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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 $x$?

2 Answers 2

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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 $m_b(T)$.

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    Dear @Marc, you are right but I never claimed that it was exceptional for a minimal polynomial to be reducible: I just wanted to warn users who are used to minimal polynomials in field extensions, where the minimal polynomial *is* irreducible, that it is no longer the case in general. Also, I thin$k$ that everyone $k$nows about minimal polynomials of matrices but it is not so usual in boo$k$s to put that in the perspective of algebraic elements of an algebra. Anyway, thanks a lot for your pertinent comment.2012-07-10
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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 $\overline{K}$.