Cardinality of a ring and ring intuition My understanding this rings and their cardinality is not very good, if I make any mistakes could someone explain the misunderstanding in simple terms?
Say I have a ring $(R,+,*)$ with three elements $1_R, 0_R,a$ and we know that it is closed under multiplication and addition.
So we start of with |R|=3. Then, from the axioms their negative and inverse should also be in the ring so then |R|=6, with $0_R$ no inverse and negative is same as it self, $1_R$ same inverse as itself.
And then further on if for each of the 6 elements, if we were to add them or multiply them in some way then the result(should also be in the ring and so be a new element) and its negative and inverse would then increase the cardinality of the ring. So this process can be repeated to continuously find new elements in the ring until the result always becomes what is already in the ring, which will then be its final cardinality.
 A: The axioms never "add in" any new elements. If you start off with a ring $R$ of cardinality $\lvert R \rvert = 3,$ then that means that $R$ has three elements, say $R=\{0_{R},1_{R},a\},$ and the set $R$ with the given binary operations satisfies the axioms as given. For example, there is an element $-1_{R}\in R;$ this element is obviously not $0_{R},$ but we have not ruled out $-1_{R}=1_{R}$ or $a=-1_{R}.$
The axioms are not telling you "take the set $\{0_{R},1_{R},a\}$ and throw in a new element $-1_{R}$", but rather are telling you that somewhere in the set $\{0_{R},1_{R},a\}$ there must already exist an element which is equal to $-1_{R}$ as it is defined - otherwise, $R$ would not be a ring.
In fact, the case $a=-1_{R}$ is realisable: take $R=\mathbb Z/3\mathbb Z,$ and note that $2\equiv -1\pmod{3}.$
With a bit more work we can see that the case $1_{R}=-1_{R}$ is not realisable (in this case!). Note that $1_{R}+a\in R$ also. We cannot have $1_{R}+a=1_{R},$ as then $a=0_{R},$ contradicting the fact that $\lvert R \rvert = 3.$ Similarly, we cannot have $1_{R}+a=a,$ as then $1_{R}=0_{R},$ with the same contradiction. It follows that $1_{R}+a=0_{R},$ i.e., $a=-1_{R},$ and we are in the case previously considered.
A: Rather, you'll have $-1_R=a$ and $x=x^{-1}$ for $x\ne 0_R,$ so no new elements are added from inverses and negatives. Moreover, $1_R+1_R=a$ and $a+a=1_R,$ so no new elements are added that way, either. The fact that it is closed under multiplication and addition means that no new elements will be included by taking products and sums.
To make it concrete, $\langle R,+,*\rangle$ is (isomorphic to) the quotient ring $\Bbb Z/(3).$
