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From Barbeau's Polynomials:

  • (a) 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 some nonzero coefficient such that $P(c)=0$ for each number $c$)?

And then I thought about 2 hypothesis:

  1. P(c-c)
  2. I thought about a polynomial such as $ax^2+bx+c=0$, then I could make a polynomial with $a=1$, $b=-x$, $c=0$ which would render $x^2+(-x)x=0$. I tested it on Mathematica with values from $-10$ to $10$ and it gave me $0$ for all these values.

When I went for the answer, I've found:

               enter image description here

When I went to the answer, I couldn't understand it, can you help me? I'm trying to know what's he doing in this answer, I guess it's a way to prove it, but It's still intractible to me. You can explain me or recommend me some thing for reading. I'll be happy if you also tell me something about my hypothesis. Thanks.

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Your idea in (2) doesn't work because the coefficients of a polynomial have to be independent of the variable. We choose $a,b,c$ and then form a polynomial $ax^2 + bx + c$; you can't choose $b=-x$ because at the time you choose $a,b,c$ there is no such thing as $x$ yet. –  Ted Sep 19 '12 at 7:17
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Your idea in (1) doesn't work either, because after substituting every instance of $c-c$ with $0$, you get the zero polynomial. –  Arthur Sep 19 '12 at 7:19
    
Incidentally, for an even better understanding of this concept and proof you might want to have a look at the polynomial $x^{p^n}-x$ over the (finite) field $GF(p^n)$ - for instance, the polynomial $x^2-x$ over the two-element field $GF(2)$ - and try to understand how things can break down there. –  Steven Stadnicki Sep 25 '12 at 1:43

2 Answers 2

I guess there is no way to forbid people to present contorted arguments to prove a simple result, but they do the reader a disservice by masking the essential point. Here the essential point is that we want a polynomial that vanishes in more points than its degree, and this can only be achieved by the zero polynomial. (In abstract algebra one should specify: polynomial in one variable over a commutative domain, but that is the case here, so you can forget this.) This is (more or less) the proof referred to at the end of your citation, with reference to the Factor Theorem: if $P$ vanishes in distinct points $a_1,a_2,\ldots,a_n$ then $P$ is divisible by the product $(x-a_1)(x-a_2)\cdots(x-a_n)$ (using an induction argument: $P$ is divisible by $x-a_n$ and $P/(x-a_n)$ still vanishes in $a_1,a_2,\ldots a_{n-1}$), so either $P=0$ or $P$ has degree at least $n$). Add to that the fact that there are more numbers one can take for the $a_i$ (namely infinitely many) than the degree of any fixed polynomial, and you are done.

This argument also immediately shows that one should not expect this result if one only has finitely many numbers to take for the $a_i$, as happens when working over a finite field, as the product taken over all possible values for $a_i$ gives a counterexample. For instance $x(x-1)=x^2-x$ is a nonzero polynomial over $\mathbf Z/2\mathbf Z$ that vanishes "everywhere", i.e., on $\{0,1\}$. Those who have a copy of Lang's Linear Algebra at hand should check how he swindles to establish the result in arbitrary fields, on which he even bases his definition of polynomials; a fine example of a textbook blunder*.

So we are left with the Factor Theorem. But in spite of the capitals, it is no big deal; the text in the question in fact already uses it for $a=0$. But the general case is similar: if $a$ is a value at which a polynomial $P$ in $x$ vanishes, write the $P$ as a polynomial in $y=x-a$ by using $x=y+a$ and expanding; now since $P$ vanishes at $y=0$ the result has a zero constant term, so it is divisible by $y$ QED.


*Upon rereading that book, I think I found that the explanation lies in the fact that in it Lang defines the term field to mean sub-field of $\Bbb C$, for which limited case the result is true. I still think that, even when writing for a specific audience, one should not so constrain established mathematical terms (the same goes for defining "polynomials" as polynomial functions), any in any case avoid talking about "arbitrary fields" if one does.

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Can you provide me an example on the product $(x-a_1)(x-a_2)\cdots(x-a_n)$, I'm stuck here. –  Vÿska Sep 25 '12 at 2:21
    
@GustavoBandeira: What do you need to know about the example? Take $(x-3)(x+5)(x-0)(x+1)=x^4+3x^3-13x^2-15x$ in $\mathbf Q[x]$, which clearly vanishes in $3,-5,0,-1$, and nowhere else. Or $x(x-1)(x-2)=x^3-x$ in $(\mathbf Z/3\mathbf Z)[x]$ which vanishes on all of $\mathbf Z/3\mathbf Z$. –  Marc van Leeuwen Sep 25 '12 at 5:15
    
