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Some days ago, I've made this question and I guess I've finally found an answer to this question:

  • (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 $c$)?

Which is mentioned in that post, I tried to proceed in the following way: I decomposed the polynomial in 2 kinds of structures - due to their similar behaviors inside the polynomial - for example:


The red structures are the $t$'s (1) with a coefficient and the green structure is a number alone, for the red structures, it's impossible to have $P(c)=0$ for every number $c$, with nonzero coefficients. Then I eliminated the reds and started to think about the green alone and I perceived that the only number which would allow me to find this polynomial was zero, thus making it impossible to find the requested polynomial.

Is this valid?

(1) - I have no idea on how to call the $t$'s in the polynomial, the book I'm reading provided me with some terms like: Leading coefficient, coefficients, leading term, constant term/constant coefficient, linear coefficient and linear term, but it mentions no name for the $t$'s, can you provide me a name for it?

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Are you perhaps trying to induct on the number of non-zero coefficients? – EuYu Oct 14 '12 at 6:32
Individual $t$ is the indeterminate of the polynomial (and I suppose the $x$ in the second term should be $t$ as well), the powers $t^i$ are monomials. Instead of "structure" most people say "term". Also your argument is fundamentally flawed, because say a sum of two part (red and green) is always zero does not imply each of the parts is always zero. Consider $\color{red}{-x}+\color{green}{x}$. – Marc van Leeuwen Oct 14 '12 at 10:59

The result is correct over infinite fields (you probably should use induction). Over finite fields you can find such polynomial: for example, you can take $x^2+x$ over $\mathbb{F}_2$

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I still don't know what are finite fields, but the book still didn't mention them - I guess there's no problem in using this answer, right? Also, I don't know how to use induction, can you recommend me something? – Voyska Oct 14 '12 at 6:13
This is a bit misleading: the result is correct over infinite fields, but Gustavo’s argument is seriously incomplete, if not downright circular. – Brian M. Scott Oct 14 '12 at 6:15
@BrianM.Scott Can you provide me more information? – Voyska Oct 14 '12 at 6:20
@Gustavo: Hang on just a bit, and I’ll finish writing something up. – Brian M. Scott Oct 14 '12 at 6:34

No. You're statement that "for the red structures, it's impossible to have $P(c)=0$ for every $c$" is exactly what you're trying to prove. The following statement is true, and it's what you're are trying to prove $$ \text{Over an infinite field, $P(x)=0$ for every $x$ in the field if and only if $P$ is the zero polynomial} $$ Based on your vocabulary, you might not know what a field is. If you don't know, all you need to know is that the real numbers are an infinite field. What you call the red structures are generally called the variables, and the green term is called a constant. Do you know what mathematical induction is? It might be a good place to start when trying to prove the above statement.

Mathematical induction is a proof technique where you show that a statement is true for a base case, then you assume it's true for every value up to $n$, and they you prove that the statement is true for $n+1$. For this proof, you want to induct on the degree of the polynomial. You wrote out a polynomial of degree $n$ in your post; that means the highest power of a variable is $n$. For the base case in the proof, we want to look at what happens when $n=0$, or when the polynomial is a constant. If the polynomial is a constant, it has the form $P(x)=a$ for some $a$ in the real numbers (we can write this more compactly as for some $a\in\mathbb{R}$). Clearly, if $a$ is anything but $0$ we can't satisfy the condition that $P(x)=0\;\forall\;x\in\mathbb{R}$ ($\forall$ means for every, or for all). Since that's what we want to show, we've proven the base case. Now we want to assume that the statement is true for every degree up to $n=k$. This is called the induction hypothesis. Finally, you need to use the induction hypothesis to show that the statement is true for polynomials of degree $n=k+1$.

