# Unity in a partial ring of quotients

I have the following question

If we let $R$ be a commutative ring and we let $T$ be a nonempty subset of $R$ closed under multiplication not containing zero nor divisors of zero, i am asked to show that $R \times T$ can be made into a partial ring of quotients for $R$, and i need to show that this partial ring of quotients $Q(R,T)$ can have unity even if $R$ does not, now i was thinking about using the element $(x,x)$ to show that $(a,b)(x,x)=(ax,bx)$ and then $(ax,bx)$ ~ $(a,b)$ by just noting that $axb=abx=bax$. Is this correct?

Thank you

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Yes; you may want to note that the choice of $x$ does not matter (that is, if $t,t'\in T$, then $(t,t)$ and $(t',t')$ give the same element of the partial ring of quotients). –  Arturo Magidin Sep 8 '11 at 19:30