Algebraic independence of $a^a$ and $b^b$ for algebraic irrationals $a,b$ Let $a,b$ be algebraic irrationals.
Are there conjectures or unconditional results about the algebraic
independence of $a^a$ and $b^b$?
Probably Schanuel's conjecture is related,
maybe only $\log{a},\log{b}$ and/or $a,b$ are linearly 
independent over the rationals.
 A: They are not always independent. There are two problematic cases: the trivial case $a=b$, and the trinomial case when $c_1x^n + c_2x^m +c_3 =0$ for rational $c_1,c_2,c_3$, $a= x^n$, and $ b= x^m$, so that $ (a^a)^{c_1 /n} (b^b)^{c_2/m} x^{c_3}=1$, giving an algebraic relation between $a^a$ and $b^b$.
In every other case they are independent conditional on Schanuel's conjecture.
First we check that $\log a,\log b, a \log a, b\log b$ are linearly independent over $\mathbb Q$. If $r_1,r_2,r_3,r_4$ are rationals, not all zero, with $r_1 \log a - r_2 \log b + r_3 a \log a - r_4 b \log b = 0$ then $\frac{r_1 + r_3 a}{r_2  + r_4 b}$ is an algebraic number with $a^{ \frac{r_1 + r_3 a}{r_2  + r_4 b} } = b$, violating Gelfond-Schneider unless $\frac{r_1 + r_3 a}{r_2  + r_4 b} $ happens to be a rational $m/n$. In this case, if $m=n$ then $a=b$ and we are in the trivial case. Otherwise, we have $a^m = b^n$ so we can find $x$ with $a=x^n$ and $b=x^m$, but then $$\frac{r_1+r_3 x^n}{r_2+r_4 x^m} = \frac{n}{m}$$ so $$ (m r_3) x^n - (n r_4) x^m + (mr_1 - nr_3)=0$$ and we are in the trinomial case.
Now apply Schanuel's conjecture. The transcendence degree of $$\overline{\mathbb Q}(\log a,\log b, a\log a,b\log b, a,b,a^a,b^b) = \overline{\mathbb Q}(\log a,\log b, a^a,b^b)$$ is at least $4$, so those elements are all independent transcendentals, so in particular $a^a$ and $b^b$ are independent transcendentals.
