Exercise
Let p be a prime number. For which values of p does there exist an isomorphism Z/p2Z -> Z/pZ[X]/(X2)?
None.
p = 2.
All odd primes p.
All p.
Correct. But can you prove
it? The cardinalities and numbers of invertible elements are the same
for all p!
No.
The additive subgroup generated by 1 is
Z/4Z in one case and
Z/2Z in the other.
No.
The additive subgroup generated by 1 is
Z/p2Z in one case and
Z/pZ in the other.
No. Reconsider!