real analysis - Are polynomials with integer coefficients uniquely determined by their coefficients?

Assume f and g are polynomials, where f = g. More explicitly;

$$ f = a_{n}z^{n} + \ldots + a_{1}z + a_{0} = b_{n}z^{n} + \ldots + b_{1}z + b_{0} = g $$

where $ \ \ a_{0} \ , \ldots, \ a_{n} \in \mathbb {Z}, \ and \ z \in \mathbb{C} $

then $$ a_{0} = b_{0} \ , \ldots, \ a_{n} = b_{n} $$

Attempted proof:

$ f - g = 0 $, thus we can now equate coefficients;

i.e. $$ a_{0} - b_{0} = 0 \ , \ldots, \ a_{n} - b_{n} = 0 $$

$$ \implies a_{0} = b_{0} \ , \ldots, \ a_{n} = b_{n} $$

$\square$

So, is this essentially stating that polynomials are uniquely determined by there coefficients?

This question is spurred as I am trying to prove that the set of all algebraic numbers is countable, and consequently my proof will in turn rely upon such a statement as this, since if each polynomial is uniquely determined in this manner, then we should be able to find some (not exactly sure if it will be an isomorphism or just a monomorphism?) correspondence between the set of all polynomials with integer coefficients of degree n and the set of all n-tuples with integer components.

I believe we can then show that this set of all n-tuples with integer coefficients will be countable, because it will be the cartesian product:

$$ \mathbb{Z} \times \ldots \times \ \mathbb{Z} \cong \mathbb{Z}^{n} $$

and the finite cartesian product of countable sets is countable. We then can see that the set of all polynomials with integer coefficients is countable, which i feel puts me a step closer to solving said problem.

I feel that given this to be true, then as all polynomials only have a finite number of zeroes, we can use that the countable union of finite sets is countable to finish the proof.

Does this seem like a feasible plan for proving the statement?

Answer

If $f(x)=g(x)$ as functions, then set $h(x)=f(x)-g(x)$, a polynomial since $f,g$ are. Since $f=g$ as functions, $h(x)$ is identically zero, i.e. zero for every $x$. If $h(x)$ isn't the zero polynomial, then it has some degree, say $m$. But then it would have at most $m$ complex zeroes. But $h(x)$ has infinitely many zeroes. Hence $h(x)$ must be the zero polynomial, so in fact the coefficients of $f,g$ must be identical.