Quibble with terminology Proposition 5.15 on page 63 in Atiyah-Macdonald goes as follows:
Let $A \subset B$ be integral domains, $A$ integrally closed, and let $x \in B$ be integral over an ideal $ \mathfrak a$ of $A$. Then $x$ is algebraic over the field of fractions $K$ of $A$ and if its minimal polynomial over $K$ is $t^n + a_1 t^{n-1} + \dots + a_n$ then $a_1, \dots , a_n$ lie in $r(\mathfrak a)$.
According to Wikipedia, an algebraic element is defined as follows: 
"If $L$ is a field extension of $K$, then an element a of $L$ is called an algebraic element over $K$, or just algebraic over $K$, if there exists some non-zero polynomial $g(x)$ with coefficients in $K$ such that $g(a)=0$."
1)Here $L$ is a field. Is it ok to call an element algebraic even if $L=B$ is just an integral domain? "algebraic" is never defined in AM.
2)Also, in the proof following the theorem: what is a conjugate of $x$?
 A: Question 1)
 If $K$ is a field and $B$ is a completely arbitrary  $K$-algebra, it makes perfectly good sense to say that an element $b\in B$ is algebraic over $K$.  
This means  that the ideal $I_b\subset   K[T]$ consisting of the $P(T)\in K[T]$ such that $P(b)=0$ is non-zero.
The monic  generator $m_b(T)$ of $I_b$ is then called the minimal polynomial of $b$ .
Beware however that, in contrast to the case where $B$ is a field, this minimal polynomial needn't be irreducible over $K$.
It may happen that all elements of $B$ are algebraic over $K$: the algebra $B$ is then called (a bit funnily) an algebraic algebra.
The best-known example of such an algebraic algebra is the algebra $M_n(K)$ of matrices over $K$.
Non-zero nilpotent matrices then have as minimal polynomials  a power $T^k\; (2\leq k\leq n)$,  which is thus an example of a non irreducible minimal polynomial.
Question 2)
The conjugates of $b$ are the  roots of $m_b(T)$ in some field extension of $K$ containing a splitting field of $m_b(T)$.
A: 1) algebraic just means integral, and integral applies to arbitrary ring extensions (even ring homomorphisms). But in this case, you should imagine that $x$ is algebraic w.r.t to the field extension $Q(A) \subseteq Q(B)$.
2) If $x$ is algebraic over a field $K$, its conjugates are the zeroes of its minimal polynomial. Equivalently, these are the images $\sigma(x)$, where $\sigma$ runs through all $K$-automorphisms of some algebraic closure $\overline{K}$.
