Proof attempt of fundamental theorem of algebra
I've been working on an elementary proof of the fundamental theorem of algebra, which doesn't even require calculus. The most advanced concepts are that of minima/maxima. I don't see this proof anywhere in the literature, so I suspect there might be a flaw or a jump in logic which requires more complicated theorems in order to justify, but I don't necessarily see what it would be. Here is what I have, I'd appreciate any comments on whether it's effective and/or what it needs.
Let be a non-constant polynomial. The first thing we need to establish is that |f| attains an absolute minimum somewhere on the complex plane (some functions, for example |e^z|, do not).
For any positive real A, the following fact holds: for some positive R. So it is possible to find a finite region D of the complex plain in which |f| attains some values that will be be lesser than any value |f| attains outside of D. Being a finite region, D contains an absolute minimum for |f|. Therefore, |f| attains an absolute minimum.
Suppose that f has no roots. Then |f| attains a non-zero absolute minimum somewhere on the complex plane. If there are multiple points at which this minimum is attained, choose 1 such point. Call it (a,b) (meaning that (f(a) = b.)
For simplicity: let . g(z) is a polynomial such that its absolute value attains an absolute minimum at (0,1). Therefore, for all z.
Let be the coefficient of z in the expansion of g. Suppose that is non-zero. Then, for sufficiently small positive , it is clear that . In particular, if is the coefficient of z^n in g, this inequality will hold if is less than http://latex.codecogs.com/gif.latex?...1%5En%7D%7C%7D.
But if |g| is less than 1, this contradicts the fact that for all complex numbers. Therefore, the coefficient of z in the expansion of g must in fact be 0.
A similar argument can be used to show that if is nonzero, a number x can be constructed such that |g(x)| < 1. To generalize, let be the lowest-power nonzero coefficient in g, and let n be its power. For sufficiently small , . (Note that ^(1/n) has n values. Any one of them can be chosen with the same consequence.)
Therefore, we must conclude that all the coefficients of g are zero except the constant term. But if g is constant, then f is constant. So the only polynomial who's absolute value can attain [an absolute minimum which is greater than zero] is a constant polynomial.
Edit - I just realized that my notation for f and g inconsistently switch between x and z, because I didn't write it all at once in this forum. It is fixed now, sorry for the confusion.
Re: Proof attempt of fundamental theorem of algebra
There is a proof in an ALGEBRA BOOK published by the Russian Mathematician A. Kurosh. There is a translated version in English try to find it...
Higher Algebra ? Kurosh | Mir Books