# Localization 1

I’m calling this “Localization 1″ since it is technically the beginning of the construction that we eventually want. But for now, I don’t even want to define what it means for a ring to be local, or why this may or may not be a (the) “localization.”

Last time we formed the field of fractions, but it came with some restrictions that we don’t need (as long as we don’t care that the end result isn’t a field). So today we will be forming the “ring of fractions.” Let’s think about what made that construction work. Essentially we didn’t want to “divide” by things that could end up as zero. This was the restriction that R be a domain. So let’s throw out that restriction, and just make sure there are no zero divisors. We also didn’t really care about the ring structure at all for the denominator. We only cared that it was closed under multiplication.

Let $S$ be a multiplicative set with 1 (it is closed under mult.) with no zero divisors. Now do the same exact construction as before on $\{(r,s)\in R\times S\}$. We call the set of equivalence classes the ring of fractions and denote it $S^{-1}R$. It is straightforward computation to check that this is still a ring.

Now the only reason I included 1 in S was so that there is a natural embedding $\phi :R\hookrightarrow S^{-1}R$ given by $r\mapsto r/1$. This embedding tends to be really nice for proving certain properties about the ring of fractions. One of these properties is exactly what we would hope: the embedding sends ideals to ideals, i.e. if $I\subset R$ an ideal such that $I\cap S=\emptyset$, then $\phi(I)\subset S^{-1}R$ is an ideal (with this being a bijection between prime ideals).

So the preimage under the embedding turns out to be fairly important. What is it on the element level? Well $\phi^{-1}(S^{-1}I)=\{x\in R : \exists s\in S, \ sx\in I\}$. And without giving too much away, we get the nice result that if $I$ is prime and $I\cap S=\emptyset$, then $\phi^{-1}(S^{-1}I)=I$. Then we get that if $P\subset R$ prime, then $S^{-1}P\subset S^{-1}R$ is prime. So maybe there will be some importance to the notion of prime later.

Now let’s backtrack and try to do this construction one more time (OK, so it will come up again next time, but…). We really, really don’t want to have to restrict the set S to have no zero divisors. This is just too pesky of a condition. Can we somehow get around it?

This time let’s suppose $S$ is multiplicative with 1, and let’s only assume $0\notin S$. There can be zero divisors, though. Let $I=\{x\in R: \exists s\in S, \ sx=0\}$. Then $I$ is an ideal. Clearly if $x,y\in I$ then pick $s,t\in S$ such that $sx=0$ and $ty=0$. Then $st\in S$ and $st(x-y)=stx-sty=0$, so $x-y\in I$. Also $x\in I$ and $r\in R$, (since we’ve assumed commutativity) we clearly have $rx\in I$.

Since we have a perfectly good ideal, and this ideal in a sense mimics the property of zero divisors, we do the natural thing and mod out by it. Thus form $R\to R/I=\overline{R}$, and $S\to \overline{S}$. Then $\overline{S}$ has no zero divisors in $\overline{R}$. Suppose there was one, say $a\in \overline{S}$. Then there exists $r\in \overline{R}$ such that $ra=0$. By definition of the “bar” we have that $a=s+I$ for some $s\in S$ and $r=r_0+I$ for some $r_0\in R$ (and $0=I$), so $(s+I)(r_0+I)=sr_0+I=I$ so $sr_0=0$ which means that s itself was a zero divisor. But trivially all zero divisors are sent to 0 under the mapping, so $a=0\in \overline{S}$.

So we form $S^{-1}R$ by first canonically factoring through $\overline{S}^{-1}\overline{R}$. Why all this trouble? In essence what we are trying to do is form a new minimal ring that contains the old ring such that every element is a unit. If the kernel of the mapping is not {0} (i.e. there were zero divisors in the set), then the mapping is not an embedding and so the new ring doesn’t contain the old ring. The astute reader will note that this almost seems like a universal construction. This is what we’ll do when I actually get to localization for real.