Commutative rings among whose proper subrings (with unity) , there exist a unique field

Question

Let $R$ be a commutative ring with unity , let us say $R$ has property P if the following holds : There exists a unique field among all proper subrings of $R$ (with the same unity as that of $R$) .

My question is : Let $R$ be a ring with property P which is not a field ; then does $R[X]$ also has property P ?

Note that it is necessary to assume $R$ is not a field . Otherwise say $R=\mathbb Q(\sqrt 2)$ has property P but clearly $R[X]$ doesn't .

If $R$ has non non -zero nilpotent then $R[X]^{\times}=R^{\times}$ and $R$ has property P and if $F$ is the unique proper subfield of $R$ , then $F$ is also a subfield of $R[X]$ . If $K$ is any other subfield of $R[X]$ , then $K \setminus \{0\} \subseteq R[X]^{\times}=R^{\times}$ implies $K \subseteq R$ and $R$ is not a field and $R$ has property P then implies $K=F$ . So if $Nil(R)=\{0\}$ and $R$ is not a field and has property P, then so does $R[X]$ .

This is what I have observed about rings with property P : $R$ properly contains $\mathbb Q$ or $\mathbb Z_p$ ( for some prime $p$ ) as a subring according as $char R=0$ or positive ; let us call this unique subfield of $R$ the prime subfield . If $R$ is an integral domain , and we consider $R$ as a subring of its field of fractions, then either every element of $R\setminus F $ ( this is non-empty) is transcendental over $F$ or some $a \in R\setminus F$ is algebraic over $F$ , in which case $R=F[a]$ is a field and so $[R:F] < \infty$ ; where $F$ is the prime subfield of $R$. If $R$ it self is a field with prime subfield $F$ , and $t \in R \setminus F$ is transcendental , then $F \subseteq F(t^2) \subseteq F(t) \subseteq R$ and $F \ne F(t^2) , F(t) \ne F(t^2)$ shows then $R$ has more than one proper subfield ; hence no element in $R$ can be transcendental over $F$ if $R$ is a field , and $R=F[b]=F(b) , \forall b \in R \setminus F$ . So moreover , if $R$ is a field of positive characteristic , then $R$ is a finite extension over $\mathbb Z_p$ , so let $R \cong F_{p^n}$ , but then property P implies $n=[R : \mathbb Z_p]$ has to be a prime . Hence a field $R$ of positive characteristic has property P if an only if $R \cong F_{p^q}$ for some primes $p$ and $q$ , where $p=char R$ .

Answer 1

Edit: OP points out that this argument only works if $R$ has no non-zero nilpotents.

Yes, because if $F\subset R[X]$ is a field with the same unity as $R[X]$ (and thus the same one as $R$), then in fact $F\subset R$ because the only invertible elements in $R[X]$ are in $R^*$ (the units of $R$) and every non-zero element in $F$ needs to be invertible. So $F$ is unique, since $R$ has property $P$.

More generally if you want your ring $R$ to have the property $P$, you want the largest field they contain to be either $\mathbb Q$ or a unique $\mathbb F_p$ a finite field.

Answer 2

I think a ring $R$ has property $P$ if and only if $R^{\times}$ is (isomorphic to) either $\mathbb{F}_p^*$ or $\mathbb{Q}^*$.

Indeed if $R$ has property $P$, denote $k$ its unique subfield. Because $k$ cannot contain any subfield, it's either $\mathbb{F}_p$ or $\mathbb{Q}$ (depending on its characteristic). Then I claim that $k^* = R^{\times}$ : we know that $k^* \subset R^{\times}$, and if $x \in R$ is an invertible element outside of $k$, then $k(x)$ is another subfield of $R$ (which is impossible, so no such $x$ exist). So we have one way.

To prove the other implication, we assume that $R^{\times}$ is either $\mathbb{F}_p^*$ or $\mathbb{Q}^*$. Now if $k$ is a subfield of $R$, then $k^*\subset R^{\times}$ but since the fields $\mathbb{F}_p$ or $\mathbb{Q}$ have no proper subfield, then $k^*=R^{\times}$ and we're done.

Now using this criterion, I think we can prove that $R[X]$ inherits property $P$ since $R[X]^{\times}=R^{\times}$.