Are all finite integral domains Euclidean?

even better than that: all finite integral domains are fields, and all fields are (vacuously) euclidean domains. to see that finite integral domains are fields, suppose A is a finite integral domain. let a in A, a not equal to 0. suppose a does not have a multiplicative inverse. then ab is not equal to 1 for all b in A. let |A|=n. since 1 is not equal to any of the n different products ab for b in A, the subset {ab | b in A} of A is of cardinality less than or equal to n-1. hence there exist two distinct elements b and c of A such that ab=ac. hence a(b-c)=0 and b-c is nonzero since b and c are distinct. this is a contradiction since A is an integral domain, so each nonzero element of A has a multiplicative inverse. thus A is a field.

I think so, but I'm not that good at algebra. Here's a (sketch?) of a proof.

Suppose there is a finite integral domain I that is not Euclidean and assume I has at least three elements (there's only one integral domain with two elements and it's Euclidean). Then there must be two elements a and b of I with b nonzero such that there is no r in I that satisfies a = rb (otherwise you could just define a norm on I by n(a) = 1 for all a in I and I would be Euclidean). Then since I is finite there must be two distinct elements r and q of I such that rb = qb. Thus (r - q)b = 0. So q = r since I is an integral domain, a contradiction.