Is the zero polynomial the only polynomial that vanishes at every point of $\mathbb C$?

Question

The zero polynomial has the property that every value it takes on $\mathbb C$ is zero.

Is the converse true, or are there other polynomials $f$ such that $ f(x)=0$, for all $x \in \mathbb{C}$?

Answer 1

Any nonzero polynomial can be written as $$ a_0+a_1x+\dots+a_nx^n $$ with $a_n\ne0$. This $n$ is called the degree of the polynomial.

The zero polynomial is usually assigned degree $-\infty$ (or no degree at all). If we multiply two nonzero polynomials, the degree of the product is the sum of the degrees of the factors.

Theorem. If $r$ is a root of the polynomial $f(x)$, then there exists a polynomial $g(x)$ such that $f(x)=(x-r)g(x)$.

Proof. This is based on the division algorithm, that allows us to write $f(x)=(x-r)g(x)+a$, where $a$ is a constant, because $x-r$ has degree $1$. From the hypothesis $f(r)=0$, we get $(r-r)g(r)+a=0$, that is $a=0$. QED

Theorem. If $r_1,r_2,\dots,r_m$ are pairwise distinct roots of the nonzero polynomial $f(x)$ and $f(x)$ has degree $n$, then $m\le n$.

Proof. This is clear if $n=1$. Suppose we have proved the thesis for all polynomials of degree $n-1$. Since $f(x)=(x-r_m)g(x)$ and $r_1,\dots,r_{m-1}$ are also roots of $f(x)$ they must be roots of $g(x)$, as $r_i-r_m\ne0$ for $i=1,2,\dots,m-1$. Since the degree of $g(x)$ is $n-1$, the induction hypothesis implies $m-1\le n-1$, that is, $m\le n$. QED

In particular, a polynomial of degree $n$ has at most $n$ roots.

Corollary If a polynomial has infinitely many roots, then it is the zero polynomial.

Proof. If the polynomial is nonzero, it has a degree and so finitely many roots. QED

Answer 2

Assume $$f(z) = a_n z^n + a_{n-1} z^{n-1} + \dots a_2 z^2 + a_1 z + a_0, \quad a_n \ne 0.$$

Then it is easy to see that $$\lim_{t \in \mathbb{R}, t \to +\infty} \frac{f(t)}{t^n} = a_n \ne 0,$$ contradicting the fact that $f(z)$ is identically zero.

Answer 3

If $f$ were a least degree nonzero polynomial with $\,f(\Bbb N) = 0,\,$ then $\,f(0)= 0\,$ hence $\,f(x) = x g(x).\,$ Thus $\,0 = g(1)=g(2) = \ldots,$ so $\,\bar f(\Bbb N) = 0\,$ for $\bar f(x) = g(x\!+\!1),\,$ contra minimality of $\deg f$.

Remark $\ $ See also the related thread Apostol proof divides by zero?

Answer 4

How about: $${\bf 0}(x) = 0 + 0x + 0x^2 + \cdots +0x^n\,?$$

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How can we prove that ONLY the zero polynomial works?

Simple. Write $p(x) = a_0 + a_1x + \cdots +a_nx^n$. You know that $p(x) = 0$ for all $x \in \Bbb C$. Take $x = 0$. Then you get $a_0 = 0$. Then take $x = 1,2 \ldots$, and solve the system for $a_1,\ldots, a_n$. The system will have the unique solution $a_i = 0, \ \forall \,i$.