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I tried reading the Wikipedia page but it's stated in terms of complex roots, and I don't really understand how that relates to the following proposition:

if a real valued polynomial:

$$\sum_{i=0}^n a_i x^i = 0$$

for all $x \in \mathbb{R}$, is it right to say the Fundamental Theorem of algebra implies that $a_i = 0$ for all $i$?

This was stated off handedly during a recent talk and I am not sure if I heard it correctly, since when I look up the theorem it is difficult for me to understand the way that page is written

symplectomorphic
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Xiaomi
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  • Yes, the fundamental theorem of algebra does imply this result. To see this, suppose otherwise that the function was not the zero function. Then by the fundamental theorem there are at most $n$ distinct roots (values of $x$ for which the function is zero). But each of $x=1,2,3,\dots,n,n+1$ are roots of the polynomial, more roots than the at most $n$ roots for a polynomial of degree $n$ allowed, a contradiction. – JMoravitz Oct 18 '18 at 05:45
  • An advanced related question on the reverse mathematics of "an identically zero polynomials is the zero polynomial": https://mathoverflow.net/questions/44410/reverse-mathematics-strength-of-identically-zero-polynomials-are-the-zero-polyn – symplectomorphic Oct 18 '18 at 06:29

2 Answers2

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A bit pedantically, it really depends on what you take to be the "Fundamental Theorem of Algebra (FTA)".

Strictly speaking, the result classically called the FTA is an existence result. It simply says that every (non-constant) polynomial over $\mathbb{C}$ has a root. Continuing inductively, an easy corollary is that a degree $n$ polynomial has at least $n$ roots.

This is sometimes combined with a distinct result, i.e., the factor theorem, to conclude that a non-zero degree $n$ polynomial over $\mathbb{C}$ cannot have more than $n$ roots, and hence has exactly $n$ roots. Overtime, this slightly strengthened corollary has also become known as the "Fundamental Theorem of Algebra".

But note that the fact that a non-zero degree $n$ polynomial over $\mathbb{C}$ cannot have more than $n$ roots technically has nothing to do with the FTA, which is a statement about the algebraic closure of $\mathbb{C}$. In fact, the upper bound on the number of roots holds for rings much more general than $\mathbb{C}$, and has a much easier proof than the FTA.

Theorem: Let $R$ be an integral domain. Then any non-zero degree $n$ polynomial over $R$ has at most $n$ roots counting multiplicity.

For a proof, see here.

The result you want would follow from applying the above theorem for $\mathbb{R}$, for example.

EuYu
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There are two questions in your post. The answers to them are different.

Is this a correction interpretation of the fundamental theorem of algebra?

No, according to the standard meaning of "fundamental theorem of algebra," which says every polynomial with real coefficients has at least one complex root. That theorem is different and harder than the theorem you need for the rest of your post.

Is it right to say the Fundamental Theorem of algebra implies that $a_i=0$ for all $i$?

Yes, the fundamental theorem implies your result. But your result follows from a weaker theorem that holds over fields that aren't necessarily algebraically closed: every polynomial of degree $n\geq 1$ has at most $n$ roots.

symplectomorphic
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  • Here is an interesting advanced question on Math Overflow that naturally arises out of this discussion -- on the reverse mathematics of the theorem that identically zero polynomials are the zero polynomial. – symplectomorphic Oct 18 '18 at 06:27