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

Question

Since $\pi$ and $e$ are transcendental, neither can be the root of a polynomial with rational coefficients. However, it is easy to construct a polynomial transcendental coefficients (with $\pi$ or $e$ as one of it's roots), namely $(x-\pi)$ and $(x-e)$. However, is it possible to construct a finite polynomial (one of whose roots is $\pi$ or $e$) with only algebraic coefficients? (i.e. something such as $y=x^5-(\sqrt 2 -1)x^2+x$ .) A polynomial with only radical coefficients would be the best.

P.S. The polynomial doesn't necessarily need to have BOTH $\pi$ and $e$ as it's roots, just one is interesting enough.

Answer 1

No. There is a theorem that says that the algebraic closure of a field is algebraically closed. In this case it means that the zeroes you can reach with rational coefficients are the same as the zeroes you can reach with algebraic coefficients, although the polynomials you would use may be very different.

Answer 2

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.

Answer 3

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.