6
$\begingroup$

I have a doubt, not about the proof, but about the nomenclature of a theorem.

I'm researching some interesting applications of the Inverse Function Theorem and found a version of the Fundamental Theorem of Algebra, the proof of which applies that theorem. The version is as follows:

Fundamental Theorem of Algebra: If $p:\mathbb{C} \to \mathbb{C}$ is the nonconstant polynomial defined by $$p(z) = a_0 + a_1z + \ldots + a_nz^n,$$ where $a_n \neq 0$ and $n \geq 1$, then $p$ is surjective. In particular, there exists $z_0 \in \mathbb{C}$ such that $p(z_0) = 0$.

I found this version a bit "strange" as it looks weaker than the following:

Fundamental Theorem of Algebra: If $p:\mathbb{C} \to \mathbb{C}$ is a nonconstant complex polynomial defined by $$p(z) = a_0 + a_1z + \ldots + a_nz^n,$$ where $a_n \neq 0$ and $n \geq 1$, then the equation $p(z) = 0$ has $n$ solutions, not necessarily distinct.

Is the first version I listed correct? Are they equivalent? To me they don't seem to be, the first seems to be weaker than the second.

Taking advantage of the question, I would like to ask for suggestions for the application of this theorem (of Inverse Function in the case).

$\endgroup$
6
  • 1
    $\begingroup$ Better write $\mathbb C$ instead of $\mathbb R^2$. $\endgroup$ Commented Oct 8, 2021 at 23:38
  • $\begingroup$ If you write $\mathbb C$ both are true, hence equivalent. $\endgroup$ Commented Oct 8, 2021 at 23:39
  • $\begingroup$ Ok. I writed of this form because my Theorem is in $\mathbb{R}^n$. $\endgroup$
    – Croos
    Commented Oct 8, 2021 at 23:39
  • 1
    $\begingroup$ @Croos: what is the product in ${\mathbb R}^n$? $\endgroup$
    – Taladris
    Commented Oct 8, 2021 at 23:53
  • $\begingroup$ Just FYI - The FTA is more accurately "the Fundamental theorem of the algebra of complex numbers", because it is a theorem about $\Bbb C$ specifically, and not algebra in general. And despite the name, It is a theorem of analysis, not algebra itself, as it requires the analytic properties of $\Bbb C$ to prove. The statement of the theorem is algebraic, but the proof is not. However, once you prove the existence of one root, the rest of it follows by simple algebra, so sometimes the FTA is considered just the existence of that first root, and the rest is just corollaries. $\endgroup$ Commented Oct 9, 2021 at 13:41

2 Answers 2

17
$\begingroup$

While the second version is logically stronger than the first version, it's not hard at all to get the second from the first: divide $p(z)$ by $z-z_0$, and it can be shown that the quotient is still a polynomial and then apply the first version to the quotient, until the quotient is a constant. This step is completely trivial compared to the hardness of establishing the first version.

$\endgroup$
0
9
$\begingroup$

The statements are equivalent: first, note that the second version obviously implies the first.

To see how the first version implies the second, it can be shown (via induction) that there are complex numbers $\lambda_1,\ldots,\lambda_n$ (not necessarily distinct) such that $$ p(z) = \prod_{i=1}^n (z - \lambda_i).$$ Indeed, if the first version holds, then $p(\lambda_1)=0$. By the Division Algorithm for polynomials, there is a polynomial $q$ such that $p(z) = (z- \lambda_1)q(z)$. The polynomial $q$ is a polynomial of degree $n-1$, so apply the induction hypothesis.

$\endgroup$
0

You must log in to answer this question.

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