Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x$?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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 $m_b(T)$.

share|improve this answer
    
Nice complete answer. However I have one remark: in $M_n(K)$ it is not at all unusual for a minimal polynomial to be reducible (nothing as exceptional as nilpotence needed): whenever a matrix has at least two distinct eigenvalues, its minimal polynomial will be reducible (also companion matrices show that all monic polynomials of degree $n$ are minimal polynomials). –  Marc van Leeuwen Jul 10 '12 at 8:33
    
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 think that everyone knows about minimal polynomials of matrices but it is not so usual in books to put that in the perspective of algebraic elements of an algebra. Anyway, thanks a lot for your pertinent comment. –  Georges Elencwajg Jul 10 '12 at 8:45

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 $\overline{K}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.