Given a ring, when can I find a field that extends it making it a subring?

Question

My question is the following one. Given a ring $A$, when can I find a field $B$ such that $A$ is a subring of $B$?

Of course, if $A$ is not a domain this is imposible because there would be two non-trivial elements in $A$ (so they would be also in $B$) with trivial product.

However, there are some examples that this can be made. For instance, if I begin with $\mathbb{Z}$ I can find a field that contains it as subring (for instance $\mathbb{Q}$).

Is this possible for any ring which is a domain? If not, is there some sufficient hypothesis to make this hold?

Thank you.

Answer 1

This may be more general than what you were searching for, but what the hey. The nicest embedding theorem that exists for rings in general says roughly this

For a ring $R$, there is a division ring in which $R_R$ embeds "densely" if and only if $R$ is a right Ore domain.

All commutative domains are Ore domains, so this characterizes the rings which can be embedded (densely even) into a field.

Unfortunately in the noncommutative case, some domains which aren't right Ore can still be embedded into division rings, just not "densely." I'm not aware of any more general results spelling out what exactly a noncommutative domain has to satisfy to be a subring of a divison ring. So it is hard to tell which ones can fit in division rings.

Indeed, there are some domains which cannot be embedded in division rings at all. The main counterexample relies on finding a monoid ring whose multiplicative group cannot be a submonoid of the multiplicative group of a division ring.

Answer 2

If the ring $R$ is an integral domain, then the field of fractions is what you're looking for.

It mimics the construction of $\mathbb Q$ from $\mathbb Z$. The elements of the field of fractions are equivalence classes of pairs $(r,s)$ where $r\in R$ and $s\in R\setminus\{0\}$ (intuitively representing $\frac rs$), under the equivalence relation $$ (r_1,s_1)\sim (r_2,s_2) \iff r_1s_2 = r_2s_1 $$ and the arithmetic operations are taken from the rules for arithmetic on fractions: $$ (r_1,s_1)+(r_2,s_2) = (r_1s_2 + r_2s_1, s_1 s_2) \\ (r_1,s_1)\cdot (r_2,s_2) = (r_1 r_2, s_1 s_2) $$ Verifying that these rules agree with the equivalence relation, and that they make the set of equivalence classes into a field is routine.

The field's $0$ is the equivalence class of $(0,1)$; $1$ is the equivalence class of $(1,1)$, and in general the map $r\mapsto (r,1)$ is an injective ring homomorphism.

Answer 3

No, in general one cannot embed a ring into a field, even if the ring is a domain, i.e., has no zero divisors. An example is the ring of real quaternions, being a non-commutative division ring. Of course, if you require in addition to no zero divisors also commutativity for the ring, i.e., require an integral domain, then there is always the field of fractions.