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This is a follow up to my previous question on the equational identities of the $(+,*,0,1)$ reducts of commutative rings. In this question, I want to consider the equational identities of the $(+,*,0,1)$ reducts of rings, not necessarily commutative. I conjecture that

  1. $x+0=x$
  2. $x+y=y+x$
  3. $(x+y)+z=x+(y+z)$
  4. $x*1=x$
  5. $1*x=x$
  6. $(x*y)*z=x*(y*z)$
  7. $x*(y+z)=(x*y)+(x*z)$
  8. $(x+y)*z=(x*z)+(y*z)$
  9. $x*0=0$
  10. $0*x=0$

is a sufficient finite basis. Is this true?

Alex Kruckman
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user107952
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  • In addition to providing the link to your previous question (I edited to add it), you should at least indicate that you tried to adapt the answer given by Keith Kearnes to the case of non-commutative rings. At what point does the argument break down? – Alex Kruckman Apr 21 '21 at 23:40
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    Axioms $1$ to $10$ are exactly the axioms of a semiring. So, you want to show that the subvariety of semirings generated by the rings contains all semirings? – Geoffrey Trang Apr 22 '21 at 03:29
  • @AlexKruckman I didn't adapt the answer from Keith Kearnes. To be honest, I didn't even understand his argument on my previous question. – user107952 May 01 '21 at 16:51
  • @user107952 then why did you accept it? – Alex Kruckman May 01 '21 at 16:51
  • @AlexKruckman Because it "looked" correct. Honestly, sometimes I don't understand the answers to some of my questions, but I accept it, because I can tell it seems to answer the question. – user107952 May 01 '21 at 19:16

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