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September 1, 2019

Order of rings: TIFR GS 2018 Part B Problem 12

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]The number of rings of order 4, up to isomorphism, is:
(a) 1
(b) 2
(c) 3
(d) 4.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR GS 2018 Part B Problem 12[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Abstract Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Dummit and Foote[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

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First, ask yourself how many groups are there of order 4. 
the answer is simple => Z/4Z and Klein’s four group (K).
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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]

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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.
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Bonus Problem:
Prove that there are only two non-ismorphic p-rings(ring with p elements) upto isomorphism.
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Watch the video

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Similar Problems

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