**33. (Preliminary for Th, April 12; finalized Sat, April 21) Let p be a prime.
a. Prove that if [0]≠[a]∈Fp, then {[a],[2a],…,[(p−1)a]}=F×p.**

If $[0]\neq [a] \in F_p$. Then that is true because if it went up to $[pa]$ Then since $p$ is just equal to 0, you need it to go up to only $[(p-1)a]$.

**b. Prove Fermat's little theorem: if [0]≠[a]∈Fp , then [a]p−1=[1]. (Hint: Try multiplying [a]⋅[2a]⋅⋯⋅[(p−1)a] and use part a.)**

we may assume that a is in the range $0 ≤ a ≤ p − 1$

it suffices to prove that

$a^{p-1}\equiv 1{\pmod {p}}$

for a in the range $1 ≤ a ≤ p − 1$.

multiplying both sides by a yields the original form of the theorem,

$a^{p}\equiv a{\pmod {p}}$

On the other hand, if a = 0, the theorem holds trivially.

**c. Consider the polynomial ring Fp[x] and the ring of polynomial functions FuncFp[x] defined in class. Prove that the rings Fp[x] and FuncFp[x] are not isomorphic. There are a couple of ways to prove this; however you do it, also find two distinct polynomials in Fp[x] that yield the same function. (Hint: use part b.)**

Using part b, we know they are not isomorphic because there is no bijective function that maps one to the other. However they are homomorphic.

Assume you have two distinct polynomials that yield the same function, then it is disproven.

**d. Prove that in contrast, the polynomial ring Q[x] is isomorphic to the ring of polynomial functions FuncQ[x]. (You only have to show injectivity of the natural map from Q[x] into FuncQ[x] since we took care of the other requirements in class.)**

It is isomorphic because we know it is already homomorphic and surjective trivially. It is injective because say for instance you have two pieces of $\mathbb{Q}[x]$, then they are equal to each other if they are mapped to the same thing. So it works.