# Other Algebraically Independent Transcendentals

I was thinking about a incomplete answer I gave earlier today to a interesting question made by the user lurker. The question was about wheter or not the sum or the product of two transcendental numbers in $\mathbb{C}/\mathbb{Q}$ could result on a algebraic number in the same set. A infinite amount of trivial examples can be shown by manipulating the same transcendental, like shown in the answers and in the comments, but I was intrigued by what I believed being the question's true aim, the non-trivial results.
After fiddling around with some algebra to try and understand the nature of transcendental extensions, then proceed to read about it during the day, I reached the known result that it depends on wheter or not the two given transcendental numbers are algebraically independent. If they are, then we cannot generate a 'non trivial algebraic' transformation between them that yields a algebraic number (i.e. using only the four basic operations). It's just a question of stating a polynomial with rational coefficients whose roots are the given transcendental numbers.
Now to the real question. From Nesterenko's results  we know the independence of $\pi$, some values of the Gamma $\Gamma$ function and some powers of $e^\pi$. From the unproved Schanuel's conjecture follows that $\pi$ and $e$ are independent.

1. Are there examples of more notable (conjectured or known) algebraically independent transcendental numbers?
2. Are there a pair of transcendental numbers, maybe notable constants, whose algebraical dependence has been proved or conjectured?

## 1 Answer

Generally speaking the answer is "no," however we have results such as Lindemann-Weierstrass and Baker's theorem and the Gelfond–Schneider theorem (some of which are just refinements of others, but you get the idea). The actual algebraic independent is usually phrased based on sets involving numbers versus their exponentials, and then a careful analysis of properties of the exponential function to show that both cannot simultaneously be algebraic.

However--generally speaking--doing this for specific numbers which don't fit immediately into that framework is a difficult order to fill. This is mostly because transcendentals have very little to work with: ordinarily we prove facts about them by contradiction because the only fact they have about them is that they have no polynomial with rational coefficients they satisfy. So one argues by assuming a number is algebraic and showing it satisfies some property (usually a bound on some invariant) and then proving that numbers which do not have that property exist in some set, and then those numbers are transcendental.

To give you an idea of the scope of the difficulty, we still do not know if the logarithms of the rational primes are algebraically independent, one of the oldest problems in Diophantine analysis and transcendence.