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Prime Ideals 
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#1
Jan713, 02:16 PM

P: 248

I have a question about the proof that I attached...
1) Since R/I is not the zero ring, we know that [tex]1 \not= 0[/tex]. What is the reason to say [tex]1 + I \not= 0 + I[/tex] instead of [tex]1 \not= 0[/tex]? 2) Also, how do we compute something like (a+I)(b+I)? Isn't this correct [tex](a+I)(b+I) = ab+aI+bI+I^2[/tex]? 3) Finally, if we have something like R/I, how do we know if the elements in R/I are of the form a+I or aI? Or is it both (since it's a ring)? Thank you in advance 


#2
Jan713, 09:01 PM

P: 772




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