Is every commutative ring contained in a field?

Question

I imagine the answer to this question is very simple, but I haven't been able to locate it.

Can every commutative ring be imbedded in a field? This seems very plausible to me, for it seems we could just take the "closure" under inverses.

Thanks!

score 21 · Accepted Answer · answered Dec 06 '17 at 17:56

21

A commutative ring can be embedded in a field iff it is an integral domain.

Indeed, if a ring can be embedded in a field then it cannot have zero divisors because fields cannot have zero divisors.

Conversely, every integral domain can be embedded in a field, namely, its field of fractions.

answered Dec 06 '17 at 17:56

lhf

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Does this generalize to the non-commutative case? I mean, is it true that a ring can be embedded in a skew field if and only if that ring has no zero divisors (neither left zero divisors nor right zero divisors)? – Jeppe Stig Nielsen Dec 07 '17 at 12:01
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@JeppeStigNielsen, perhaps ask a separate question. But first see https://math.stackexchange.com/questions/336893/can-we-embed-cancellative-rings-into-division-rings. – lhf Dec 07 '17 at 12:25
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That is a nice reference, and it seems there is no need for me to repeat their Math SE question. Reading the two answers currently there, I see (1) that Maltsev (Malcev, Mal'cev, Ма́льцев) already proved the answer is "no", (2) that the Ore condition (after Ø. Ore) is what extra hypothesis we need to make this work. The Wikipedia Ore condition article I link, under section Application, also gives a clear answer to the question I raised. – Jeppe Stig Nielsen Dec 07 '17 at 13:34

anomaly · Answer 2 · 2017-12-06T23:11:02.427

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No. Every nonzero element of a field is invertible; commutative rings can contain zero divisors.

The "closure" operation you're talking about is localization. In a (commutative) domain $A$, localizing at $0$ gives you the field of fractions $\operatorname{Frac}(A)$ of $A$, and the canonical map $A \to \operatorname{Frac}(A)$ is an embedding. In general, the localization is a more complicated ring than just a field.

edited Dec 06 '17 at 23:11

answered Dec 06 '17 at 17:53

anomaly

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ah, that makes sense. Thank you. – Van Latimer Dec 06 '17 at 17:54
Why the downvote? – anomaly Dec 08 '17 at 17:52

score 3 · Answer 3 · answered Dec 07 '17 at 09:24

Let $R$ be a commutative ring. Define $\bar Q(R)=R\times (R-\{0\})$ as the set of pairs $(r,s)$ with $r,s\in R$ and $s\not=0$. We define the equivalence relation $$(r,s)\sim(r',s')\quad:\Leftrightarrow\quad rs'=sr'.$$ We define $Q(R):=\bar Q(R)/\!\!\sim$ and give it a ring structure (denoting by $x$ the equivalence class that contains $x$) via

$$(r,s)+(r',s')=(rs'+sr',ss'),\qquad (r,s)\cdot (r',s')=(rr',ss').$$

This turns out to be well-defined if and only if $R$ was an integral domain. Then $Q(R)$ will be a field $-$ the field of fractions. Think of $(r,s)\in Q(R)$ as the fraction $r/s$ or the element $rs^{-1}$. $R$ can be embedded into $Q(R)$ via $r\mapsto(rx,x)$ for arbitrary $x$. The zero and unity of $Q(R)$ are $(0,x)$ and $(x,x)$ respectively. A non-zero element $(r,s)$ has the inverse $(s,r)$.

Something seems to be not okay with my answer (therefore the downvote?). I can improve or remove it if it will be explained what is the problem. — M. Winter, Dec 08 '17 at 13:46
There's a downvote for each of these answers, though they all seem perfectly fine to me. Maybe just a wandering troll. — anomaly, Dec 08 '17 at 17:53

Is every commutative ring contained in a field?

3 Answers3