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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.

user102580
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  • $P(\sqrt 2,y)$ = 0 for any real $y$ so that doesn't tell you much about $e$. Shouldn't $P$ be irreducible ? – mercio Oct 22 '13 at 19:10
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    The polynomial must be "nontrivial," certainly the fact that it factors easily makes it trivial here. – Will Jagy Oct 22 '13 at 19:10
  • $e$ and $1$ are also algebraically dependent over $\Bbb Q$, as per your definition. Take $P(x, y) = xy - x$. – Balarka Sen Jul 10 '14 at 18:18

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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

rrogers
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