I Can Fundamental Theorem Algebra be proven by simple DOF?

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A monic polynomial of degree N has N number of coefficients. The product of N number of linear factors has N number of free terms. A complex number has 2 DOF. Therefore, both a monic polynomial and the product of free terms have 2N number of DOF of real values. Thus, it must be possible to have a monic polynomial to be resolved into a set of roots whose cardinality is the degree of the polynomial.

What is wrong with this statement?
 

.Scott

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The polynomial of degree N has N independent coefficients. So 4x+2=0 is equivalent to 2x+1=0. Both are degree 1, both have 2 coefficients.
The phrase "DOF of real values" is equivalent to just "DOF".
It is possible to "have a monic polynomial to be resolved into a set of roots whose cardinality is the degree of the polynomial". But I do not thing your logic is sequitur.

Certainly you can show that the product of any set of N roots expressed as ##(x-a_i)## is a polynomial of degree N. But you need more than what we've stated to show that the reverse it true - that every polynomial of degree N has N roots.
 

StoneTemplePython

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Also, OP doesn't show why scalars in ##\mathbb C## are enough -- why don't we need some 'hypercomplex number' above and beyond ##\mathbb C##? (Incidentally Leibniz thought this was the case.)
 
Also, OP doesn't show why scalars in ##\mathbb C## are enough -- why don't we need some 'hypercomplex number' above and beyond ##\mathbb C##? (Incidentally Leibniz thought this was the case.)
It seems that there are a set of symmetric functions, each of which is the sum of products of combinations of roots, that end up being the coefficients. Since addition & multiplication are closed in the complex numbers, there should be not be the possibility of hypercomplex numbers.

( x - a ) ( x - b ) ( x - c ) = x3 - ( a + b + c ) x2 + ( a b + a c + b c ) x - ( a b c )
 

StoneTemplePython

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It seems that there are a set of symmetric functions, each of which is the sum of products of combinations of roots, that end up being the coefficients. Since addition & multiplication are closed in the complex numbers, there should be not be the possibility of hypercomplex numbers.

( x - a ) ( x - b ) ( x - c ) = x3 - ( a + b + c ) x2 + ( a b + a c + b c ) x - ( a b c )
I really don't know what you are getting at here. Addition and multiplication are closed in all fields. This seems to be a detour best to be avoided.

A better and closely related point is your original argument fails break for reals. You may say that well you found a counterexample of why not all roots are real. (Insert example.) But then how do you know you haven't found a counterexample yet for complex coefficient polynomials having a root? In effect your original argument is fine if you assume fundamental theorem of algebra holds, but if you assume that, then why bother with the original argument?

If you want your original argument to hold on its own, then you in effect need to construct a basis or invertible map to prove that every polynomial can be factored this way via elementary symmetric functions. Or prove that the lack of such a mapping creates a contradiction.

Your original argument doesn't do this, it just talks about "degrees of freedom" without concretely spelling out how this mapping is supposed to work. The fact that you aren't explicitly calling on continuity in here is something of a red flag that you'll hit a dead end with your argument at some point.
 

mathwonk

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your argument, say in the case of degree two polynomials, seems to be the following: since pairs of roots have 2 degrees of freedom, and degree two polynomials have also the same number, the map taking a pair of roots to the polynomial having those roots, must be surjective. E.g. we have (r,s)-->(x-r)(x-s) = x^2-(r+s)x + rs --> (r+s, rs). so we may ask if in fact the map taking (r,s)-->(r+s,rs) = (b,c), is indeed surjective? But note that in this case we have b^2-4c = r^2 - 2rs + s^2 = (r-s)^2 ≥ 0. so the image of this map is the subset of those pairs of reals (b,c) satisfying b^2-4c ≥ 0, a proper subset of R^2, so the map is definitely not surjective, since it misses e.g. the pair (0,1).

In general there is no good reason to expect a map from k^n-->k^n to be surjective unless we know more about both the map and the field k. In particular this example apparently shows it is not enough that the map be finite to one (only (r,s) and (s,r) have the same image), and given by polynomials.

Notice however that although r^2 is an "open set" with no boundary, the image is a closed set with boundary the curve b^2 = 4c. This is allowed by the fact that the map is generally 2 to 1, except over that curve, so it folds R^2 up along a curve, and hence maps it all on one side of (and onto) that parabola.
 
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WWGD

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A monic polynomial of degree N has N number of coefficients. The product of N number of linear factors has N number of free terms. A complex number has 2 DOF. Therefore, both a monic polynomial and the product of free terms have 2N number of DOF of real values. Thus, it must be possible to have a monic polynomial to be resolved into a set of roots whose cardinality is the degree of the polynomial.

What is wrong with this statement?
Once you set the poly. to be Monic you lose a dof, since the coefficient of the leading term has been fixed to be 1. This restricts the values of the other terms.
 
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WWGD

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Also, if I understood your argument correctly, any discontinuous function at a finite collection of points can be redefined to be continuous
, or function undefined at a finite collection can be extended to be continuous. But there is more to it than the dof.
 

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