I just proved that a commutative ring $R$ is an integral domain iff $R$ is isomorphic to a subring of a field.

My question is why can't $R$ be a field with these conditions? Aren't we satisfying all of the field's properties?


  • 3
    $\begingroup$ Being a sub-thing of a field is pretty different from being a field. The integers are subring of the reals, but can you invert 2 and stay in the integers? $\endgroup$
    – Randall
    Commented Sep 1, 2017 at 13:42
  • $\begingroup$ It is not excluded that under these conditions $R$ is a field. But also not guaranteed. It might be that elements of $R$ have no inverse. Looking at it as subring of field $F$ that means that their inverse is in $F-R$. $\endgroup$
    – drhab
    Commented Sep 1, 2017 at 13:45

3 Answers 3


For a counter-example, let's have a look at $\mathbb{Z} \subseteq \mathbb{Q}$.
Here $\mathbb{Z}$ is an integral domain which is not a field;
also you can check that $\mathbb{Z}$ is a sub-ring of the field of rational numbers $\mathbb{Q}$.

Note that $\mathbb{Z}$ satisfies all of the field's properties; except the property which concerns the
existence of multiplicative inverses for non-zero elements. For example $2^{-1} \notin \mathbb{Z}$ .

  • $\begingroup$ I overlooked that example I guess. I think it is because I kept thinking about finite integral domains for some reason :(. Don't know why I couldn't wring myself out of there... Thanks. $\endgroup$
    – Friedman
    Commented Sep 1, 2017 at 14:07
  • 6
    $\begingroup$ @Friedman By the way, in case you don't already know, a finite integral domain $\textit{is}$ always a field. $\endgroup$
    – wgrenard
    Commented Sep 1, 2017 at 14:14

No, subring of a field does not satisfy all the field's axioms. Namely, the problem is twofold: the subring doesn't have to contain $1$ and even when it does, there is trouble with inverses.

Let $1\in R\subseteq \mathbb F$, where $\mathbb F$ is a field and let $0\neq r\in R$. Sure, $r$ is invertible in $\mathbb F$, but what guarantees that $r^{-1}\in R$? Well, nothing. See the answer by Famke where they give example of $\mathbb Z\subseteq \mathbb Q$.

  • $\begingroup$ "The subring doesn't have to contain $1$" is not necessarily a problem. If you work with unitary rings such problem doesn't exist. $\endgroup$
    – Xam
    Commented Sep 1, 2017 at 20:05
  • $\begingroup$ But if you don't it does. Anyway, it's obvious that the emphasis was put on inverses, @Xam. $\endgroup$
    – Ennar
    Commented Sep 1, 2017 at 20:07

A yet another example are polynomials $R[x]$ over a field $R$. While (obviously) all other properties are fine, there are much less inverses than there are required for a field.

Now formal power series offer somewhat a remedy...


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