Let $R$ be a nonzero commutative ring, and let $T$ be a nonempty subset of $R$ closed under multiplication and containing neither $0$ nor zero divisors. Then, we can show that the ring $R$ can be enlarged to a partial ring of quotients, called $Q(R,T)$. The following are also true:

A) $Q(R,T)$ has unity even if $R$ does not (unity is [($a, a$)]), and

B) In $Q(R,T)$, every nonzero element of $T$ is a unit.

Show that every nonzero commutative ring containing an element $a$ that is not a zero divisor can be enlarged to a commutative ring with unity.

What I did was:

Take $T$ = {$a^n$ | $n \in \mathbb{Z}^+$}, which "enlarges." This is because first, it is impossible for $a^n = 0$ since $a$ is a nonzero element, and $a^n = 0$ has no solutions. Secondly, [($a^n, a^n$)] can be reduced down to [($a, a)$] since $a*a*...*a$ is commutative in ring $R$. And finally, every element [($a^n a^n$, $a^n$)] has an inverse [($a^n, a^n a^n$)] which gives unity when multiplied (and are thus units). Is this a sufficient application of A) and B) to prove what is necessary? Am I missing anything, or misunderstanding anything? Thank you!

  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/2014883/… $\endgroup$
    – user26857
    Nov 16, 2016 at 17:46
  • $\begingroup$ I think enlarge here means $Q$ contains an isomorphic copy of $R$ $\endgroup$
    – hermes
    Nov 16, 2016 at 20:35
  • $\begingroup$ Oh, that "enlarged," yes @User1006 is correct. $\endgroup$
    – Max
    Nov 16, 2016 at 20:36
  • $\begingroup$ @Max So, have you proved that there is an injective ring homomorphism $R\to Q(R,T)$? $\endgroup$
    – user26857
    Nov 16, 2016 at 20:37
  • $\begingroup$ @user26857 It was given to us by the professor, so I haven't personally proved it. $\endgroup$
    – Max
    Nov 16, 2016 at 20:42

1 Answer 1


Define $Q(R,T)$ as follow: $$ (a,b)\in Q(R,T)\iff a\in R,b\in T,\: b\neq0\tag1 $$ For any $a,c\in R, b,d\in T,\:b\neq0,d\neq0$ ($T$ contains no zero divisor of $R$) $$ (a,b)=(c,d)\iff ad=bc\tag2 $$ $$ (a,b)+(c,d)=(ad+bc, bd)\tag3 $$ $$ (a,b)\cdot(c,d)=(ac, bd)\tag4 $$ Since $T$ contains no zero divisor of $R$, $bd\neq0$ as long as $b\neq0$ and $d\neq0$. So $(3)$ and $(4)$ always hold.

We can prove $Q(R,T)$ be a commutative ring with unity, where $(0,a)$ is zero and $(a,a)$ is unity ($a\in T,\:a\neq0$), by proving all ring axioms hold for $Q(R,T)$. For example,

  1. Commutativity of addition: $$ (a,b)+(c,d)=(ad+bc, bd)=(cb+da, db)=(c,d)+(a,b) $$
  2. Commutativity of multiplication: $$ (a,b)\cdot(c,d)=(ac, bd)=(ca,db)=(c,d)\cdot(a,b) $$
  3. Additive zero: $$ (a,b)+(0,d)=(ad+b0,bd)=(ad,bd)=(a,b) $$
  4. Multiplication unity: $$ (a,b)\cdot(d,d) = (ad,bd)=(a,b) $$ and rest follow.

In addition, for any $a\neq0,\:b\neq0$ $$ (a,b)\cdot(b,a)=(ab,ba)=(c,c) $$ Thus every nonzero element of $T$ is a unit.


To prove that $R$ can be enlarged to $Q(R,T)$, consider $Q(R,a), \:a\in T$. Clearly $$ Q(R,a)\cong R\quad\text{and}\quad Q(R,a)\subset Q(R,T) $$ Thus $R$ can be enlarged to a partial ring of quotients $Q(R,T)$.


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