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A nit: "reals are a field extension of ℚ. They could be considered an algebraic number field..." This is not an algebraic extension. Pi is a "useful real numb
by doomrobo 7y ago
A nit:
"reals are a field extension of ℚ. They could be considered an algebraic number field..."
This is not an algebraic extension. Pi is a "useful real number" and it is not algebraic over Q.
- klodolph 7y agoYes—and to elaborate, the reason why an algebraic field extension of ℚ cannot contain π is because: - If it is a field, it contains π, π², π³, … which are linearly independent. - By definition, an algebraic field extension is finite dimensional.
- arberavdullahu 7y agoYou are wrong! The algebraic field extension ℚ[π] contains π.
- OscarCunningham 7y agoI think people would normally call that a transcendental extension and not an algebraic one.
- klodolph 7y agoℚ[π] is not an algebraic extension of ℚ.
- jopolous 7y agoI guarantee that is not an algebraic extension. It's not even a finite extension
- lonelappde 7y agoThat's not a valid critique, as other commenters explained
- doomrobo 7y agoFinite implies algebraic. The other comment says that the converse isn't true.
- jopolous 7y agoYou're totally right, I could have worded that differently. I actually posted an infinite field extension that is algebraic elsewhere in this thread, but usually those are tricky and I haven't seen them pop up as often as finite algebraic extensions
- OscarCunningham 7y agoI don't think that's the usual definition of algebric extension. Wikipedia (https://en.wikipedia.org/wiki/Algebraic_extension https://en.wikipedia.org/wiki/Algebraic_extension) says that an algebraic extension is one where every element is the root of some nonzero polynomial over the base field. So for example the algebraic numbers would be algebraic over the rationals, even though they're infinite dimensional over the rationals.
- jopolous 7y agoAnother nit, not all algebraic extensions are finite dimensional. Just adjoin an infinite number of linearly independent roots (square roots, cube roots, etc). Algebraic, but not finite
- impendia 7y agoSlight correction. Your first bullet point, and hence your proof that ℚ cannot contain π, is correct. A number x is algebraic over ℚ if and only if it generates a finite field extension, i.e. if x, x^2, x^3, etc. have a linear dependence relation. However, as jopolous pointed out, you can get infinite dimensional algebraic field extensions by adjoining infinitely many algebraic numbers. For example, the set of all numbers which are algebraic over ℚ is a field, and this field is an infinite degree extension of ℚ.
- Someone 7y ago”π, π², π³, … which are linearly independent.” Do we know that? My search doesn’t get more than https://www.encyclopediaofmath.org/index.php/Lindemann_theorem https://www.encyclopediaofmath.org/index.php/Lindemann_theor..., which proves it for “𝑒, 𝑒², 𝑒³, …“.
- klodolph 7y agoThere are two ways we can go about this. We can either take a closer look at the Lindemann theorem, or we can talk about the definition of “transcendental number”. I’m not going to put a full proof here. If you look at the Lindemann theorem, you can transform the equation so that it uses π instead of e. Multiply all of the exponents by i (which is algebraic!) and then use Euler’s identity. You end up with the same formula, but with π instead of e. However, if we already know that π is transcendental (which is proven by the Lindemann theorem using the above technique), we can rewrite any linear combination of B = {1, π, π², π³, …} as P(π) where P is a polynomial with coefficients in ℚ. Because π is transcendental, we know that P(π)=0 only if P is the zero polynomial (that is the definition of transcendental number). In general, one of the big tricks here is that the set of polynomials is a vector space, and the powers B = {1, x, x², x³, …} span the entire vector space.