{Q}[\sqrt{d}]$. By convention, one says that an element $a \in K$ is integral (without reference to a base ring) if $a$ is integral over $\mathbb{Z}$. In other words, we say that $a \in K$ is integral if $p(a) = 0$ for some monic $p(x) \in\mathbb{Z}[x]$. As it turns out, the set $\mathfrak{O}_K$ of integral elements of $K$ is a ring (called the "ring of integers of $K$"), and the relationship between $\mathfrak{O}_K$ and $K$ is very much analogous to the relationship between $\mathbb{Z}$ and $\mathbb{Q}$. This motivates the calling the elements of $\mathfrak{O}_K$ the integers of $K$ (or at least thinking about them this way).
Regarding your particular example, the ring $K = \mathbb{Q}[\sqrt d]$ for $d \in\mathbb{Z}$ is not integral over $\mathbb{Z}$, but its subring $\mathbb{Z}[\sqrt d]$ is. This can actually be seen because even $\mathbb{Q}$ is not integral over $\mathbb{Z}$; if $q \in\mathbb{Q}$ is not an integer, then neither is $p(q)$ for any monic polynomial $p(x)$ with integer coefficients, and hence $p(q) \neq 0$ for any such $p(x)$.
On the other hand, every element $r \in \mathbb{Z}[\sqrt d]$ is of the for $r = a+b\sqrt{d}$ for $a,b\in\mathbb{Z}$, and hence satisfies
$$
    (r-a)^2 - b^2d = 0.
$$
Therefore $r$ is a root of the following monic polynomial $p(x) \in \mathbb{Z}[x]$:
$$
    p(x) = (x-a)^2 - b^2d = x^2 - 2ax + (a^2 - b^2d).
$$
It's also worth noting that $\mathbb{Z}[\sqrt d]$ is not always equal to $\mathfrak{O}_K$, instead being a proper subring of $\mathfrak{O}_K$ for certain values of $d$. Specifically, for any squarefree $d \neq 0$ and $K = \mathbb{Q}[\sqrt{d}]$, we have that
$$
    \mathfrak{O}_K = \begin{cases}\mathbb{Z}[\sqrt{d}] &\text{if }d\equiv 1 \pmod{4}\\
                                  \mathbb{Z}\left[\frac{1+\sqrt{d}}{2}\right] &\text{if }d\equiv 2,3 \pmod{4}.
                     \end{cases}
$$
This accounts for all such $d$, since $d$ is assumed to be squarefree (and hence not divisible by $4$).
A: A ring $S$ is an integral overring of ring $R$ iff each element of $S \supset R$ satisfies a monic polynomial with coefficients in $R$.
Now $\mathbb{Q}$ is not an integral overring of $\mathbb{Z}$, nor is the bigger overring $\mathbb{Q}[\sqrt{d}]$ an integral overring of the integers.  For example, $\frac{1}{2}$ does not satisfy a monic polynomial with integer coefficients.
As Bill Dubuque points out (see Comment above), this is a consequence of the Rational Root Theorem,  which indeed implies that if a rational number $r \in \mathbb{Q}$ is integral over $\mathbb{Z}$, then $r \in \mathbb{Z}$.  There are no elements of $\mathbb{Q}$ integral over $\mathbb{Z}$ that are not already in $\mathbb{Z}$.
When this happens for the inclusion of a domain $R$ in its quotient field $F$, we say the domain is integrally closed, meaning that the formation of the quotient field doesn't produce any new integral elements (over the base domain).
However $\mathbb{Q}[\sqrt{d}]$ for $d \in \mathbb{Z}$ that is not a perfect square does create new integral elements (over $\mathbb{Z}$).  In particular, $\sqrt{d}$ is integral over $\mathbb{Z}$ because it satisfies $x^2 - d = 0$, but $\sqrt{d}$ does not belong to $\mathbb{Z}$.
Happily all the elements of overring $S$ that are integral over $R$ will form an overring $\overline{R}$ that we call the integral closure of $R$ in $S$.  So perhaps the Question was prompted by a discussion (elsewhere) about the integral closure of $\mathbb{Z}$ in $\mathbb{Q}[\sqrt{d}]$, which would be a classic problem of algebraic integers.
