polynomial polynomialfunction unique polynomialfunction is unambigiously determined by polynomial but polynomial is not unambigiously determined by polynomialfunction.
for example $f:\mathbb{Z}_2\rightarrow\mathbb{Z}_2,f(x)=x^2-x$ is the zerofunction. I.e. we also could have chosen the zeropolynomial. It was used as an example to show the second statement.
I find it difficult to understand the first statement: it says given a polynomial then the function is unambigiously determined.
Why can one polynomial not define two functions?
 A: "a polynomial function is unambiguously determined by polynomial" - This means that if you give me a polynomial, then it defines a unique function.
"a polynomial is not unambiguously determined by a polynomial function" - This means that if you give me a polynomial function, e.g. $f:\mathbb{Z}_2\to\mathbb{Z}_2$, then $f$ does not necessarily come from a single (unique) polynomial. The example you gave is correct: Let $p_1(X)$ and $p_2(X)$ be polynomials in $\mathbb{Z}_2[X]$, where
$$
p_1(X)=0\text{ (i.e. the zero polynomial) and }p_2(X)=X^2-X.
$$
Clearly, they define the same polynomial function $f:\mathbb{Z}_2\to\mathbb{Z}_2$, but $p_1(X)$ and $p_2(X)$ are not the same polynomials.
A: Consider a commutative ring $R$ (such as $\Bbb Z, \Bbb Q, \Bbb R, \Bbb Z_2$ etc.) and recall that the polynomial ring $R[X]$ is given by the ring of formal polynomials, by which I mean formal linear combinations of powers of an indeterminate $X$ with coefficients in $R$. In other words an element of $R[X]$ hence is a term $a_nX^n + ... + a_1X^1 + a_0X^0$ where all $a_k\in R$. About $X$ we only know that it is not an element of $R$, but it satisfies the distributive law with respect to elements of $R$ and moreover no two distinct powers of $X$ agree.
Now consider a given polynomial $p = a_nX^n + ... + a_1X + a_0 \in R[X]$. Intuitively, given an element $r\in R$ we can substitute it for $X$ and evaluate the expression $a_nr^n + ... a_1r + a_0 \in R$. Let’s denote the resulting element of $R$ by $p(r)$. But as we can do this for any choice of $r\in R$ we have indeed defined a function $\tilde p: R \rightarrow R, r \mapsto p(r)$. As functions are determined by their values, this is the unique function obtained by „evaluating“ the polynomial $p$ at elements of $R$.
It turns out that functions $f:R \rightarrow R$ obtained in this way are quite nice, so let us define that $f$ is a polynomial function, if there exists a polynomial $p \in R[X]$ st. $f=\tilde{p}$. Now two questions should come to mind:

*

*Is every function $R \rightarrow R$ a polynomial function? In general not, since for $R=\Bbb R$ every nonconstant polynomial function is unbounded (tends to $\pm \infty$ as $x$ goes to $\pm infty$), but there sure are nonconstant bounded functions (eg. $\sin x$)

*Given a polynomial function $f$, is the polynomial $p\in R[X]$ satisfying $f = \tilde{p}$ unique? Again it won’t be the case in general, as you noted both $0 \in \Bbb Z_2[X]$ and $X^2-X\in \Bbb Z_2[X]$ evaluate at every element $r\in \Bbb Z_2$ to 0 and hence define the constant function at 0. So formally (as linear combinations of powers of $X$) the polynomials are distinct, but their induced polynomial functions behave the same.

