The fundamental theorem of algebra 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.
 A: 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.
A: 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 $\mathbb{R}$.
