Do I Have to Show All Axioms to Prove a Set is a Ring?

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SUMMARY

To prove that a set \( S \) is a ring, it is necessary to demonstrate all the ring axioms. While many of these axioms are straightforward, each must be verified as they can sometimes yield unexpected results. In contrast, to establish that a set is a subring, it suffices to show only the criteria specific to subrings, leveraging the fact that a subring is a subset of a ring where all ring axioms are already satisfied.

PREREQUISITES
  • Understanding of ring theory and its axioms
  • Familiarity with the definitions of rings and subrings
  • Basic knowledge of set theory
  • Experience with mathematical proofs
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  • Study the complete set of ring axioms in abstract algebra
  • Learn the criteria for identifying subrings
  • Explore examples of rings and subrings in various mathematical contexts
  • Practice constructing proofs for sets being rings or subrings
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Mathematics students, educators, and anyone studying abstract algebra who seeks to deepen their understanding of ring structures and their properties.

evinda
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Hey again! :)
I have a question..
If I have to show that a set $S$ is a ring,do I have to show all the axioms or is it enough to show the criteria:
$s_1,s_2 \in S$ and
  • $s_1-s_2 \in S$
  • $s_1 \cdot s_2 \in S$
?
 
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evinda said:
Hey again! :)
I have a question..
If I have to show that a set $S$ is a ring,do I have to show all the axioms or is it enough to show the criteria:
$s_1,s_2 \in S$ and
  • $s_1-s_2 \in S$
  • $s_1 \cdot s_2 \in S$
?

Welcome back! ;)

I'm afraid you have to show all the axioms.
Luckily most axioms are often enough fairly trivial.
Still, you have to show all of them, and every now and then one of them holds a surprise, turning everything upside down.

It's a different matter if you have to show something is a subring.
Then you only have to show what is required for a subring.
Effectively you are borrowing from the fact that it is a subset of a ring for which all the ring axioms are already satisfied.
 
I like Serena said:
Welcome back! ;)

I'm afraid you have to show all the axioms.
Luckily most axioms are often enough fairly trivial.
Still, you have to show all of them, and every now and then one of them holds a surprise, turning everything upside down.

It's a different matter if you have to show something is a subring.
Then you only have to show what is required for a subring.
Effectively you are borrowing from the fact that it is a subset of a ring for which all the ring axioms are already satisfied.

I understand...Thanks! :)
 

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