Let $\alpha,\beta$ be transcendental numbers. Which of the followings are true?

1)$\alpha\beta\ \text{ is transcendental}$.

2)$\mathbb{Q}(\alpha)\ \text{is isomorphic to }\mathbb{Q}(\beta)$

3)$\alpha^\beta\ \text{is transcendental }$

4)$\alpha^2\ \text{is transcendental}$

I know option 4 is true. And I feel option 1 is also true, but I don't know the exact reason. While I have no idea for the remaining two.

  • 4
    $\begingroup$ What about $\pi$ and $1/\pi$? (Or replace $\pi$ with any number you know is transcendental.) $\endgroup$ – David Feb 10 '16 at 6:33
  • $\begingroup$ For 3), $e$ and $\ln 2$ are transcendental (the latter by Lindemann-Weierstrass), but $e^{\ln 2}$ is not. $\endgroup$ – André Nicolas Feb 10 '16 at 6:47
  • $\begingroup$ 1) isn't true because we can get $\alpha\beta = a$ (a algebraic) by simply setting $\beta = a/\alpha$. 4) is true because if $\alpha^2$ solves $P(x)$ than $\alpha$ solves $P(x)^2$. 2) is true because... 3) isn't true because for any algebraic $a$ and any transendental $b$ $a= b^{\log_b a}$ and $\log_b a$ is trans. (can replace $b$ with $e$ if you want.) $\endgroup$ – fleablood Feb 10 '16 at 6:48
  • $\begingroup$ Regarding 2): In general, you can think of any algebraic dependencies that $\alpha$ has over a ground field $K$ as obstructions to $K(\alpha) \simeq K(X)$, where $X$ is an indeterminate. $\endgroup$ – mad_algebraist Feb 10 '16 at 6:57
  1. Let $\alpha$ be any transcendental. Then $\beta := \frac{1}{\alpha}$ is transcendental and $\alpha\cdot \beta =1$. Thus 1. is false.
  2. Consider $f \colon \mathbb Q(\alpha) \to \mathbb Q(\beta), \frac{x_0 + x_1 \alpha + \ldots + x_n \alpha^n}{y_0 + y_1 \alpha + \ldots + y_m \alpha^m} \mapsto \frac{x_0 + x_1 \beta + \ldots + x_n \beta^n}{y_0 + y_1 \beta + \ldots + y_m \beta^m}$, where the $x_i,y_j$ are rationals. Verify that $f$ is indeed an isomorphism.
  3. Notice that $e$ and $\ln 2$ are transcendental, but $e^{\ln 2} = 2$. So this is false.
  4. If $\alpha$ is transcendental, then $\alpha^n$ is transcendental for every $n \in \mathbb N$ (if there were a nontrivial polynomial $p(x) \in \mathbb Z[x]$ with $p(\alpha^n) = 0$, then let $q(x)$ be the polynomial obtained from $p(x)$ by replacing every $x^k$ in $p(x)$ with $x^{n\cdot k}$. Then $q(x)$ is nontrivial and $q(\alpha) = p(\alpha^n) = 0$, so $\alpha$ is algebraic - Contradiction!)
  • $\begingroup$ $\ln 2$ is transcendental by the Hermite-Lindemann Transcendence Theorem, which is difficult. There do exist trans. numbers $a,b$ with $a^b$ trans., because the algebraic numbers form a countable set, whereas the set of reals is uncountable. $\endgroup$ – DanielWainfleet Feb 10 '16 at 6:57
  • $\begingroup$ @user254665 I never claimed that $\alpha^\beta$ couldn't be transcendental for transcendental $\alpha$ and $\beta$. Instead I provided an example for transcendentals $\alpha$ and $\beta$ such that $\alpha^\beta$ is algebraic (or in my case - an integer). $\endgroup$ – Stefan Mesken Feb 10 '16 at 7:01
  • $\begingroup$ .No of course you didn't. I was adding something else. $\endgroup$ – DanielWainfleet Feb 10 '16 at 13:01

Here is a proof that point 3 is false that doesn't use any difficult theorems.

As $\gamma$ ranges through all transcendental numbers, the number $2^\gamma$ takes on uncountably many values. Since only countably many of these values can be algebraic, there must exist a transcendental $\gamma$ for which $2^{\gamma}$ is also transcendental.

Now let $\alpha = 2^{\gamma}$ and $\beta = 1/\gamma$.

  • $\begingroup$ This is probably what @user254665 had in mind. $\endgroup$ – Stefan Mesken Feb 10 '16 at 7:17
  • $\begingroup$ @Stefan Possibly, although I wasn't able to tell precisely what proof user254665 was referring to. $\endgroup$ – David Feb 10 '16 at 7:22

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