Can $\pi$ or $e$ be a root of a polynomial with algebraic coefficients?

Solution 1:

The answer is NO. If there was a polynomial with algebraic coefficients, there would also be a polynomial with rational coefficient (with a larger degree). That's because $\bar{\mathbb Q}$ is algebraically closed.

Solution 2:

Suppose that $\pi$ were the root of a polynomial $f(x)=x^n+a_{n-1}x^{n-1}+\dots+a_0$ with the $a_i$ being algebraic numbers.

Then $F=\mathbb{Q}(a_0,a_1,\dots,a_{n-1})$ is a finite extension of $\mathbb{Q}$ since the $a_i$ are algebraic, and $K=\mathbb{Q}(a_0,a_1,\dots,a_{n-1},\pi)=F(\pi)$ is a finite extension of $F$ since $\pi$ satisfies a polynomial over $F$ of degree $n$, hence by the tower rule $K$ is also a finite extension of $\mathbb{Q}$.

But then $\pi$ must be algebraic over $\mathbb{Q}$, which is a contradiction. The same proof works for $e$ of course.