# Given an integral domain $R$ and a subset $S$ of $R$, does there exist a smallest subfield of $R$ that contains $S$?

Given an integral domain $R$ and a subset $S$ of $R$, does there exist a smallest subring $T$ of $R$ such that $T$ is a field and $S \subseteq T$ ? ($S$ need not be a subring of $R$.)

I proved that this is true if there exists at least one subring $F$ of $R$ such that $F$ is a field and $S \subseteq F$. In that case, the smallest subfield $T$ is $\cap_{F \in \mathcal{F}} F$, where $\mathcal{F} = \{F | F$ is a subring of $R$, $F$ is a field, and $S \subseteq F$$\}$. But I don't know if this is true even when there is no subring $F$ of $R$ such that $F$ is a field and $S \subseteq F$.

I need the answer to this question because of the following reason: Let $R$ be an integral domain and let $F$ and $S$ be a subfield and a subset of $R$, respectively. I learned that $F(S)$ means the smallest subfield of $R$ that contains $F$ and $S$. But I want to know if $F(S)$ always exists.

• Tried the integers? – Pedro Tamaroff Jul 25 '18 at 3:12
• @PedroTamaroff Oh, I got it. If R is Z and S is {2}, then there is no such subfield. Thank you! – zxcv Jul 25 '18 at 3:19

There does not always exist such a subring $T$. For example, take $R = \mathbb{Z}$ and $S$ to be any subset, say $S = \{ 2 \}$. There is clearly no subring of $\mathbb{Z}$ that is a field, and in particular no subring of $\mathbb{Z}$ that is a field and contains $2$.