Are numbers $\sqrt{2}$ and $e$ algebraically dependent over $\mathbb{Q}$?
If yes, they belong to the same Mahler class. However, $\sqrt{2}$ is A-number, while $e$ is S-number.
On the other hand, if we consider non-zero polynomial $P(x,y) = x^2y - 2y$, then clearly $P(\sqrt{2},e) = 0$, hence they are algebraically dependent.
What is wrong?
Thank you in advance.
That's a factoring mistake; but to answer your question: no. y(x^2-1)=0 has two solutions but e != 0 so only the second need apply. It's easy to prove that this approach doesn't work; given that e is transcendental. I am sure there are stronger proofs but here is an elementary one:
Suppose they were algebraically dependent over R:
say for P(e,sqrt(2))
now let m=sqrt(2)
Separate even and odd powers of m in P(e,m)
= Q(e,m^(2n)) + R(e,m^(2p+1); here n,p can be zero.
Rewrite it as
Q(e,m^(2n)) + m*S(e,m^(2p))
Form Q(e,m^(2n)) - m*S(e,m^(2p)).
Then multiply and we get:
Q()^2 - m^2 * S()^2. and all terms in m are even and thus multiples of 2.
But this would be an algebraic polynomial in e over R .
Which doesn't exist. QED