# Polynomial root of algebraic number

According to Wikipedia, an algebraic number is any complex number (including real numbers) that is a root of a non-zero polynomial in one variable with rational coefficients.

The polynomial has many roots, how do I know which of the roots does the algebraic number denotes?

Is this where e.g. "isolating interval representation" comes in, that is, I need to somehow bound which root I'm interested in? What other ways there are to identify a given root?

As you could surely tell, I'm very new to this kind of math...

• All the roots are algebraic. Phrased differently, $\alpha \in \mathbb C$ is algebraic iff $\exists \{\lambda_i\}_{i=0}^n\in \mathbb Q$, not all $0$, such that $\sum \lambda_i\alpha^i=0$. – lulu Feb 14 '18 at 18:40
• For $n$-th order polynomial, in general, all roots have the equal footings. However, an algebraic number has a minimal polynomial defining itself. – Ng Chung Tak Feb 14 '18 at 18:45
• Somewhat related: math.stackexchange.com/questions/672795/… – law-of-fives Feb 14 '18 at 18:51

It follows directly from the definition that every root of a non-zero polynomial with rational coefficients is an algebraic number.

• What confuses me is that when I try to visualise it, two real roots are at different distances from the zero, so I though the two algebraic "numbers" means these two distances. So the right way to think about an algebraic number is that it denotes all the distances from zero at the same time? – Faaf Feb 14 '18 at 19:28
• @Faaf They have nothing to do with distances. Take $-\sqrt2$, for instance. Or $i$. Both of them are algebraic, but none of them is a distance. – José Carlos Santos Feb 14 '18 at 21:02
• Yes, I meant real roots. But the point is, that one algebraic number represent all the roots, not just one root, correct? – Faaf Feb 14 '18 at 22:00
• @Faaf Wong. Each root is an algebraic number. For instance, $x^2+1$ has $2$ roots: $i$ and $-i$. Therefore, each one of them is an algebraic number. – José Carlos Santos Feb 14 '18 at 22:06
• I'm very sorry for being slow on this one, but this is where I get confused. Ok, both $i$ and $-i$ are algebraic numbers. But since they are the roots of the same minimal polynomial, how to distinguish between them? My original question was, given $x^2+1$, how do I "talk" about $i$, instead of $-i$? – Faaf Feb 14 '18 at 22:15

If I've understood them correctly,

1. Each algebraic number has a unique minimal polynomial.
2. More than one algebraic number may have the same minimal polynomial. The set of numbers having the same minimal polynomial are the roots of that polynomial.

One method of identifying the boundaries in the complex plane that contain a given root of a minimal polynomial is the Kollins-Crandick algorithm.

Those rectangles provide a means of uniquely identifying which root of the minimal polynomial you are interested in once the rectangles have been refined enough to be disjoint.

Algebraic numbers that are roots of the same minimal polynomial are indistinguishable algebraically: There is always an automorphism of the field of algebraic numbers that interchanges them. However, these automorphisms are not order automorphisms, therefore you can tell algebraic numbers apart by comparing them using the ordering of the rational numbers, if you can induce an ordering on the algebraic numbers from that. To do this, you don't have to explicitly specify for each possible minimal polynomial the ordering of its roots in a way that consistently turns the algebraic numbers into an ordered superfield of the rationals. You can specify a root using rational numbers to specify an interval, or for complex numbers a circle/rectangle, that contains exactly one root of the minimal polynomial. This of course means you have to be able to tell whether such an interval/rectangle/circle contains a single root, [and there are algorithms for that]. Note that the number of roots in an interval is independent from the concrete ordering of the algebraic numbers.