1.4 Algebraic numbers and number fields
An algebraic number is a complex number which is algebraic over \(\mathbb {Q}\). Meaning, it is a root of a polynomial \(f(x) \in \mathbb {Q}[x]\).
If \(\alpha \) is an algebraic number, then it has a minimal polynomial which we denoted by \(m_{\alpha ,\mathbb {Q}}\) in Proposition 1.3.10. Since from now on we will be working over \(\mathbb {Q}\), we will make the notational convention that \(m_{\alpha }:=m_{\alpha ,\mathbb {Q}}\)
If \(\alpha \) is an algebraic number and \(m_\alpha (x)\) is its minimal polynomial, then the set of roots of \(m_\alpha (x)\) in \(\mathbb {C}\) are called the conjugates of \(\alpha \).
\(0,47,\sqrt{2},\sqrt{-1},3/4, \sqrt[10]{5}\) are all algebraic numbers
If \(d\) is a square-free integer, \(\sqrt{d}\) is algebraic and its conjugate root is \(-\sqrt{d}\).
\(\alpha =\sqrt{2+\sqrt{2}}\) is an algebraic number. Lets prove it by finding its minimal polynomial:
\begin{align} & \alpha =\sqrt{2+\sqrt{2}}\\ & \alpha ^2-2= \sqrt{2}\\ & (\alpha ^2-2)^2=2\\ & \alpha ^4-4\alpha ^2+2=0 \end{align}So we see that \(\alpha \) satisfies \(x^4-4x^2+2\), which means its algebraic.
Is this the minimal polynomial? well lets check. By Eisensteins Criterion 1.2.15 with \(p=2\) we see that this is irreducible. Moreover its monic, so it must be the minimal polynomial.
Now, what are the conjugate roots of \(\alpha \)? If you work backwards through the computation above you’ll see that \(\pm \sqrt{2 \pm \sqrt{2}}\) are all the roots of \(m_\alpha \) and thus are the conjugates of \(\alpha \).
\(\pi =3.1415\dots \) is NOT an algebraic number (although this isn’t easy to prove).
A Number field is a subfield of \(\mathbb {C}\) of finite degree over \(\mathbb {Q}\).
Let \(d\) be a square-free integer then \(\mathbb {Q}(\sqrt{d})=\{ a+b\sqrt{d}| a,b \in \mathbb {Q}\} \) is a number field of degree \(2\) over \(\mathbb {Q}\).
By the Fundamental Theorem of algebra, if we take any polynomial \(f(x) \in \mathbb {Q}[x]\), then it has a root \(\alpha \) in \(\mathbb {C}\). Therefore \(\mathbb {Q}(\alpha )\) is a number field. This gives us a great supply of number fields. Similarly, using Theorem 1.3.20 we can take any finite set \(\{ \alpha _1,\dots ,\alpha _n\} \) of algebraic numbers and then \(\mathbb {Q}(\alpha _1,\dots ,\alpha _n)\) will again be a number field.
Let \(\overline{\mathbb {Q}}\) denote the set of all algebraic numbers. Prove that \(\overline{\mathbb {Q}}\) is actually a field. Is it a number field? explain your answer.
Let \(\alpha \) be an algebraic number with minimal polynomial \(m_\alpha (x)=\sum _i a_i x^i\). Then using this, write down the minimal of \(1/\alpha \).