# A doubt about the isomorphism between an integral domain and its field of quotients.

I am currently reading the proof of the fact that An integral domain can be imbedded into a field.

Let $a,b,c,d\in D$, where $D$ is an integral domain.

1. $(a,b)=(c,d)$ iff $ad=bc$
2. $(a,b)+(c,d)=(ad+bc,bd)$

$(a,b)$ here stands for $a/b$

The book (Topics in Algebra by Herstein) states that the mapping $\phi (a)=[a,1]$ is an isomorphism from $D$, which is the integral domain, to the field $F$, which is the field of quotients of $D$.

I don't understand how. For example, the field of quotients of $\Bbb{Z}$ would be $\Bbb{Q}$. Clearly one equivalence class in $\Bbb{Q}$ is $[1,2]$. What element of $\Bbb{Z}$ is mapped to $[1,2]$ through this mapping?

Thanks in advance!

## 1 Answer

I'm not sure what Herstein's definitions are, but yeah, you're right that the mapping is not an isomorphism of $D$ with $F$ since it is not onto $F$. It is just an injective ring homomorphism of $D$ into $F$. (A real, surjective isomorphism would be nonsensical anyway since rings isomorphic to fields are fields, and your domain will not always be a field.)

Some authors are in the habit of calling these "isomorphisms into" with the intended meaning that they are not surjective maps. If you (like many of us) prefer to reserve the word "isomorphism" only for bijective ring homomorphisms, then you'll have to dodge Herstein's sentence :)