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!
Best Answer
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,
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.
Edit:
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 $Q(R,T)$.