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Is the following equivalent to the usual statement of the fundamental theorem of algebra:


Let $$f(z)=c_nz^n+\cdots+c_1z+c_0$$

be a polynomial with complex coefficients. For all but finitely many $w \in \mathbb C$, $f(z)-w$ has $n$ distinct roots in $\mathbb C$.


This seems different to just saying that $f(z)$ has, including multiplicities, $n$ roots. Because this statement does not leave the possibility that there could be finitely many points $w\in\mathbb C$ such that $f(z)=c_nz^n+\cdots+c_1z+(c_0-w)$ will not have $n$ roots. Any insight is appreciated!

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  • $\begingroup$ @MichaelHardy My bad, I edited it to "include multiplicities" and forgot to delete the "distinct". Thanks! $\endgroup$ – Mike May 19 '16 at 21:51
  • $\begingroup$ Then you should remove "distinct". $\endgroup$ – marty cohen May 19 '16 at 21:52
  • $\begingroup$ @martycohen Unfortunately, I can't comment about my editing it and edit it at the same exact moment! $\endgroup$ – Mike May 19 '16 at 21:57
  • $\begingroup$ It reads that way to me as very similar to the fundamental theorem of algebra. Or, in my interpretation. If f(z) - w has has some number of multiple roots, giving w a small perturbation, will break up the multiplicities. Yah, it is a little fuzzy. $\endgroup$ – Doug M May 19 '16 at 21:57
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    $\begingroup$ The two are equivalent in the uninteresting way that all true propositions are equivalent. For a more interesting notion of "equivalent", what is allowed to deduce one version from the other? $\endgroup$ – Daniel Fischer May 19 '16 at 21:58
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A polynomial is separable (has distinct roots) if it shares no zeroes with its formal derivative. If $ P(X) - w $ is a polynomial of degree $ n $ for $ w $ a constant, then its formal derivative $P'(X) $ is a polynomial of degree at most $ n-1 $ and does not depend on $ w $. Let the distinct zeroes of the formal derivative be $ z_1, z_2, \ldots, z_r $ where $r \leq n-1 $; then $ P $ can only be inseparable if it has one of these as a root. But this requires $ w \in \{P(z_1), P(z_2), \ldots, P(z_r) \}$, and there are only finitely many elements of this set; which means $ P(X) - w $ has $ n $ distinct roots for all $ w $ outside of this set.

For the reverse implication, note that if $ P(X) $ was a polynomial with no roots, then $|P(X)| $ would have a nonzero minimum value (this is guaranteed by the growth lemma), say $ r $. In that case, we would have that $ |P(X) + q| \geq ||P(X)| - q| \geq r/2 $ for all $ 0 \leq q \leq r/2 $, meaning that none of these polynomials can have any roots in $ \mathbb{C} $. This violates our hypothesis, since there were only finitely many such polynomials which did not have $ n $ distinct roots.

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  • $\begingroup$ Your firsts paragraph is not a proof that «if the FTA holds then $P(X)-w$ has distinct roots for all but finitely many values of $w$», it is simply a proof of «$P(X)-w$ has distinct roots for all but finitely many values of $w$». You never used the funtamental theorem of algebra, only that a polynomial has finitely many roots which is a much weaker statement. $\endgroup$ – Mariano Suárez-Álvarez May 19 '16 at 22:03
  • $\begingroup$ Thanks, I'll have to do a bit of reading to follow your reverse implication. $\endgroup$ – Mike May 19 '16 at 22:03
  • $\begingroup$ @Mariano Suarez-Alvarez: The fundamental theorem of algebra is needed to guarantee that the roots are in $ \mathbb{C} $. $\endgroup$ – Starfall May 19 '16 at 22:04
  • $\begingroup$ A way to phrase this is that the discriminant of $P(X)-w$ is a polynomial in $w$, so it vanishes for finitely many values of $w$ only. $\endgroup$ – Mariano Suárez-Álvarez May 19 '16 at 22:05
  • $\begingroup$ This is a nice argument, though it uses rather strong properties of $\mathbb{C}$ (in particular, that it has a norm which makes it locally compact). I would still be very interested to see whether the same result holds if you assume only that you have a field of characteristic $0$. $\endgroup$ – Eric Wofsey May 19 '16 at 22:06
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Since the two statements are true, they are clearly wquivalent, but this is not what is being asked, in all probability. What the OP is asking, surely, is if the two statements are equivalent in the sense that proving one is almost the same as proving the other.

The statement

the polynomial $P(X)-w$ has $n$ distinct roots for all but finitely many values of $w$

is trivially equivalent to

the discriminant of $P(X)-w$ has finitely many roots.

SInce the discriminant of $P(X)-w$ is a polynomial of $w$, this last claim is true, but reasons much weaker than the fundamental theorem of algebra.

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    $\begingroup$ You are assuming that the field is algebraically closed... How is that any weaker than the FTA? $\endgroup$ – Starfall May 19 '16 at 22:10
  • $\begingroup$ By «has $n$ distinct roots» is meant here is separable, that is, its roots in an algebraic closure are distinct., $\endgroup$ – Mariano Suárez-Álvarez May 19 '16 at 22:13
  • $\begingroup$ It is trivial that if you know any degree $n$ polynomial has $n$ roots (counting multiplicity), then you can use the discriminant to show $P(X)-w$ has $n$ distinct roots for all but finitely many values of $w$. But this only shows that FTA implies OP's statment, not the reverse implication. $\endgroup$ – Eric Wofsey May 19 '16 at 22:13
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    $\begingroup$ The question says "$n$ distinct roots in $C$", not "$n$ distinct roots in an algebraic closure". $\endgroup$ – Eric Wofsey May 19 '16 at 22:13
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ok, Let's look at the statemt that says

For all but finitely many $w \in C$, $f(z)-w$ has $n$ distinct roots in $\mathbb C$.

As written, that statement doesn't say there are any roots at all when $w$ is one of those finitely many exceptions. Thus it falls short of entailing the fundamental theorem, which says there is always at least one root.

You wrote "including multiplicities, $n$ dinstinct roots". Maybe you're missing the meaning of the word "distinct". If any of the roots has multiplicity more than $1$, and the sum of the multiplicities is $n$, then the $n$ roots that you have (including multiplicities) are not distinct.

Regardless of the value of $w$, there will be $n$ roots if you count them by multiplicities, i.e. the sum of the multiplicities will always be $n$. But for finitely many values of $w$, the roots will not be distinct.

You ask whether the statement is "equivalent" to the "fundamental theorem of algebra". If "equivalent" means they are either both true or both false, then they are. If equivalence is to be defined relative to a specified axiom system in which the axioms are not enough to decide whether either statement is true or false, then it can make sense to say what "equivalent" means.

But if "equivalent" means the deduction of either as a corollary of the other is quick and simple, then "equivalent" is not all that precisely defined. Sometimes it means somebody has proved that either both are true or both are false, without saying which.

Obviously if there are $n$ roots, counted by multiplicities, then there is at least one root, and that's what the fundamental theorem says.

On the other hand, if the fundamental theorem is true (and it's not really a theorem of algebra as we usually understand that term today), then a simple theorem in algebra says we can write $$ f(z) = (z-z_1) g(z) $$ where $z_1$ is a zero of $f(z)$, and then applying the fundamental theorem and that simple theorem of algebra again, we get $$ f(z) = (z-z_1)(z-z_2)h(z) $$ and so on. So you get $n$ roots, counted by multiplicities. You can make this into a precisely stated proof by mathematical induction.

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