Exponential extension of $\mathbb{Q}$

Question

A non-trivial exponential function $E:\mathbb{K} \rightarrow \mathbb{K}$ in a field $\mathbb{K}$ is a function such that

\begin{split} E(x+y)=E(x)E(y) \quad \forall x,y \in \mathbb{K} \\ E(x)=1 \iff x=0 \end{split}

For the exponential function such that $E(1)=a \in \mathbb{K}$ write $E_a(X)=a^x$. We know that if $\mathbb{K}=\mathbb{Q}$ such a function can not be defined as $a^{\frac{m}{n}} $ can be irrational, and $E_a(x) \notin \mathbb{Q}$.

Call exponential extension of $\mathbb{Q}$ an extension of $\mathbb{Q}$ in which we can define an exponential function. We know that $\mathbb{R}/\mathbb{Q}$ and $\mathbb{C}/\mathbb{Q}$ are such exponential extension. But it can be shown that there is no exponential extension $\mathbb{E}/\mathbb{Q}$ whith $ \mathbb{E}\subset \mathbb{R}$ and $\mathbb{E}\ne \mathbb{R}$? I can not find such a demonstration.

(Sorry for my bad English).

Aren't the algebraic numbers such an extension? If $a,q \in \mathbb{Q}$ and $a \geq 0$ then $a^q$ is an algebraic number. — Ian, Oct 26 '14 at 21:44
I don't think $\mathbb C/\mathbb Q$ is such an extension, because $\exp(2\pi i)=1$ while $2\pi i\neq0$. — mr_e_man, May 03 '22 at 16:06

Martín-Blas Pérez Pinilla · Accepted Answer · 2014-10-27T09:11:04.533

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Let be $$\Bbb E_0=\Bbb Q$$ $$B_1=\{2^r\,|\, r\in\Bbb E_0\}$$ $$\Bbb E_1=\Bbb E_0(B_1)$$ $$B_2=\{2^r\,|\, r\in\Bbb E_1\}$$ $$\Bbb E_2=\Bbb E_1(B_2)$$ $$\cdots$$ $$\Bbb E=\bigcup_{n\in\Bbb N}\Bbb E_n$$ and $\Bbb E\ne\Bbb R$ because is countable.

edited Oct 27 '14 at 09:11

answered Oct 26 '14 at 21:57

Martín-Blas Pérez Pinilla

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Pérez Pinilla: but you can show that $\mathbb{E}$ is a field? Is $2+2^\sqrt{2}$ in $\mathbb{E}$? – Emilio Novati Oct 27 '14 at 08:58
@EmilioNovati, is a increasing union of fields. If $x,y\in\Bbb E$, then $x,y\in\Bbb E_n$ for $n$ large enough. – Martín-Blas Pérez Pinilla Oct 27 '14 at 09:08
@EmilioNovati, I've corrected my post. – Martín-Blas Pérez Pinilla Oct 27 '14 at 09:10
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@EmilioNovati Just in case, $\mathbb{E}_1(B_2)$ is the field extension of $\mathbb{E}_1$ by $B_2$, in other words it's the smallest subfield of (say) $\mathbb{C}$ containing $\mathbb{E}_1$ and $B_2$. In particular it contains, by definition, $2 + 2^{\sqrt{2}}$, because $2 \in \mathbb{E}_1$ and $2^\sqrt{2} \in B_2$ (and this because $\sqrt{2} \in \mathbb{E}_1$). – Najib Idrissi Oct 27 '14 at 09:38
Seem you need to specify which choice of $2^r$ you want when $r$ is not an integer. – Kevin Carlson Nov 04 '14 at 19:03
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@KevinCarlson, true. Taking $2^r>0$ is the easier choice. – Martín-Blas Pérez Pinilla Nov 05 '14 at 07:50

Emilio Novati · Answer 2 · 2014-11-01T23:02:56.393

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Thank you very much! If I have well understood: for each $a \in \mathbb{Q}$ we can build a field $$ \mathbb{E}_a=\bigcup_{n=0}^{\infty}\mathbb{E}_{a,n} $$ such that $E: \mathbb{E}_a \rightarrow \mathbb{E}_a$, $E_a(x)=a^x \quad \forall x \in \mathbb{E}_a$ is a well defined exponential function. Now the problem becomes: the field closures of the sets $$ \mathbb{E}=\bigcup_{a \in\mathbb{Q}}\mathbb{E}_a \varsubsetneq \mathbb{C} \qquad \mbox{ and } \qquad \mathbb{E}^+=\bigcup_{a \in\mathbb{Q}^+}\mathbb{E}_a \varsubsetneq \mathbb{R} $$ are are proper subsets of $\mathbb{C}, \mathbb{R}$ ?

And I get a further question: the Napier's constant $e$ is an element of $\mathbb{E}^+$? And, more generally, there is a method to determine if a transcendental number $ \in \mathbb{E}$?

edited Nov 01 '14 at 23:02

answered Oct 27 '14 at 14:09

Emilio Novati

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Verifying the proof I have another question. How can I prove that $\mathbb{E}_i$ is countable? – Emilio Novati Oct 27 '14 at 16:01
About countability: every element of $\Bbb K(x_1,x_2,\cdots)$ can be written as a quotient of polynomials with coefficients in $\Bbb K$. – Martín-Blas Pérez Pinilla Oct 29 '14 at 07:24
@Martin-Blas Pérez Pinilla Yes !!! Thankyou – Emilio Novati Oct 29 '14 at 09:08
I've corrected my post because $ {\mathbb{E}_a}$ is not a chain of fields. – Emilio Novati Nov 01 '14 at 22:53
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Please ask your new question as a new question, with a link to this one if it helps provide context. – Nov 01 '14 at 23:20
Sorry... the new question is at link. – Emilio Novati Nov 04 '14 at 18:49

Exponential extension of $\mathbb{Q}$

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