TITLE: Proving a multivariable function zero
QUESTION [0 upvotes]: Let $f(x_1,x_2,...,x_m)$ be a polynomial of indeterminates $x_1$, $x_2$, ..., $x_m$ with complex coefficients.
I need to show that if $f(a_1,a_2,...,a_m)=0$ for all nonegative $a_1$, $a_2$,..., $a_m$ then $f(x_1,x_2,...,x_m)=0$
I started by induction on m=1.It is true  for one variable using fundamantal theorem of algebra then i suppose my supposition true for m-1 and try to prove it true for m where i am stuck!

REPLY [1 votes]: The hypotheses imply $f\equiv 0$ in an open ball. By translating $f,$ we can assume this open ball is $B(0,r).$ Then for each unit vector $u\in \mathbb R^m,$ the function $p_u(t)= f(tu)$ is a polynomial on $\mathbb R$ that vanishes on $(-1,1).$ Hence $p_u$ vanishes identically. This is the same as saying $f\equiv 0$ on every line through the origin. Hence $f\equiv 0$ on $\mathbb R^m.$
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