By doing that, you manage to build a sort of staircase that works for every possible polynomial. Since you proved the statement was true for a base case, you built the first step. With the induction hypothesis, we manage to show that there is a next step. Do you have any ideas how to go about showing that $$ a_{k+1}t^{k+1} + a_k t^k+\dotsb+a_1 t+a_0 =0 \;\forall\;t\in\mathbb{R}\Leftrightarrow a_0=a_1=\dotsb=a_k=a_{k+1}=0 $$ assuming that $a_k t^k+\dotsb+a_1 t+a_0 =0 \;\forall\;t\in\mathbb{R}\Leftrightarrow a_0=a_1=\dotsb=a_k=0$ ($\Leftrightarrow$ means if and only if)?

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I've already read somethings about induction, but it wasn't a guide on how to use it, it was just some examples. – Voyska Oct 14 '12 at 6:18

This really ought to be cast as a proof by induction, but to make the idea a little clearer, I’m not going to do that.

Let $p(x)=a_0+a_1x+\ldots+a_nx^n$ be a polynomial of degree $n>0$, and $c_1$ be any real number. Using ordinary polynomial long division to divide $p(x)$ by $x-c_1$, we can write $$p(x)=(x-c_1)q_1(x)+r\tag{1}$$ for some polynomial $q_1(x)$ of degree $n-1$ and some constant $r$. If $p(c_1)=0$, $(1)$ becomes $$0=p(c_1)=(c_1-c_1)q_1(c_1)+r=r\;,$$ and therefore $p(x)=(x-c_1)q_1(x)$. For future reference note that the leading coefficient of $q_1(x)$ is $a_n$.

Now suppose that $c_2$ is another real number such that $p(c_2)=0$. Then


and $c_2-c_1\ne 0$, so $q_1(c_2)=0$, and we can repeat the argument to discover that there is a polynomial $q_2(x)$ of degree $n-2$ such that $q_1(x)=(x-c_2)q_2(x)$, and the leading coefficient of $q_2(x)$ is $a_n$; clearly $$p(x)=(x-c_1)q_1(x)=(x-c_1)(x-c_2)q_2(x)\;.$$

If there are at least $n$ real numbers $c_1,\dots,c_n$ such that $p(c_k)=0$ for $k=1,\dots,n$, we can repeat this process a total of $n-1$ times to write $p(x)$ in the form


where $q_n(x)$ is a polynomial of degree $0$ with leading coefficient $a_n$. In other words, $q_n(x)$ is the constant polynomial $q_n(x)=a_n$, and $$p(x)=a_n(x-c_1)(x-c_2)\ldots(x-c_n)\;.$$

Now let $c$ be any real number different from $c_1,\dots,c_n$; then $$p(c)=a_n(c-c_1)(c-c_2)\ldots(c-c_n)$$ is a product of non-zero real numbers and is therefore not $0$. That is, if there are distinct real numbers $c_1,\dots,c_n$ such that $p(c_k)=0$ for $k=1,\dots,n$, then for any real number $c\notin\{c_1,\dots,c_n\}$ we must have $p(c)\ne 0$, and therefore $p(x)$ cannot be identically $0$. And of course if there are not $n$ distinct real numbers $c_1,\dots,c_n$ such that $p(c_k)=0$ for $k=1,\dots,n$, then $p(x)$ certainly isn’t identically $0$. It follows that no polynomial of positive degree is identically $0$, and of course the only constant polynomial that is identically $0$ is the zero polynomial.

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If the polynomials are viewed as having real coefficients, then I think something like Gustavo's argument might work. If we let $a_nt^n$ be the highest term of the polynomial (no further powers higher than the nth, and $a_n$ not $0$, we can arrive at a contradiction in case $n>=1$ by finding the limit of $p(t)/t^n$, which is $a_n$. If $a_n$ is not zero we could then pick a $t$ large enough to dominate the earlier terms of the polynomial and guarantee a nonzero value of $p(t)$.

Maybe Gustavo's idea was one of successively peeling off the higher powers one by one until he arrived at just the constant term, which then would have to be zero, finishing the argument. But I agree with those above that as it stands he has not presented a full proof.

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