# The Field of Quotients of An Integral Domain

according to Fraleigh,

in the Field of Quotients of An Integral Domain,

( Let D be an integral domain and form the Cartesian product D x D = {(a,b)} | a,b $\in$ D} and S = {(a,b) | a,b $\in$ D, b is not 0},

let (a,b) , (c,d) are equivalent iff ad= bc and define F tobe the set of all equivalence classes [(a,b)] for (a,b) $\in$ S.)

[(-a,b)] is an additive inverse for [(a,b)] in F.

But, [(-a,b)] + [(a,b)] = [ (ab-ba, b$^2$ ] = [ (0, b$^2$ ]

how is this equal to [(0.1])?

Also, I'm not sure how to show the distributive laws hold in F.

I tried to show [(a,b)] ( [(c,d)] + [(e,f)] ) = [(ac, bd)] + [(ae, bf)] but they are clearly different, since the left side equals to [(adf+bcf+bde,bdf)] while the right side equls to [(acbf+bdae, bdbdf)]

• Please define your notation: What is $F$, and what is $[(a, b)]$? – user61527 Aug 1 '13 at 17:22
• @T.Bongers I just edited my question! – InfimumMaximum Aug 1 '13 at 17:26
• recall that $(a,s) = (b,t)$ precisely when the usual cross multiplication works (for a field of fractions). that is to say, $at = bs$. in your case, $(0, b^2) = (0, 1)$, as $0 \cdot 1 = 0 \cdot b^2$ – citedcorpse Aug 1 '13 at 17:26
• for clarity: putting the square brackets around the tuple means you want things understood up to the equivalence relation. so certainly $(0, b^2) \neq (0,1)$ as tuples, but they are equal up to the equivalence relation, and that's what we care about – citedcorpse Aug 1 '13 at 17:27

The definition of $(a,b)\sim(c,d)$ you were given here for domains was probably $ad=bc$. When two pairs are similar, the classes are the same: $[(a,b)]=[(c,d)]$.

So, is $(0,s)\sim(0,1)$ for any $s\in D\setminus \{0\}$?

Another good thing to establish is that $(a,b)\sim(at,bt)$ for any $b,t\neq 0$, and any $a$, giving you the equality $[(a,b)]=[(at,bt)]$.

For distributivity, you need to check your work. It looks like there were some weird mistakes. You should have $acf+ade$ in the "numerator" of the left-hand side. You also have one too many $d$'s in the "denominator" of the right-hand side.

After you've fixed the mistakes, the "another good thing" I mentioned should carry you home.

The equivalence relation in the field of fractions as you are defining it is

$$(a,b)\sim (c,d)\Longleftrightarrow ad=bc$$

Notice this shows that $[(0,b)]=[(0,1)]$ for any $b\ne 0$. As far as the distributive law, we have:

$$[(a,b)]([(c,d)]+[(e,f)])=[(a,b)][(cf+ed,df)]=[(acf+aed,bdf)]$$

and

$$[(ac,bd)]+[(ae,bf)]=[(acbf+abde,b^2df)]$$

See if you can show, using the equivalence relation, that the two right hand sides are equivalent.