Finding the kernel of a homomorphism mapping polynomials to polynomial functions Consider the ring homomorphism $\phi: A[t]\to A^A$ where $A$ is an integral domain and $A^A$ is the ring of all functions on $A$ with values in $A$. Determine the kernel of $\phi$. Be as explicit as possible.
I reasoned that if $f$ has $n$ distinct roots, that is, if every element in $A$ is a root of $f$, then $f\in\ker\phi$. However, my professor disagreed with me and asked me to be more explicit. What am I missing?
 A: Hint $\ $ I presume you mean that $\rm\:t\:$ maps to the identity function. Suppose $\rm\:f(t)\:$ maps to a zero function, i.e. $\rm\:f(A) = 0.\:$ If $\rm\:A\:$ is infinite then $\rm\:f = 0\:$ since a polynomial $\ne 0$ over a domain has no more roots than its degree. If $\rm\:A = \{a_1,\ldots,a_n\}$ is finite, then by the Factor Theorem 
$$\rm f(A)=0\iff\forall\: i:\ t\!-\!a_i\ |\ f(t)\iff lcm(t\!-\!a_i) = (t\!-\!a_1)\cdots(t\!-\!a_n)\ |\ f(t),\:$$
Thus the kernel $\rm\:I\:$ is generated by $\rm\:(t-a_1)\cdots(t-a_n).\:$ But $\rm\:t^n - t\in I\:$ since $\rm\:a^{n-1} = 1\:$ for $\rm\:a\ne 0,\:$ by applying Lagrange's theorem to the multiplicative group of the finite field $\rm\:A.\:$ So both polynomials are equal (else their difference is a  polynomial $\ne 0$ of degree $\rm < n$ with $\rm\:n\:$ roots).
A: Your question isn't so clear so I'll try to answer in the more coplete way.
There are two cases to consider: 


*

*$A$ is a finite integral domain;

*$A$ is an infinite integral domain.


In case 1. $A$ is a field, this can easily be proven observing that for every $a \in A$ because $A$ is an integral domain the map
$$x \in A \mapsto a x \in A$$
is a injective endo-function on a finite set, so bijective.
In this case clearly we know that for every polinomial $f \in A[x]$ is such that $\varphi(f)=0$ if and only if for every $a \in A$ we have $f(a)=0$ (or more correctely $\phi(f)(a)=0$). This, for the polynomial remainder theorem, imply that for each $a \in A$ have to be $(x-a) \mid f$ and so $\prod_{a \in A}(x-a) \mid f$ (because the polynomial $x-a$ are coprime).
Let's pass to the more interesting case, the case 2: $A$ is infinite.
Let $K$ be the field of fractions of $A$, clearly $K$ is infinite.
If $f \in A[x]$ is such that $\phi(f)=0$ this implies that for every $a \in A$ we have that $(x-a) \mid f$ as polynomial in $K[x]$, but from this would follow that $f$ should have infinite degree, that's possible if and only if $f=0$ in $K[x]$ and so in $A[x]$ too.
