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Is it possible to find a Polynomial, apart from the constant $0$ itself, which is identically equal to $0$ (i.e a polynomial $P(t)$ with same nonzero coefficient such that $P(c) = 0$ for each number $c$

This problem is an exercise on a textbook on Polynomials I am currently self-studying, the solution was provided by the author but I do not understand it and I need someone to help explain it to me properly.

Solution.

Let $P(t) = a_nt^n + . . . + a_1t + a_0$ be such a Polynomial. Then $a_0 = P(0) = 0$. For any real nonzero $c, a_nc^{n - 1} + . . . + a_1 = P(c)/c = 0$. (We do not know that the left side vanishes at $c = 0$ without further development; in order to avoid this issue, we need to make a more elaborate argument at this point.) Suppose, if possible, $a_1 \neq 0$, choose $c$ such that $0 < c < 1$ and

$2c(|a_2| + . . . + |a_n|) \lt |a_1|$ Then $a_nc^{n - 1} + a_{n-1}c^{n-2} + . . . + a_1| \ge |a_1| - [|a_n|c^{n-1} + |a_{n-1}|c^{n - 2} + . . . + |a_2|c] \ge |a_1| - c[|a_n| + |a_{n-1} + . . . + |a_2|] \gt |a_1 - (1/2)|a_1| \gt 0$,

A contradiction, hence $a_1 = 0$.

I understand the idea behind the proof, the author wishes to prove that such a Polynomial cannot exist by showing that all the coefficients must be zero. But however, I do not understand his method above, can someone please help with explanations?

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    $\begingroup$ It depends on the base field (or ring). $\endgroup$ – Randall Aug 9 '17 at 19:33
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    $\begingroup$ You have to specify the underlying field. Over finite fields there are examples. Over infinite fields there are not, as a polynomial of degree $n$ can have at most $n$ roots. $\endgroup$ – lulu Aug 9 '17 at 19:33
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The idea is to show that each coefficient $a_i$ is $0$, starting with $a_0$, and then proceeding to $a_1$, and so on. Since any polynomial has finite degree, this process will have terminated after a finite number of steps, and the claim will be proved.

To show $a_1 = 0$, we do the following: Since $a_0 = 0$, we have that $$ P(t) = t(a_nt^{n-1}+\dots+a_2t+a_1), $$ and that $P(t)=0$ for every value of $t$. Let's call $Q(t) = P(t)/t$ whenever $t$ is nonzero. Since $P$ is $0$ for every $t$, $Q$ is necessarily $0$ for every nonzero $t$.

The idea now is to show, by way of contradiction, that $a_1$ is $0$. To accomplish this, the author assumes that $a_1$ is nonzero, and uses this fact to choose some particularly clever value $0<c<1$ such that $|Q(c)| > 0$. This is a contradiction because since $c$ is positive, we must have $Q(c) = 0$.

Assuming you understand the details of the author's proof by contradiction, now you want to show that $a_2$ is $0$. To do this, write $$ Q(t) = t(a_nt^{n-2} + \dots + a_3t + a_2). $$ Now, use the same idea we used to show $a_1 = 0$ to show that $a_2 = 0$. Since any polynomial has finite degree, this process of showing that each $a_i = 0$ must terminate after some finite number of steps, and hence we will have shown that $P$ is identically $0$.


Since you're having some trouble with the inequalities, I will try to break them down for you, piece by piece. The goal here is to show, first of all, that $a_1 = 0$. Once that's done, we can show that $a_2 = 0$, and so on.

