Is this set a ring? I am studying Ring Theory for the first time in my life- so the following question may be a very silly one. While trying to solve an (unrelated) exercise, this question clicked in me, and it has been bugging me since then. The idea is, if a general set can be compared to a ring, then whether it is indeed a ring or not. Formally framed,

Let us have a ring $(R,+,\cdot)$ and a set $S$ equipped with two operations defined on it, namely $\oplus$ and $\odot$. We also have a bijection $f:R\to S$ which satisfies the two following properties-

*

*$f(x+y)=f(x)\oplus f(y)$

*$f(x\cdot y)=f(x)\odot f(y)$
Is $S$ also a ring?

The first thing that pops up on a closer inspection is that the way the question is framed immediately adds closure of the operations $\oplus$ and $\odot$ to the hypothesis. This is because without these operations being closed, the RHS of the two properties wont make any sense. So, the first question is, can we loosen the hypothesis up a little bit more, or frame it in a little more different way so that closure is no longer required for the very definitions to make sense (although it may be implied- afterall, that's what we want)$^*$.
The next question is whether $S$ is indeed a ring or not. If it isn't, then what is the minimum amount of information that has to added to the hypothesis for $S$ to be a ring? If it is, then can we trim some information off from the original formulation, and still keep it true?
If we interpret an isomorphism as a relabelling, it intuitively looks like $S$ should be a ring. But, the relabelling analogy only works when the ring structure of $S$ is given a priori. So, it's hard to say whether it will still work where the structure of $S$ is not immediately evident.
I'm sorry if the question sounds a little vague- I tried my best to make it as formal as I could. Also, feel free to edit the title if you can find something more suitable.

$*$ One thing I should add here which I had in mind but forgot to mention is my answer to Thorgott's comment. Instead of defining the operations from $S\times S\to S$, I planned to define them from $S\times S\to S^\prime \supset S$ so as not to force the closedness, and see whether it follows only from the axioms of $R$ and $f$.
 A: The $f$ that you defined, a priori, can only be a set-theoretic bijection $R\to S$,  because even if you named two operations on $S$, we don't know the structure of $S$ (you didn't give axioms to define how the operations work on $S$).
This situation could be what you were thinking about: suppose that you have a ring $(R,\cdot,+,1_R,0_R)$ and a set $S$. If $f:S\to R$ is a bijection (viewing $R$ as a set), we can actually give $S$ the structure of a ring. We can define the ring $(S,\otimes,\oplus,1_S,0_S)$ like follows ($s,s'\in S$):
$s\otimes s'$ is the (unique because $f$ is bijective) element $t\in S$ such that $f(t)=f(s)\cdot f(s')$; similarly, $s\oplus s'$ is the element $t\in S$ such that $f(t)=f(s)+ f(s')$; and obviously $0_S:=f(0_R)$ and $1_S:=f(1_R)$.
You may verify that, with this definition for the operations, $S$ is a ring. For example, let us verify the distributive property; we will denote with $t_r$, for any $r\in R$, the element of $S$ for which $f(t_r)=r$.  $$s_1\otimes(s_2\oplus s_3)=t_{f(s_1)\cdot f(s_2\oplus s_3)}=t_{f(s_1)\cdot f(t_{f(s_2)+f(s_3)})}=t_{f(s_1)\cdot (f(s_2)+f(s_3))}=t_{(f(s_1)\cdot f(s_2))+(f(s_1)\cdot f(s_3))}$$ $$=t_{f(s_1)\cdot f(s_2)}\oplus t_{f(s_1)\cdot f(s_3)}=(s_1\otimes s_2)\oplus(s_1\otimes s_3).$$ (Justifying all the equalities may seem confusing at first, but they are all immediate from the definitons and the distributivity of $R$).
PS: notice that $f$ must be defined $S\to R$, and not viceversa; intuitively, you can convince yourself that it is the right direction thinking that $S$ can be given a ring structure if it can be "inserted" in a ring (in fact the hypothesis of $f$ surjective could be replaced with a weaker one, while the injectivity is fundamental).
