Fermat's little analogue for euclidean domains ?? Consider euclidean domains.
When we take an element of that euclidean domain and divide it by a prime ( in that euclidean domain ) we Get a remainder.
This remainder is the analogue of $ \mod p$ for the ring of integers.
Fermat's little then tells us $x^p = x \mod p $.
Now back to the Euclidean domains.
What are the theorems or analogues to Fermat little for remainders from an euclidean domain ?
Does it matter if the Euclidean domain is norm-euclidean ??
For instance What can we say about the remainders for the euclidean domains : gaussian integers and eisenstein integers ??
 A: The Euclidean assumption is irrelevant. Suppose $R$ is any commutative ring and $P$ is any prime ideal of it such that $R/P$ is finite; define $N(P) = |R/P|$ to be the norm of $P$. Then $R/P$ must be the finite field of size $N(P)$ (by Wedderburn's little theorem), and thanks to general facts about finite fields it follows that
$$r^{N(P)} \equiv r \bmod P$$
for all $r \in R$. What this means is just that $r^{N(P)} - r$ is in the ideal $P$ and you don't need a division algorithm to make this definition. If $P = (p)$ is the principal ideal generated by some prime element this just means that $r^{N(P)} - r$ is divisible by $p$, and again you don't need a division algorithm to make this definition.
Example. Let $R = \mathbb{Z}[i]$ be the Gaussian integers; this is a Euclidean domain but we will make no use of that fact. The prime ideals have the form $(p)$ where $p \in \mathbb{N}$ is a prime congruent to $3 \bmod 4$ and $(a + bi)$ where $p = a^2 + b^2$ is a prime congruent to $1 \bmod 4$. The norms of these prime ideals are $N((p)) = p^2$ and $N((a + bi)) = a^2 + b^2$. Thus, for example, we have
$$z^{49} \equiv z \bmod 7$$
and
$$z^5 \equiv z \bmod (1 + 2i)$$
for all $z \in \mathbb{Z}[i]$. 
