Commutative ring $R\neq \{0\}$ in which every subring is finite is a field question on proof

Question

Assume we are given a commutative ring $R \neq \{0\}$ with no zero divisors (not necessarily with a unit element) in which every proper subring only has finitely many elements. Show that $R$ is a field.

I am aware of the fact that there is a similar question found here, but my questions are of a different nature. First of all, the formulation "...in which every subring only has finitely many elements..." could be simply substituted with $R$ is finite. Am I right?

Then in the prove, we construct a mapping $$\rho_y : \begin{cases} R \to (y)\\x \mapsto xy\end{cases}$$ for some $y \in R\setminus \{0\}$. This is clearly a bijection (kernel is trivial by cancellation law). By $(y) \subseteq R$ we get that $R = (y)$. In the solutions in my book it is argumented that this is since $(y)$ is finite by assumption. But does this not also hold if $R$ or any of its subrings is necessarily finite?

Answer 1

Assuming you're talking about possibly nonunital rings and that the statement is about proper subrings being finite, we can observe that each ideal is a subring. Also I assume $R\ne\{0\}$.

Suppose $xR$ is proper for every $x\in R$, $x\ne0$. Being a finite commutative ring with no nonzero divisors, $xR$ is a field, so it has an identity $xe\ne0$. Then $xexr=xr$, for every $r\in R$, so $exr=r\in xR$ and $xR=R$. A contradiction.

Hence, for some $x\ne0$, we have $xR=R$. Then there is $y\in R$ with $xy=x$. In particular, $xR\subseteq yR$, so also $yR=R$ and there exists $z$ with $yz=y$.

If $r\in R$, then $ry=ryz$, so $r=rz$. Hence $R$ has an identity $1=z$.

Now, let $r\in R$, $r\ne0$. Then the minimal subring $S$ of $R$ containing $1$ and $r$ is either finite or $R$. In the first case $S$ is a field, so $r$ is invertible in $S$ and hence in $R$ (they share the identity).

Assume $S=R$. Let $P$ be the prime subring of $R$; then there is a surjective homomorphism $P[X]\to R$, sending $X$ to $r$. If this homomorphism is injective, then the image of $P[X^2]$ is a proper subring of $R$ and is infinite: a contradiction. Then the homomorphism has a nontrivial kernel and so $R$ is a finite field.

Answer 2

$R\ne \{0\}$ is a commutative ring (not necessarily with unity ) such that $R$ has no non-zero zero divisors . We show that if every proper ideal of $R$ is finite , then $R$ is a field .

Note that this is a more general statement , as every ideal is necessarily a subring (a subset of the ring which has $0$ , is closed under addition , subtraction and multiplication ) but not conversely . So we are assuming finiteness for a smaller class of structures .

Proof : If $R$ is finite , then $R$ has a unity , so $R$ becomes an integral domain and it follows that $R$ is a field . So assume $R$ is infinite . Let $S= R \setminus \{0\} $ . We know that $S$ is a commutative semigroup under the ring multiplication , so we only now need to show that $S$ is a group . It is well-known that a semigroup $(A,.)$ is a group iff it is a quasi group i.e. for every $a,b \in A , \exists$ unique $x,y \in A$ such that $ax=b , ya=b$ ; see https://groupprops.subwiki.org/wiki/Nonempty_associative_quasigroup_equals_group ( we don't actually even need the uniqueness in the quasi group condition ) . Now the uniqueness for our case is immediate from the absence of non-zero zero divisors ; so we only need to show existence , moreover we only need to show existence for one kind of equation as our semigroup is commutative . So we have to show that for every $a,b \in S , \exists x \in S$ such that $ax=b$ . So let $a,b \in S$ ; consider the map $f : R \to R$ as $f(r)=ar , \forall r \in R$ . Now $ar_1=ar_2$ implies $a(r_1-r_2)=0$ and $a \ne 0$ and $R$ has no non-zero zero divisors implies $r_1=r_2$ , hence $f$ is an injective map . So $R$ and $f(R)=aR$ are bijective as sets , and then since $R$ is infinite , the ideal $aR=f(R)$ of $R$ must also be infinite , hence $aR$ cannot be a proper ideal . So $aR=R$ , and hence $b \in R=aR$ , so $\exists x \in S$ such that $b=ax$ , and since $b \ne 0$ , so $x \ne 0$ , so $\exists x \in S$ such that $ax=b$ , and this is what we wanted to show .