Apologies. My definition of a ring need not include the multiplicative identity.
I had this problem asking if there exists rings $R_1, R_2$ such that they both have order $2$ but they are not isomorphic. So I constructed $R_1$ and $R_2$ as follows:
Let $R_1=\{0,a\}$, $R_2=\{0,b\}$. Since they are both additive groups, for a start, we must have $a+a=0_{R_1}$ and $b+b=0_{R_2}$. Then in order to avoid they being isomorphic to one another, I must have $a\times a=0_{R_1}$ and $b\times b=b.$ At this point I was trying to catch a contradiction but without any luck so I verified the axioms for being a ring and I realised they both satisfy axioms. Hence I concluded that there can be $2$ rings, both order $2$ and they are not isomorphic.
Have I gone wrong somewhere? My gut instinct is saying that there must exists some sort of contradiction. If my gut instinct happens to be false, is there a more elegant way of showing both $R_1$ and $R_2$ I constructed above are legit, rather than checking axioms?
Best Answer
If both rings are required to be unital, then of course they are isomorphic. If they are not required to be unital, then this is worth mentioning explicitly, as this is rather uncommon. In this case, you correctly note that the multiplication table of such a ring is almost entirely determined by the ring axioms; only the square of the nonzero element is not determined. It is then routine to verify that both choices satisfy the ring axioms.
You could also note that $\Bbb{F}_2=\Bbb{Z}/2\Bbb{Z}$ is a ring of order $2$, as well as the ideal $(x)\subset\Bbb{F}_2[x]/(x^2)$. The first ring is unital and the second is not, so they are not isomorphic.