2
$\begingroup$

How is arbitrary member of $\mathbb{Q}(\pi)$ defined? $\mathbb{Q}(\pi)$ means the extension of $\mathbb{Q}$ by $\pi$ , thanks

I thought maybe it's

$$x+y\pi,\quad x,y\in\mathbb{Q}$$

but how to get $$\pi^2=x+y\pi$$ or $$1/\pi$$

with that?

$\endgroup$
  • 2
    $\begingroup$ Since $\pi $ is transcendental, $\Bbb Q (\pi) $ is an infinite extension. So I think your approach is wrong. $\endgroup$ – Thomas Shelby Mar 19 at 0:18
  • 2
    $\begingroup$ It simply comprises all rational fractions in $\pi$ with integer coefficients. $\endgroup$ – Bernard Mar 19 at 0:20
  • $\begingroup$ @Bernard can you give examples of this please? in there a general expression? $\endgroup$ – Loli Mar 19 at 0:38
  • 1
    $\begingroup$ Do you have any reason to believe that $\pi^2$ and $\frac{1}{\pi}$ are in Q(π)? $\endgroup$ – user247327 Mar 19 at 0:48
  • 1
    $\begingroup$ @Thomas Shelby I think a typical element is $p(\pi) / q(\pi)$, $p, q \in Q[x]$ $\endgroup$ – Loli Mar 19 at 1:38
2
$\begingroup$

$\Bbb Q(x)$ is isomorphic to $\Bbb Q(π)$, by sending $x$ to $π$. So, no typical element is of the form $x+πy$, as that'd mean we are dealing with a degree $2$ extension, but $\Bbb Q(π)$ is an infinite extension. This is always the case when you adjoin a transcedental number to $\Bbb Q$.

$\endgroup$
  • $\begingroup$ make sense! thanks $\endgroup$ – Loli Mar 19 at 1:09
  • 1
    $\begingroup$ I think a typical element is $p(\pi) / q(\pi)$, $p, q \in Q[x]$ $\endgroup$ – Loli Mar 19 at 1:37

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .