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I am confused with regards to the proof Axler gives for the claim that if a polynomial is the zero function, that the coefficients must all be zero. Here is his proof:

Axler’s proof

I understand the application of the triangle inequality in the first line of inequalities, but I don’t see why the second line of the inequalities is true (namely what it has to do with the first line).

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    $\begingroup$ It follows from $z^j \leq z^{m-1}$ and $|a_m|z = |a_0| + |a_1| + \cdots +|a_{m-1}| + 1$ $\endgroup$ Jan 4, 2020 at 1:24
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    $\begingroup$ The idea is that $x^n$ grows faster the higher the power $n$. You just plug in a number that is so large, the highest nonzero term can’t cancel with the lower order terms. The theorem is slightly misinformative. If you coefficients are from a finite field there are nonzero polynomials that define the zero function. $\endgroup$ Jan 4, 2020 at 1:25
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    $\begingroup$ What is $F$ meant to be? $\endgroup$
    – lulu
    Jan 4, 2020 at 1:36
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    $\begingroup$ The usual argument over a field is to show first that $p(a)=0\implies p(x)=(x-a)q(x)$ where the degree of $q(x)$ is one less than the degree of $p(x)$. It follows that $p(x)$ can not have more roots than its degree...so as long as your field has more than $\deg p(x)$ elements, $p(x)$ can not be identically $0$. This argument works even over finite fields, at least so long as the fields are not too small. $\endgroup$
    – lulu
    Jan 4, 2020 at 1:44
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    $\begingroup$ @lulu question is important. This result is false for finite fields, for example. $\endgroup$
    – GEdgar
    Jan 4, 2020 at 1:45

3 Answers 3

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It is a sort of backward way to develop the theorem; this sort of "write down the construction and then check that it works" style is common in textbooks because it is clean, but it is not really how you would think through the proof if you were developing it from scratch.

The actual goal of the proof is to show that if $a_m \neq 0$ and $|z|$ is large enough then $|a_m z^m|>\left | \sum_{j=0}^{m-1} a_j z^j \right |$. The question for a constructive proof is then "how big must $|z|$ be?". To figure this out, first you can assume $|z| \geq 1$, so that $|z|^m$ itself will be bigger than each of the lower $|z|^j$. In general in analysis this is a common start to a "there exists a sufficiently large thing" proof: pick some initial threshold such that if the thing is bigger than that threshold, the situation as a whole becomes simpler.

Using this assumption, $\left | \sum_{j=0}^{m-1} a_j z^j \right | \leq |z|^{m-1} \sum_{j=0}^{m-1} |a_j|$. So it suffices to have $|a_m z^m| > |z|^{m-1} \sum_{j=0}^{m-1} |a_j|$, for which it suffices to have $|a_m z|>\sum_{j=0}^{m-1} |a_j|$ or $|z|>\frac{\sum_{j=0}^{m-1} |a_j|}{|a_m|}$. As long as $|z|$ is larger than that and also $\geq 1$, you're good.

This requires some things of $\mathbb{F}$...I haven't read Axler but my guess is that $\mathbb{F}$ is a subfield of $\mathbb{C}$ containing $\mathbb{Q}$ at least.

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From the equation $$z = \frac{|a_0| + |a_1| + \cdots + |a_{m-1}|}{|a_m|} +1$$ we can get $$|a_0| + |a_1| + \cdots + |a_{m-1}| = |a_m|(z-1)$$ and hence $$|a_0| + |a_1| + \cdots + |a_{m-1}| < |a_m|z.$$ So, multiplying both sides of the preceding inequality by $z^{m-1}$ gives us $$(|a_0| + |a_1| + \cdots + |a_{m-1}|)z^{m-1} < |a_m|z^m.$$

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since, F is a field. F contains a zero. let the polynomial be p(z). Then, $p(0)=a_0 = 0$

Then, $\thinspace p'(0) = 0 \Rightarrow a_1 = 0$, $\thinspace p"(0) = 0 \Rightarrow a_3 = 0$

and so on, by this inductive method it is trivial that all the coeficcients are zero

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  • $\begingroup$ In the field with two elements, $x^2+x$ is always $0$. $\endgroup$
    – lulu
    Jan 4, 2020 at 1:36
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    $\begingroup$ I agree this solution appears to be only valid for infinite fields. $\endgroup$ Jan 4, 2020 at 1:39
  • $\begingroup$ $p^{(k)}(0)$ is not $a_k$, but $k!a_k$. $\endgroup$
    – user239203
    Jan 4, 2020 at 1:39
  • $\begingroup$ you are right I rectified my mistake. $\endgroup$ Jan 4, 2020 at 1:42

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