A function $p$ is a polynomial with coefficients in F if $p$: F $ \to$ F as $p(z) = a_0 + a_1z + a_2z^2 + ··· + a_mz^m$ for some $a_0,\ldots,a_m$ and all $z \in$ F.
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Your definition of polynomial doesn't agree with what usually people want to model.
Note that taking $\mathbb F=\{0,1\}$, according to your definition, setting $p(x)=x^2+1$ and $q(x)=x+1$, for all $x\in \{0,1\}$, one has $p=q$. But usually it is wished that $p\neq q$.
What you've defined is a polynomial function.
The operator that takes polynomials to its polynomial functions isn't injective in general (see example above), therefore you can't equate polynomial function with polynomial.
But it works sometimes, namely in fields of characteristic $0$.
Git Gud
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4It works for all infinite fields even if they don't hace characteristic 0. – Omar Antolín-Camarena May 29 '14 at 15:35
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@OmarAntolín-Camarena Thank you for your comment. Please keep it there for future reference. – Git Gud May 29 '14 at 15:37
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2Also for infinite integral domains. – drhab Aug 10 '14 at 18:51