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# Understand the problem

The number of rings of order 4, up to isomorphism, is:
(a) 1
(b) 2
(c) 3
(d) 4.
##### Source of the problem
TIFR GS 2018 Part B Problem 12
Abstract Algebra
Easy
##### Suggested Book
Dummit and Foote

Do you really need a hint? Try it first!

First, ask yourself how many groups are there of order 4. the answer is simple => Z/4Z and Klein’s four group (K).
So intuitively there should be two rings with 4 elements.
Okay, I will give you two rings or same order=> (R,+,.), (R,+,*) . see that order of the rings are same but I have changed the multiplication, and

I define a*b=0 for all a,b in R. [(R,+,*) is called zero ring]

Question: Prove that (R,+,.) and (R,+,*) are not isomorphic! (easy)

So, we had (Z/4Z,+,.) and (K,+,.) as our answers. But if you change the multiplication to “*” then there will be 4 different rings *upto isomorphism* right?

Hence the answer is 4.
Bonus Problem:Prove that there are only two non-ismorphic p-rings(ring with p elements) upto isomorphism.

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