# The Unapologetic Mathematician

## Do I smell cotton candy?

While I slept, The carnival came to town! (at The Geomblog)

## Fractions

One thing we haven’t given good examples of is fields. We can get some from factoring out a maximal ideal from a commutative ring with unit, but the most familiar example — rational numbers — comes from a different construction.

First we define a multiplicatively closed set. This is a subset $S$ of a commutative ring with unit $R$ which is, predictably enough, closed under the ring multiplication. We also require for technicality’s sake that $S$ contains the unit $1$. A good place to get such multiplicatively closed sets is as complements of prime ideals — given two elements $a$ and $b$ in $R$ but not in the prime ideal $P$, their product $ab$ must also be outside $P$. Another good way is to start with some collection of elements and take the submonoid they generate under multiplication.

In general not all the elements of $S$ will be invertible in $R$. What we want to do is make a bigger ring that properly contains (a homomorphic image of) $R$ in which all elements of $S$ do have inverses. We’ll do this sort of like how we built the integers by adding negatives to the natural numbers.

Consider the set of all elements $(r,s)$ with $r\in R$ and $s\in S$. We’ll think of this as the “fraction” $\frac{r}{s}$. Now of course we have too many elements. For example, $(s,s)$ should be “the same” as $(1,1)$ for all $s\in S$. We introduce the following equivalence relation: $(r_1,s_1)\sim(r_2,s_2)$ if and only if there is a $t\in S$ with $t(r_1s_2-r_2s_1)=0$. Notice that if $S$ contained no zero-divisors we could do away with the “there is a $t$” clause, but we might need it in general.

So as usual we pass to the set of equivalence classes and assert that the result is a ring. The definitions of addition and multiplication are exactly what we expect if we remember fractions from elementary school. Choose representatives $(r_1,s_1)$ and $(r_2,s_2)$, and define $(r_1,s_1)+(r_2,s_2)=(r_1s_2+r_2s_1,s_1s_2)$ and $(r_1,s_1)(r_2,s_2)=(r_1r_2,s_1s_2)$. From here it’s a straightforward-but-tedious verification that these operations are independent of the choices of representatives and that they satisfy the ring axioms.

We call the resulting ring by a number of names. Two of the most common are $S^{-1}R$ and $R_S$. If $S$ is generated by some collection of elements $\{x_1,...,x_n\}$ we sometimes write $R[x_1^{-1},...,x_n^{-1}]$. There are a few more, but I’ll leave them alone for now.

It comes with a homomorphism $\iota:R\rightarrow R_S$, sending $r$ to $(r,1)$. If $S$ contains no zero-divisors then this is an isomorphism onto its image, since then $(r_1,1)\sim(r_2,1)$ would imply that $r_1-r_2=0$. That is, a copy of $R$ sits inside $R_S$. This homomorphism has a nice universal property: if $f:R\rightarrow R'$ is any homomorphism of commutative rings with units sending each element of $S$ to a unit, then $f$ factors uniquely as $\bar{f}\circ\iota$. That is, $\iota:R\rightarrow R_S$ is the “most general” such homomorphism.

Now let’s say we start with an integral domain $D$. This means that the ideal $\mathbf{0}$ consisting of only the zero element is prime. Then its complement — all nonzero elements of $D$ — is a multiplicatively closed set $D^{\times}$. We construct the field of fractions $D_{D^{\times}}$ by adding inverses to all the nonzero elements. Now every nonzero element has an inverse, so this really is a field. In fact, it’s the “most general” field containing $D$.

And, finally, let’s apply this construction to the integers. They are an integral domain, so it applies. Now the field of fractions consists of all fractions $\frac{m}{n}$ with $m,n\in\mathbb{Z}$, with the above-defined sum and product. That is, it consists of the fractions we all know from elementary school. We call this field $\mathbb{Q}$: the field of rational numbers.

May 19, 2007 Posted by | Ring theory | 3 Comments