Sachchidanand Prasad

10-05-2023

Problem: A number $\alpha \in \mathbb{C}$ is said to be algebraic if there exist a polynomial $p(x)\in \mathbb{Z} [x]$ such that $p(\alpha )=0$. Show that the following numbers are algebraic.

  1. $\cos \left( \frac{2\pi}{n} \right) $, $n\in \mathbb{N} $
  2. $5^{\frac{1}{7}} + 7^{\frac{1}{5}}$
  3. $\sqrt{2} + \sqrt[7]{7} $
  4. $\iota \sqrt[5]{5} + \sqrt[3]{3} + \sqrt[7]{7} + \sqrt[11]{11}$

Solution: We recall that the sum and product of algebraic numbers is again an algebraic number. For a proof we observe the following:

If $K$ is a finite extension of $F$ and $a\in K$, then $a$ is algebraic over $F$.

Proof: Since $K$ is a finite extension of $F$, which implies $[K:F]<\infty $, say $[K:F]=n$. Since, $a\in K$, and the degree of the extension is $n$, the elements $1,a,a^2, \cdots ,a^n$ must be linearly dependent. So, there exists $\alpha _i\in F,~i=0,1,\cdots ,n$ with not all zero, such that \[ \alpha _0 + \sum_{i=1} ^n \alpha _i a^i = 0. \] Take $f(x) = \sum_{i=0} ^n \alpha _ix^i\in F[x]$ We have $f(a) = 0$, thus $\alpha $ is an algebraic number over $F$.

The next result is also useful.

Let $a,b\in F$, then \[ [F(a,b):F(a)] \leq [F(b):F]. \]

Using the above two results, we note that if $a$ and $b$ be two algebraic numbers over $F$, then $[F(a):F] \lt \infty $ and $[F(b):F]\lt\infty $, thus, \[ [F(a,b):F] = [F(a,b):F(a)][F(a):F] \leq [F(b):F][F(a):F] \lt \infty , \] and hence using Result 1, we conclude that all the elements of $F(a,b)$ is algebraic. In particular, $a+b,~a\cdot b$ are algebraic.

Now we will show that each of the given numbers are algebraic.

  • $\cos \left( \frac{2\pi}{n} \right) $. Note that $x^{n}+1 $ has roots $\cos \left( \frac{2\pi}{n} \right) \pm \sin \left( \frac{2\pi}{n} \right) $. Therefore, these two are algebraic and hence there sum is also algebraic. Therefore, $\cos \left( \frac{2\pi}{n} \right) $ is algebraic.
  • Note that the numbers $5^{\frac{1}{7}}$ and $7^{\frac{1}{5}}$ are algebraic as the satisfy the polynomials $x^7 - 5$ and $x^5 - 7$ respectively. Therefore, the sum will also be an algebraic number.
  • Note that $\sqrt{2} $ satisfies $x^2 - 2$ and $\sqrt[7]{7} $ satisfies $x^7 - 7$. Therefore, each of the numbers is algebraic and hence the sum.
  • Here also, each of the numbers is algebraic and therefore the given number is algebraic.