2
$\begingroup$

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 $e$?

$\endgroup$
  • 9
    $\begingroup$ Try $f(x)=x-\pi$. $\endgroup$ – vadim123 Mar 6 '15 at 14:46
  • 5
    $\begingroup$ $\pi$ and $e$ are real numbers $\endgroup$ – Jack Yoon Mar 6 '15 at 14:46
  • 1
    $\begingroup$ Every element of a field $K$ is algebraic over $K$. $\endgroup$ – Marm Mar 6 '15 at 18:11
6
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ Or $f(x) = x - \pi$ $\endgroup$ – SubSevn Mar 6 '15 at 14:48
  • $\begingroup$ @SubSevn Exactly! $\endgroup$ – Aaron Maroja Mar 6 '15 at 14:48
  • 5
    $\begingroup$ (+1) Might as well kill both birds with one stone: $f(x) = (x - e)(x - \pi)$...? :) $\endgroup$ – Andrew D. Hwang Mar 6 '15 at 15:00
  • $\begingroup$ @user86418 That's right!! $\endgroup$ – Aaron Maroja Mar 6 '15 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.