Is $\pi$ or $e$ algebraic over $\mathbb R$?

I'm reading some basic introduction on fields and Galois theory.

By definition, let $F$ be an extension field of $K$ -- An element $u$ of $F$ is said to be algebraic over $K$ provided that $u$ is a root of some nonzero polynomial $f \in K[x]$. If $u$ is not a root of any nonzero $f \in K[x]$, $u$ is said to be transcendental over $K$.

Now, this book says: $\pi$ and $e$ are algebraic over $\mathbb R$, while transcendental over $\mathbb Q$.

But is this true? How could a polynomial $f \in R[x]$ has a root as $\pi$ or $e$?

• Try $f(x)=x-\pi$. – vadim123 Mar 6 '15 at 14:46
• $\pi$ and $e$ are real numbers – Jack Yoon Mar 6 '15 at 14:46
• Every element of a field $K$ is algebraic over $K$. – Marm Mar 6 '15 at 18:11

Hint: What about $f(x) = x - e$.
• Or $f(x) = x - \pi$ – SubSevn Mar 6 '15 at 14:48
• (+1) Might as well kill both birds with one stone: $f(x) = (x - e)(x - \pi)$...? :) – Andrew D. Hwang Mar 6 '15 at 15:00