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In some places (including this), the fundamental theorem of algebra is said to state that for a polynomial of degree d, there are at most d roots.

Now, how is this equivalent to stating the existence of at least one root-the way the fundamental theorem of algebra is commonly known to be.

kalkanistovinko
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  • One distinction is that the linked page is talking about roots in $\mathbb{R}$ whereas the FTA you refer to would be over $\mathbb{C}$. – Jason Knapp Aug 01 '14 at 19:36
  • In what way, then, does the linked page dare call it the FTA? – kalkanistovinko Aug 01 '14 at 19:38
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    Haha, well, I suppose its not the best practice to call two things by the same name, but the statement on that page is certainly an implication of FTA for polynomials on $\mathbb{R}$. I guess for the first many years, most people's education in math is really just people lying to them progressively less about the ultimate origin of basic mathematics. – Jason Knapp Aug 01 '14 at 19:41
  • according to wikipedia, the FTA states that for a polynomial of degree d, that counting multiplicities, there are exactly d complex roots. As a consequence, in the worst case scenario, there is a single complex root of multiplicity d. – John Joy Aug 02 '14 at 10:07

2 Answers2

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I've never seen the statement

A polynomial of degree $d$ has at most $d$ roots

called “fundamental theorem of algebra”; this name usually refers to the statement

Every polynomial of positive degree with coefficients in $\mathbb{C}$ has at least one root.

The first statement holds for polynomials with coefficients in an arbitrary integral domain, so in cases where existence of roots is by no means guaranteed. On some fields we can find polynomials of arbitrary degree $>1$ which have no roots.

For instance, the polynomials $x^2+1$ and $x^3-x-1$ have no roots in $\mathbb{Q}$ and it's easy to find from them a polynomial with arbitrary degree having no roots. If $d>1$ is even, then $(x^2+1)^{d/2}$ has no roots; if $d>1$ is odd, then $d\ge3$ and $(x^2+1)^{(d-3)/2}(x^3-x-1)$ has no roots.

If you consider the finite field with $p$ elements $F$, for every $d>0$ there exists an irreducible polynomial $f$ with degree $d$; for $d>1$, this polynomial clearly has no roots.

Of course, this depends on the base integral domain (or field). If the coefficients are in $\mathbb{R}$, then every polynomial with odd degree has at least one root, a fact that can be proved with the second statement, but that actually is needed for its proof.

The first statement, for polynomials with coefficients in an integral domain $R$, can be easily proved by induction. It's clear that a polynomial of degree $0$ (that is, a nonzero constant) has no roots. Suppose that $f$ has degree $d>0$; if $r$ is a root of $f$, then $f(x)=(x-r)g(x)$, where $g$ has degree $d-1$, hence at most $d-1$ roots by the induction hypothesis. A root of $f$ is either a root of $x-r$ (so it is $r$) or a root of $g$, because the ring is an integral domain.

The proof of the second statement is much more difficult and has only generic connections with the first statement. After all, saying “at most $d$ roots” and “at least one root” are very different conditions.

egreg
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They're not equivalent at all. But the linked page considers only polynomials over $\mathbb{R}$, and it is not true that every polynomial of degree $d$ has exactly $d$ roots with multiplicity over $\mathbb{R}$. It is not even true that there is at least one. For example, the polynomial $x^2 + 1$ has no roots over $\mathbb{R}$.

davidlowryduda
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