0
$\begingroup$

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)]

Give me some help please!

$\endgroup$
  • $\begingroup$ Please define your notation: What is $F$, and what is $[(a, b)]$? $\endgroup$ – user61527 Aug 1 '13 at 17:22
  • $\begingroup$ @T.Bongers I just edited my question! $\endgroup$ – InfimumMaximum Aug 1 '13 at 17:26
  • $\begingroup$ 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 $0 \cdot 1 = 0 \cdot b^2$ $\endgroup$ – citedcorpse Aug 1 '13 at 17:26
  • 1
    $\begingroup$ 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 $\endgroup$ – citedcorpse Aug 1 '13 at 17:27
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.