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

Question

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 $e$?

$\pi$ and $e$ are real numbers – Jack Yoon Mar 06 '15 at 14:46 — Jack Yoon, Mar 06 '15 at 14:46
Every element of a field $K$ is algebraic over $K$. – Marm Mar 06 '15 at 18:11 — Marm, Mar 06 '15 at 18:11

score 6 · Accepted Answer · answered Mar 06 '15 at 14:47

6

Hint: What about $f(x) = x - e$.

answered Mar 06 '15 at 14:47

Aaron Maroja

17,571

3

Or $f(x) = x - \pi$ – SubSevn Mar 06 '15 at 14:48
@SubSevn Exactly! – Aaron Maroja Mar 06 '15 at 14:48
5

(+1) Might as well kill both birds with one stone: $f(x) = (x - e)(x - \pi)$...? :) – Andrew D. Hwang Mar 06 '15 at 15:00
@user86418 That's right!! – Aaron Maroja Mar 06 '15 at 15:01

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

1 Answers1