Why a root for an integer polynomial in definition of transcendental number?

Question

I've read many definitions for a transcendental number and some of them say that a transcendental number is a number that is not the root of any integer polynomial, while other say is a number that is not the root of any rational polynomial.

My question is why such restrictions? Can't we just give algebraic polynomial instead?

Answer 1

The set of all roots of integer polynomials (let's call it $S_{\mathbb Z}$) and the set of all roots of rational polynomials (let's call it $S_{\mathbb Q}$) are identical.

This can easily be shown. The fact that $S_{\mathbb Z}\subseteq S_{\mathbb Q}$ is obvious, since any integer polynomial is also an integer polynomial.

The inverse is not obvious, but not hard to show.

Let $x\in S_{\mathbb Q}$. Then, there exists some rational polynomial $p$ such that $p(x)=0$. Let $a_0,\dots a_n$ be the coefficients of $p$. Further, for each $i$, let $a_i=\frac{n_i}{m_i}$ where $n_i, m_i$ are coprime integers.

Now, let $L$ be a common multiple of all $m_i$ (could be the least common multiple of $m_0,\dots m_n$).

Then, $L\cdot p$ is an integer polynomial, and $(L\cdot p)(x)=L\cdot p(x)=L\cdot 0 = 0$, which means $x\in S_{\mathbb Z}$, which means $$S_{\mathbb Q}\subseteq S_{\mathbb Z}.$$

Since $S_{\mathbb Q}\subseteq S_{\mathbb Z}$ and $S_{\mathbb Z}\subseteq S_{\mathbb Q}$, we conclude $S_{\mathbb Q}= S_{\mathbb Z}$

Answer 2

COMMENT.- If $\displaystyle f(x)=\sum_{i=0}^n\frac{n_i}{d_i}x^i$ then $\displaystyle Mf(x)=\sum_{i=0}^nA_ix^i$ where $M$ is the least common denominator of the $d_i's$ so $f(x_0)=0$ only when $\displaystyle\sum_{i=0}^nA_ix_0^i=0$.

A trascendental number is simply a number that is not algebraic and the set of all the algebraic numbers is a field which contains very ugly numbers. For example, take an arbitrary rational function $F(x,y,z)= \dfrac{P(x,y,z)}{Q(x,y,z)}$ where $P$ and $Q$ are polynomials of $\mathbb Q[x,y,z]$. Then for $\alpha,\beta,\gamma$ arbitrary algebraic numbers, the number $F(\alpha,\beta,\gamma)=\delta$, which can be extremely ugly, is an algebraic number so it exists a polynomial one of its roots is $\delta$ (try to find the polynomial corresponding to $\delta=\sqrt[3]2+\sqrt[5]3$ and wonder about the difficulty for another more complicated $\delta$.... and this still stand if you take $F$ with a finite number of variables instead of only three).

Despite this, it is proved that the field of all the algebraic numbers is countable while the set of trascendentals is not countable..

Answer 3

It seems to me that your question is whether allowing the polynomials to have algebraic (instead of rational or integer) coefficients would allow certain transcendental numbers to occur as roots. But the answer is no, every polynomial with algebraic coefficients has only algebraic roots.

Here is one way to see this.

Let $p(x) = \alpha_n x^n + \ldots + \alpha_1 x + \alpha_0$ be a polynomial where the $\alpha_i$ are algebraic (possibly irrational). Let $K = \mathbb{Q}(\alpha_0, \ldots, \alpha_n)$. Then $K/\mathbb{Q}$ is a finite extension (you can view it as a tower of extensions $K_{i-1} \subseteq K_{i-1}(\alpha_i)$ each of degree $\leq d_i$ where $d_i$ is the degree of the minimal polynomial of $\alpha_i$ over $\mathbb{Q}$, so its degree is $\leq d_0\cdots d_n$). Then let $G = Gal(\bar{K}/\mathbb{Q})$ be the Galois group of the normal closure of $K$ over $\mathbb{Q}$ (which is also a finite extension).

Now define the polynomial $$Q(x) = \prod\limits_{\sigma \in G} p^{\sigma}(x)$$ where $p^{\sigma}(x) = \sigma(\alpha_n)x^n + \ldots + \sigma(\alpha_1)x + \sigma(\alpha_0)$ for all $\sigma \in Gal(\bar{K}/\mathbb{Q})$.

It is easy to see that the coefficients of $Q$ are invariant under each element of $G$, so they must lie in $\mathbb{Q}$. Moreover, $p$ occurs as a factor of $Q$ so all its roots are also roots of $Q$. Therefore, they are algebraic.