ENT - Algebraic Numbers And Minimal Polynomial


This document goes over the basic definitions that are needed to understand algebraic numbers and the minimal polynomial.

Algebraic numbers are a good stepping stone understanding units in $Z[\sqrt{d}]$ to which you can use Pelle’s equation!

Algebraic Number


  • $x^2 + 4x + 4 = (x+2)(x+2)$

Observe; x = -2 is a root of the polynomial.

  • $x^2 + 1 = (x-i)(x+i)$

Observe; x = i or -i are roots of the polynomial. Things are getting complex!

  • $3x^2 -x -4 = (3x-4)(x+1)$

Observe; x = $\frac{4}{3}$ or -1 are roots of the polynomial. Further, observe that one of the roots is rational.

Definition : Monic Polynomial

A polynomial is monic, if it’s leading coefficient is 1. ie given $f(x)=\sum_{i=0}^n c_ix^i$, f(x) is monic if $c_n =1$.

Definition: Algebraic Number

Given a non-zero polynomial $f(x)=\sum_{i=0}^n c_ix^i$ where $c_i \in \mathbb{Q}$.

Let $\alpha \in \mathbb{C}$, we say that $\alpha$ is an algebraic number if it is a root of $f(x)$. ie $f(\alpha) =0$.

Definition : Algebraic Integer

Given a non-zero monic polynomial $f(x)=\sum_{i=0}^n c_ix^i$ where $c_i \in \mathbb{Z}$.

Let $\alpha \in \mathbb{C}$, we say that $\alpha$ is an algebraic integer if it is a root of $f(x)$. ie $f(\alpha) = 0$.

Note from our last example, that if the polynomial is not monic, then we have roots which are not integers.

Algebraic Number vs Algebraic Integer

  • Algebraic numbers and algebraic integers can both be complex numbers.
  • Algebraic Integers are also Algebraic Numbers because $\mathbb{Z} \subset \mathbb{Q}$
  • There are still numbers, which are not algebraic numbers.

Example 0 - $a \in \mathbb{Q}$ can be an Algebraic number or an Integer.

Consider the polynomial $f(x) = x - a$.

$f(a) = 0$ so a is a root of this polynomial.

Case 1: a is an integer.

Then $f(x)$ will have integer coefficients, hence a is an algebraic integer.

Case 2: a is an rational number but it is not an integer.

Then $f(x)$ will have a rational coefficient, in particular $a$. Hence a is an algebraic number.

Example 1 - Algebraic Integers can be complex

$x^2 + 1 = (x-i)(x+i)$

  • The coefficients of this polynomial are the integers (1,1).
  • The polynomial is non-zero.
  • Therefore $i, -i$ are algebraic integers.

Example 2 - Algebraic Integers can be irrational

$x^2 - 2 = (x-\sqrt{2})(x+\sqrt{2})$

  • The coefficients of this polynomial are the integers (1,-2).
  • The polynomial is non-zero.
  • Therefore $\sqrt{2}, -\sqrt{2}$ are algebraic integers.

Example 3 - $\pi$ is neither

$x^2 - \pi^2 = (x-\pi)(x+\pi)$

  • The last coefficients of this polynomial is not in $\mathbb{Z}$ or $\mathbb{Q}$. Note, $\pi^2 \in \mathbb{R}$ but it is an irrational number.
  • $\pi$ is therefore not an algebraic number, which immediately means that it is not an algebraic integers, as all algebraic integers are algebraic numbers.
  • So what do we call an irrational number that is not an algebraic number? Transcendental!

Example 4 - All surds of the form $m + a\sqrt[n]{k}$ are Algebraic Numbers

Assume $m,a,k \in \mathbb{Q}$

Let’s work backwards.

$x = m +a\sqrt[n]{k}$

$x-m =a\sqrt[n]{k}$

$(x-m)^n = a^nk$

$(x-m)^n -a^nk=0$

The question is now whether $(x-m)^n -a^nk$ has rational/integer coefficients.

Case 1: $k$ is reducible.

If $k$ is the reducible, ie the n’th root of $k$ is rational. Then we can write $d = \sqrt[\leftroot{2}\uproot{2}n]{k}$ and the surd immediately reduces to a rational integer $m + a\times d$, which reduces to Example 0.

Case 2: $k$ is non-reducible.

If k is none reducible, then we have an irrational number. It would be wrong to now say claim that it is an Algebraic integer as this example differs from Example 2. It is more generalised.

Case 2a: $m,a,k \in \mathbb{Z}$

It then follows that $(x-m)^n -a^nk$ will have integer coefficients. The polynomial will also be monic from the binomial theorem. We therefore have an Algebraic Integer.

Case 2b: $m,a,k \in \mathbb{Q} / \mathbb{Z}$

If one of these numbers are rational, but is not an integer. Then it follows that one or more of the coefficients in $(x-m)^n -a^nk$ may be in $\mathbb{Q} / \mathbb{Z}$.

Notice that:

$(x-m)^n = {n \choose 0}x^n m^0 + {n \choose 1}x^{n-1}m^1 + {n \choose 2}x^{n-2}m^2 + \cdots + {n \choose n-1}x^1 m^{n-1} + {n \choose n}x^0 m^n$

If $\forall k \in [0,n-1]$ we have$::{n\choose k}m^k \in \mathbb{Z}$ then we know that all terms except for the constant term is an integer.

Now focusing on the constant term:

If ${n \choose n}x^0 m^n - a^nk \in \mathbb{Z}$ then all of our coefficients are integers.

If either of the checks fail, then we know we immediately have an algebraic.

An example is the minimal polynomial for the golden ration number:

$(x-\frac{1}{2})^2 - \frac{5}{4}$

We only need to check two terms :

  • ${2\choose1} {\frac{1}{2}} = 1 \in \mathbb{Z}$
  • ${2\choose2}({\frac{1}{2}})^2 - \frac{5}{4} = -1 \in \mathbb{Z}$

Although this is an application of the proof, it is quite cumbersome and there should be an easier way to check by using something like the gcd, or make some relationship between the structure of the binomial theorem and $m$.

Silver lining

For degree-2 polynomials, The 2 checks are quite easy!

  • Check1 : $2 \times m \in \mathbb{Z}$
  • Check2: $m^2 - a^2k \in \mathbb{Z}$

As mentioned above, if either of these checks fail, then it is an algebraic number.

Minor note

We did not mention it explicitly. But there is no problem, if k is some fraction $\frac{c}{d}$ where only one of $c$ or $d$ can be reduced.

Observe, that this example is a generalisation of most of the known examples that we are familiar with. To see this, set m or k to be zero for example.

Example 5 - All surds $\frac{m +a\sqrt[n]{k}}{w}$ are algebraic numbers.

First note that: $\frac{m +a\sqrt[n]{k}}{w}$ = $\frac{m}{w} + \frac{a}{w}\sqrt[n]{k}$

Let $r = \frac{m}{w}$ and let h = $\frac{a}{w}$

Our equation now looks like:

$r + h \sqrt[n]{k}$

Our problem now reduces to Example 4

Definition: Minimal Polynomial

Let $\alpha$ be an algebraic number. The minimal polynomial of $\alpha$ is the monic polynomial $f(x)$ with rational coefficients with the smallest possible degree such that $f(\alpha)=0$.

(For polynomials of degree <= 2 you can do it without this long method.) I usually use this to check my answers.

Example 1

Using a surd of the form: $m +a\sqrt[n]{k}$ . We can find the minimal polynomial using the equation derived in Example 4. $(x-m)^n -a^nk$

  • $5+\sqrt2$

The minimal polynomial is $(x-5)^2 -2$