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Osmium


Joined: Fri May 05, 2017 7:25 pm Posts: 7

Are there any examples of an inifinite, but algebraic field extension? I cannot imagine how this is supposed to be possible. Because, assume an infinite field extension, which means that I can find an infinite amount of numbers $ \displaystyle a_{i}$ as base for the extension as vector space over my original field $ \displaystyle \mathcal{F}$. Then assume that I build an infinite sum out of all those base vectors,
$ \displaystyle \sum_{i=1}^{\infty} \a_{i}$ , possibly with some coefficients in front of the $ \displaystyle a_{i}$. Wouldn't that be then possibly a transcendental number over $ \displaystyle \mathcal{F}$?
Thanks in advance.





Olinguito


Joined: Mon Aug 19, 2013 3:56 pm Posts: 157

An example would be the extension $[\mathbb A\,:\,\mathbb Q]$ where $\mathbb A$ is the field of all real numbers that are algebraic over the rationals.





Osmium


Joined: Fri May 05, 2017 7:25 pm Posts: 7

Thanks for the reply, Olinguito. But how can it be that all of the elements of $ \displaystyle \mathbb{A}$ are algebraic? For example, isn't it possible to find, e.g., an infinite amount of linearly independent, algebraic numbers with respect to $ \displaystyle \mathbb{Q}$ in the interval [0,1], then take the infinite sum (ordering the numbers from large to small), so I have a finite sum (assuming to have a sequence of numbers that get sufficiently fast smaller), and the resulting number then being transcendent over $ \displaystyle \mathbb{Q}$?
Or in other words, isn't it at least sometimes possible, having an extension field which is infinite over the original field, to construct a number which is not algebraic over the original field?





Olinguito


Joined: Mon Aug 19, 2013 3:56 pm Posts: 157

AFAIK the axioms for a field only apply to finite sums, not infinite ones: finite sums of elements of a field are elements of the field, but an infinite sum may not be. Take $\mathbb Q$ for instance: there are plenty of examples of infinite series of rational numbers converging to an irrational number. One example is
$ \displaystyle 1 + \frac1{2^2} + \frac1{3^2} + \cdots\ =\ \frac{\pi^2}6$
So taking an infnite sum of algebraic numbers won't necessary give you an algebraic number; only finite sums can be guaranteed to do so. In any case, algebraic numbers (by definition) are solutions to polynomial equations – and polynomials only involve finitely many terms, not infinite sums.





iNow


Original Member
Joined: Thu Aug 04, 2011 11:40 pm Posts: 5728 Location: Iowa

Apologies that latex is all borked here. Not sure why. Will see if SkinWalker can help.
_________________ iNow
"[Time] is one of those concepts that is profoundly resistant to a simple definition." ~C. Sagan





Olinguito


Joined: Mon Aug 19, 2013 3:56 pm Posts: 157

It is working – you just have to wait for it to load. This may take a while – maybe five minutes.





iNow


Original Member
Joined: Thu Aug 04, 2011 11:40 pm Posts: 5728 Location: Iowa

Olinguito wrote: It is working – you just have to wait for it to load. This may take a while – maybe five minutes. Fascinating. Unsure if I just missed that or if SW already made s change. Thx, Olinguito!
_________________ iNow
"[Time] is one of those concepts that is profoundly resistant to a simple definition." ~C. Sagan





Osmium


Joined: Fri May 05, 2017 7:25 pm Posts: 7

Yes, Latex is working, only after a while of loading. Thanks @Olinguito for the reply. I also considered that and this is probably the correct answer and makes life much easier.





