Uniqueness of a Polynomial Evaluated at $\pi$

Question

Short question, yet interesting (I think). Suppose $a$ and $b$ are each polynomials with integer coefficients. Let's use the notation $a[x]$ to denote the "number" which results from evaluating a polynomial at $x$. (see definition 7.6.1 of this excellent resource). Is it possible that $a \neq b$ as polynomials while $$ a[\pi] = b[\pi]? $$

For context, I'm an undergrad and my knowledge of abstract algebra reaches approximately up to, but does not include, the theory of field extensions. Thank you all.

EDIT: Wow, thank you for the quick response. I guess I might as well add: is this also true when $a$ and $b$ are rational functions? For that matter, are two rational functions considered equal if their numerators are each the zero polynomial while the denominators are distinct?

I suspect that we still have $a[\pi] = b[\pi]\implies a = b$, since (according to wikipedia) the set of rational functions is defined in terms of equivalences classes, like rational numbers, is that accurate?

Answer 1

No, that would imply

$$(a-b)[\pi]=0,$$

making $\pi$ an algebraic number, which is know to be false by the Lindemann-Weierstrass theorem.

Answer 2

No, $a(x) - b(x) $ is a polynomial with integer coefficients with $\pi$ as it's root which would mean $\pi$ is algebraic contradiction (except if $a(x) \equiv b(x) $)