I'm thinking something really naive, look: Take this polynomial $a_3n^3+a_2n^2+a_1n+a_0$, considering it has no zero coefficients, it's impossible to find a polynomial which will be $0$ for all values of $n$, because the multiplication $a \cdot n$ can't render the same number for every number $n$. –  Vÿska Sep 28 '12 at 17:41
    
@GustavoBandeira I don't get it. Why should $a\cdot n$ render the same number for all $n$, we're talking about some polynomial that vanishes everywhere. And what is your $a$ anyway? –  Marc van Leeuwen Sep 28 '12 at 17:48
    
$a$ is the coefficient. It shouldn't render, that's what I'm using as a proof. –  Vÿska Sep 28 '12 at 17:54

I don't know if this will answer all your questions, but what the author is doing is to assume there is a non-zero polynomial which evaluates to zero at all points, then reach a contradiction. He does this by analyzing the hypothetical polynomial in a number of different ways:


Assume the polynomial $p(x)$ evaluates to $0$ at all points, but is not equal to zero. Then it has some definite, positive degree $n$. Since $p(0) = 0$, we must have $p(x) = x\cdot q_0(x)$ for some non-zero polynomial $q_0$.

Since $p(1) = 0$, we must have $q_0(1) = 0$, so $q_0(x) = (x-1)q_1(x)$. Therefore we must have $p(x) = x(x-1)q_1(x)$ for some non-zero polynomial $q_1$.

Now he continues this until he reaches $p(x) = x(x-1)(x-2)(x-3)\cdots (x-n + 1)(x-n)q_n(x)$ for some polynomial $q_n$. But the left side has degree $n$, while the right side has degree at least $(n+1)$, which is a contradiction.


The linear equation he mentions goes like this:

Assume $p(x)$ has positive degree $n$. I'm going to illustrate with $n = 2$, but it's the same for all positive integers. We can then write $p(x) = ax^2 + bx + c$. To find $a, b$ and $c$, we make use of the following: $$ 0 = p(0) = a\cdot 0^2 + b\cdot 0 + c = c \quad\Longrightarrow \quad c=0 $$ as well as $$ 0 = p(1) = a\cdot 1^2 + b\cdot 1 + c = a + b\quad\Longrightarrow \quad a = -b $$ and $$ 0 = p(2) = a\cdot 2^2 + b\cdot 2 + c = 4a + 2b \quad\Longrightarrow \quad 2a = -b $$

to which the only solutions are $a = b = c = 0$, and $p$ is the zero polynomial.


The last one he mentions, Taylor's theorem, depends on the fact that many functions are completely determined by its value and all its derivatives at a single point. Polynomials are among these functions. Since $p$ always evaluates to zero, all its derivatives do too, and the only Taylor series which satisfies this is the Taylor series for the zero function.


Edit: Somehow I completely misread the first paragraph. What he does here is a variation on the first explanation I gave, only he proves that all the coefficients has to be zero the following way:

Assume $p(x) = a_nx^n + \cdots + a_1x + a_0$. Since $p(0) = 0$, we know that $a_0 = 0$. Now examine the function $a_nx^{n-1} + \ldots + a_2x + a_1 = \frac{p(x)}{x}$, and suppose $a_1 \neq 0$. Then the inequalities listed in the text arrives at a contradiction, so $a_1 = 0$. Now proceed similarily with $\frac{p(x)}{x^2} = a_nx^{n-2} + \ldots + a_3x + a_2$, and $a_2$ has to be zero. And so on.

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Since $p(0) = 0$ could we say that this is a polynomial of $deg-\infty$? –  Vÿska Sep 19 '12 at 13:49
    
I got stuck here: Since $p(0) = 0$, we must have $p(x) = x\cdot q_0(x)$ for some non-zero polynomial $q_0$. We have $p(0) = 0$, why we must have $p(x) = x\cdot q_0(x)$? –  Vÿska Sep 19 '12 at 13:52
    
@GustavoBandeira: Substituting $0$ for $x$ makes all the terms of $P$ vanish except the constant term (the one of $x^0$; this is because $0^0=1$). So if this gives $0$, the constant term was zero in the first place. But then all terms contain a factor $x$, and we can single it out (by distributivity). As I remark in my answer, this is a special case of the Factor Theorem. –  Marc van Leeuwen Sep 19 '12 at 14:00
    
@GustavoBandeira The degree of the zero polynomial is undefined. Degrees $0$, $-1$ and $-\infty$ can all be argued for. $0$ reflects that there is only a constant term, which happens to be $0$. $-1$ reflects that it has no terms at all, and $-\infty$ is the only logical degree if you want $\deg(pq) = \deg (p) + \deg (q)$ –  Arthur Sep 19 '12 at 15:00

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