Cardinality of Ring of functions $F^F$

Question

Let $F$ be a field and let $F^F$ be the ring of all functions from $F$ to $F$. Is it valid to say that $\vert F^F \vert = {\vert F \vert} ^ {\vert F \vert}$ ? This is in context to an Algebra question that is asking me to prove facts about ring homomorphisms between $F[x]$ and $F^F$. I need to show when the homomorphism is onto and I also need to show when the map is one-to-one depending on the cardinality of $F$. In my proof, I want to use $\vert F^F \vert = {\vert F \vert} ^ {\vert F \vert}$, but I am not entirely sure that this is true. Any insight would be helpful.

Answer 1

The map $F[x]\to F^F$ that sends a polynomial to the associated polynomial function is a ring homomorphism (easy proof).

If $F$ is finite, then the map is onto.

This easily follows from the Lagrange interpolation formula. If $|F|=n$, then given any $n$-tuple $(a_1,a_2,\dots,a_n)$ of elements of $F$ we can find a polynomial $f(x)$ of degree at most $n-1$ such that $f(e_i)=a_i$, where $F=\{e_1,e_2,\dots,e_n\}$.

The kernel is generated by the polynomial $(x-e_1)(x-e_2)\dotsm(x-e_n)$.

If $F$ is infinite, then the map is one-to-one. Indeed, if $f(x)$ belongs to the kernel, it is a polynomial having infinitely many roots and therefore it is the zero polynomial.

In this case the map cannot be onto, because the ring $F^F$ has zero-divisors.