# Understand the problem

The number of rings of order 4, up to isomorphism, is:

(a) 1

(b) 2

(c) 3

(d) 4.

(a) 1

(b) 2

(c) 3

(d) 4.

##### Source of the problem

TIFR GS 2018 Part B Problem 12

##### Topic

Abstract Algebra

##### Difficulty Level

Easy

##### Suggested Book

Dummit and Foote

# Start with hints

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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