Number theory - 2018-01-01

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Leyla Alkan

03:06

I'm readig this theorem: Lindemann–Weierstrass theorem — if $α_1, ..., α_n$ are algebraic numbers which are linearly independent over the rational numbers ℚ, then $e^α_1, ..., e^α_n$ are algebraically independent over ℚ. What does being "linearly independent over the rational numbers ℚ" mean here?

Leyla Alkan

03:24

Does it mean that none of $α_1, ..., α_n$ can be written as a linear combination of the other numbers?

Leyla Alkan

04:22

Okay I got it, as I encountered the following definition on Wikipedia: In abstract algebra, a subset S of a field L is algebraically independent over a subfield K if the elements of S do not satisfy any non-trivial polynomial equation with coefficients in K.In particular, a one element set {α} is algebraically independent over K if and only if α is transcendental over K. In general, all the elements of an algebraically independent set S over K are by necessity transcendental over K

7 hours later…

Martin Sleziak

11:50

@LeylaAlkan Yes that's exactly it. With the addition that coefficients have to be rational numbers.

It's the usual notion of linear (in)dependence from linear algebra applied to $\mathbb R$ as a vector space over $\mathbb Q$.

Oh, now I see that this has already been answered on the main site.

Q: The linear and algebraic dependence of numbers in Lindemann–Weierstrass theorem

I am reading Lindemann–Weierstrass theorem, which is: — if $α_1, ..., α_n$ are algebraic numbers which are linearly independent over the rational numbers ℚ, then $e^{α_1}, ..., e^{α_n}$ are algebraically independent over ℚ; in other words the extension field ℚ($e^{α_1}, ..., e^{α_n}$) has ...