How can i prove that $\mathbb{Q}(e)\cong \mathbb{Q}(\pi)$?

Also, is this isomorphism a valid one?

We have that $\mathbb{Q}(e)=\left \{ \cfrac{f(e)}{g(e)} \mid f(x),g(x)\in\mathbb{Q}[x], g(e)\neq0 \right \} $ and $\mathbb{Q}(\pi)=\left \{ \cfrac{f(\pi)}{g(\pi)} \mid f(x),g(x)\in\mathbb{Q}[x], g(\pi)\neq0 \right \} $.

Then I consider the homomorphism $\phi:\mathbb{Q}(e)\to\mathbb{Q}(\pi)$, where $\cfrac{f(e)}{g(e)}\mapsto \cfrac{f(\pi)}{g(\pi)}$.

The upper homomorphism is obviously surjective by the definition of the fields.

Moreover, $\ker\phi=\left \{ 0 \right \}$, because if $\cfrac{f(e)}{g(e)}\in\ker\phi$ we have that $\cfrac{f(\pi)}{g(\pi)}=0\Rightarrow f(\pi)=0$. So if $f(x)$ is a non-zero polynomial we have that $\pi$ is not transcendental over $\mathbb{Q}$, which is a contradiction and so $f(x)=0\Rightarrow \cfrac{f(\pi)}{g(\pi)}=0$

If $\phi$ is an isomorphism, is there another proof, more general and abstract?


It follows easily from the definition that a number $\alpha \in \mathbb C$ is a transcendental number iff $\mathbb{Q}(\alpha) \cong \mathbb{Q}(X)$, the field of rational functions over $\mathbb{Q}$. Therefore, $\mathbb{Q}(e) \cong \mathbb{Q}(X)\cong \mathbb{Q}(\pi) $.

  • 2
    $\begingroup$ That's not immediate by definition. The proof is what @Richarddedekind did in the question after replacing $e$ and $\pi$ with $X$ as well. $\endgroup$ – AHusain Jun 25 '16 at 0:37
  • $\begingroup$ I don’t think it’s a proof by definition, ’cause it depends on the theorems that $e$ and $\pi$ are both transcendental over $\Bbb Q$. $\endgroup$ – Lubin Jun 25 '16 at 3:20
  • $\begingroup$ In fact, for me, this is a satisfying observation, but the fact that $\mathbb{Q}(e)\cong\mathbb{Q}(X)$ follows my argument. $\endgroup$ – richarddedekind Jun 25 '16 at 17:32

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