TITLE: Nonzero root of equation
QUESTION [0 upvotes]: Is there a condition so that a polynomial has at least one nonzero root? 
Suppose we have the equation $$\sum_{i=0}^m \alpha_i x^i=0$$ 
Do the coefficients have to satisfy a specific condition so that there is a $y \neq 0$ that satisfies the equation?

REPLY [1 votes]: To answer this and make it answered, here is a statement.
Fact
A polynomial $p(x)=\sum_{0=1}^ma_ix^i$ has a nonzero root, i.e. there exists a nonzero complex root $y\neq0:p(y)=0$, iff there exist $i,j\in\{0,\dotsc,m\}$ with $i\neq j$ such that $a_i$ and $a_j$ are both nonzero.
Proof.
Suppose the condition holds. Suppose wlog $i<j$, and that $i$ is the least integer for which $a_i\neq0$. Then:
$$p(x)=x^i\cdot\sum_{k=i}^ma_kx^{k-i}.$$
For $k=i$, the term in the sum is $a_i\neq0$, so that polinomial doesn't have 0 as a root. Yet by the Fundamental Theorem of Algebra $\mathbb{C}$ is algebraically closed, and since the polinomial in that sum is nonconstant (as $a_j\neq0$ and $j\neq i$) it must have a complex root, in fact, it must have $m-i+1$ complex roots counted with multiplicity. Hence, $p$ has at least a nonzero root.
Suppose now that the condition does not hold, so for any two $i\neq j$ you have either $a_i=0$ or $a_j=0$. This means the polynomial is of the form:
$$p(x)=ax^k,$$
for some $a\in\mathbb{C},k\in\{0,\dotsc,m\}$. If $k=0$, then $p$ is constant and has no roots. Otherwise, the only root is 0.
$$\tag*{$\square$}$$
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