Students studying the fundamental theorem of algebra in high school are probably familiar with the statement that goes something like the following.

Every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots.

However, another statement I often see (and comes first in the Wikipedia article) is the following.

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.

These statements are equivalent and Wikipedia says that this can be proven by successive polynomial long division. However, Wikipedia does not give this proof. It is absent from Wolfram MathWorld as well.

My question is, what is the proof that these statements are equivalent? I know that the fundamental theorem of algebra cannot be proven algebraically, but can this equivalence be proven with only algebra, i.e., polynomial division?

  • 3
    $\begingroup$ I prefer the second statement, which is really the bulk of the question. $\endgroup$
    – egreg
    Feb 22, 2019 at 22:52
  • 1
    $\begingroup$ The first statement is a corollary of the second. The second is also simpler to grasp: "If $p$ is a polynomial, then $p$ has at least one root." $\endgroup$
    – MasB
    Feb 22, 2019 at 22:55
  • $\begingroup$ The real-analysis tag seems inappropriate, but I'm not confident enough of this judgement to delete it. I've added a couple more tags that do seem appropriate. $\endgroup$ Feb 23, 2019 at 1:03

1 Answer 1


Induction (on the degree of the polynomial) suffices.

As it's clear that the first implies the second, we need only argue that the second implies the first.

This is clear for degree $1$.

Inductively suppose it for degree $n-1$.

Let $P(x)$ have degree $n$. By the second definition it has at least one root, $\alpha$. Then, by standard polynomial division we may write $P(x)=(x-\alpha)\times Q(x)$ where $Q(x)$ has degree $n-1$. Applying the inductive hypothesis to $Q(x)$ shows that the second definition implies the first.

Note: An issue has been raised in the comments, namely the fact that the above assumes that the number of roots (with multiplicity) is an additive function. That is to say, for polynomials $g,h$ if $g(x)$ has exactly $a$ roots and $h(x)$ has exactly $b$ then $f(x)=g(x)\times h(x)$ has exactly $a+b$ roots. This plausible sounding claim is not true for general rings. indeed for $\mathbb Z\big / 4\mathbb Z$ we could take $g(x)=x, h(x)=x$. Then it is clear that both $g,h$ have exactly one root but their product $x^2$ has three ($0$ twice, counting multiplicity, and $2$).

For a field however, like $\mathbb C$, this situation can not happen. Note that $\mathbb Z\big / 4\mathbb Z$ has what are called zero divisors. Non zero elements that multiply to $0$. That's what we used to make the counterexample. $2\neq 0 $ but $2\times 2=0$ in that ring. Fields do not contain such elements. Indeed, if $xy=0$ in a field and $x\neq 0$ we can multiply both sides by $x^{-1}$ to see that $y=0$. (of course, in the prior example, $2$ is not invertible in that ring).

Thus we can establish the Lemma:

Lemma: if $g(x),h(x)\in \mathbb F[x]$ where $\mathbb F$ is a field then the number of roots of $f=g\times h$ is the sum of the number of roots of $g$ and $h$ (taken with multiplicity of course).

Pf: Clearly each zero of $g,h$ gives rise to a root of $f$. We must also show that every root of $f$ arises from a root of either $g$ or $h$ (or both). But if $f(\alpha)=0$ then $g(\alpha)\times h(\alpha)=0$ and since there are no zero divisors in $\mathbb F$ we must have at least one of the factors $g(\alpha),h(\alpha)=0$ and we are done.

  • $\begingroup$ The proof is incomplete. You need to prove that it has exactly $n$ roots. $\endgroup$ Feb 23, 2019 at 0:30
  • $\begingroup$ For example, over the ring $\,\Bbb Z/8 = $ integers $\bmod 8\,$ both $x-1$ and $x+1$ have exacty $1$ root, but their product $\,x^2-1\,$ has $4$ roots $\,\pm1,\pm3.\ $ $\endgroup$ Feb 23, 2019 at 1:16
  • $\begingroup$ @BillDubuque but over $\mathbb{Z}/8$ we have things like $x^2+1$ has no roots, 'violating' the second statement of the fundamental theorem of $\mathbb{Z}/8$-algebra, so that's not a counterexample (because you can't prove the first statement using the second statement if the second statement isn't true). (I do agree with your initial comment.) $\endgroup$
    – boboquack
    Feb 23, 2019 at 1:19
  • $\begingroup$ @boboquack My point was to show how additivity of number of roots fails in general rings, so it requires proof in this special case. Your remark has nothing to do with that so it may confuse readers. $\endgroup$ Feb 23, 2019 at 1:22
  • $\begingroup$ So you need to apply that $\Bbb C$ is free of (non-trivial) zero-divisors, which implies that $\Bbb C[x]$ is an UFD. $\endgroup$ Feb 23, 2019 at 9:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .