A ring embeds in a field iff every finitely generated sub-ring does it I want to prove that: if $R$ is a ring, $R$ embeds in a field iff every finitely generated sub-ring of $R$ embeds in a field. 
I think that the first implication is trivial because if I have an embedding and I take a finitely generated sub-ring of $R$, I can take the restriction of the embedding to the sub-ring and obtain an embedding of the sub-ring in the field.
For the other implication I think I should use the Compactness theorem (as we are talking about finitely something then something) but I don't know how to use it!  
Is my proof of the first implication correct? Any advice for the other? 
 A: Let $T$ be a first-order $\mathcal{L}$-theory, and let $M$ be an $\mathcal{L}$-structure. 
If $M$ embeds in a model of $T$ by $f\colon M\to K$, then every finitely generated substructure $M_0\subseteq M$ embeds in $K$ by composing $f$ with the inclusion $M_0\to M$.
Conversely, suppose every finitely generated substructure of $M$ embeds in a model of $T$. Then for every finitely generated substructure $M_0\subseteq M$, $\text{Diag}(M_0)\cup T$ is consistent. By Compactness, $\text{Diag}(M)\cup T$ is consistent, since every finite subset $\Delta\subseteq \text{Diag}(M)$ is contained in $\text{Diag}(M_0)$, where $M_0$ is the substructure of $M$ generated by the finitely many elements of $M$ mentioned in $\Delta$. So $M$ embeds in a model of $T$.
Now your problem is just the special case where $\mathcal{L} = \{0,1,+,-,\times\}$, $T$ is the theory of fields, and $M$ is a ring. As Qiaochu notes in the comments, the example of fields and rings is not such a good one, since you can solve this special case directly without using Compactness. Every integral domain embeds in its field of fractions. And conversely, since fields don't have zero divisors, neither do their subrings. So if a ring $M$ fails to embed in a field, then there exist $a\neq 0$ and $b\neq 0$ in $M$ such that $ab = 0$. This is also true in the subring $M_0\subseteq M$ generated by $a$ and $b$, so $M_0$ is not an integral domain, so $M_0$ also fails to embed in a field.