So, suppose that $a_1 \ne 0$. (If it is already $0$, great! We could then show that $a_2$ is $0$, and proceed.) Consider the fraction $$ F = \frac{|a_1|}{2\,(|a_2| + \dots + |a_n|)}. $$ Since $a_1\ne 0$, $|a_1|>0$, so $F$ is some positive number. We can, of course, choose another positive number that is less than $F$. We will call this $c$. Thus $c<F$, or, equivalently, $$ c<\frac{|a_1|}{2\,(|a_2| + \dots + |a_n|)}\iff 2c\,(|a_2| + \dots + |a_n|) < |a_1| \iff c\,(|a_2| + \dots + |a_n|) < \frac{|a_1|}{2}. \tag{1} $$ Here is a little lemma for you to help with the next inequality we're dealing with: (I won't prove it for you, so consider it an exercise to either prove it or read a proof somewhere.)

Lemma. If $\alpha$ and $\beta$ are any two real numbers, then $$ |\alpha + \beta| \ge |\alpha| - |\beta|. $$ Applying our lemma to $|Q(c)| = |a_1 + \big(Q(c)-a_1\big)|$, we have $$ |a_1 + \big(Q(c)-a_1\big)| \ge |a_1| - |Q(c)-a_1| = |a_1| - |a_nc^{n-1} + \dots + a_2c|. $$ By the triangle inequality, $|a_nc^{n-1} + \dots + a_2c| \le |a_nc^{n-1}| + \dots + |a_2c|$, and negatives flip the inequality, so we obtain: $$ |Q(c)| \ge |a_1| - (|a_nc^{n-1}| + \dots + |a_2c|) = |a_1| - (|a_n|c^{n-1} + \dots + |a_2|c). $$ Now, since $0 < c < 1$, if $k\ge 1$, then $c^k \le c$, so we get the next inequality: \begin{align*} |Q(c)| &\ge |a_1| - (|a_n|c^{n-1} + \dots + |a_2|c) \\ &\ge |a_1| - (|a_n|c + \dots + |a_2|c) \\ &= |a_1| - c(|a_n| + \dots + |a_2|). \end{align*} Using the last equivalence in (1), we have $$ |Q(c)| \ge |a_1| - c(|a_n| + \dots + |a_2|) > |a_1| - \frac{|a_1|}{2} = \frac{|a_1|}{2} > 0. $$ Now what does this mean? Since $|Q(c)|>0$, it follows that $Q(c)$ is either positive or negative, but not $0$, which is a contradiction because $c$ is positive and $Q(c) = 0$, as we said earlier. Thus $a_1$ must be equal to $0$ after all, and you can proceed in the manner I outlined above.

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  • $\begingroup$ Thanks a bunch man, I appreciate, but I still want to understand how comes the absolute values and the Inequalities above $\endgroup$ – Icosahedron Aug 9 '17 at 22:39
  • $\begingroup$ @Icosahedron: I added a section discussing the inequalities. If it isn't enough detail, let me know. $\endgroup$ – Alex Ortiz Aug 10 '17 at 3:06
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You can look at this a different way.

Since $x^n-y^n$ is divisible by $(x-y)$ for all $n\in \mathbb Z^+$, if $p(x)$ is a polynomial, $p(x)-p(a)$ has a factor $(x-a)$ whenever $a$ is an element in the ground field for which $p(a)=0$ [note that $p(x)=p(x)+0=p(x)-p(a)$].

This can be used to show that a polynomial of degree $n$ over a field has at most $n$ roots. So if the field is infinite, a polynomial which evaluates to zero at every point is the zero polynomial.

If the field is finite, the product of all the factors $(x-a)$ for $a$ running through the elements of the field clearly evaluates to zero at every point. The polynomial $p(x)=x^p-x$ over the field with $p$ elements is an example (Fermat's little theorem). But this is not the same as saying that the polynomial is the zero polynomial.

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    $\begingroup$ Is this supposed to be an explanation of the proof under question, or a completely different proof? If the latter, then you should mention that at the start of the answer to help avoid confusion. $\endgroup$ – Bill Dubuque Aug 9 '17 at 20:35
  • $\begingroup$ @BillDubuque I have edited accordingly - thanks for the comment. It was mainly a result of things said in other comments, which might have been misleading. $\endgroup$ – Mark Bennet Aug 9 '17 at 20:37

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