We have discovered today the notion of a field of fractions $(R\setminus \{0\})^{-1}R$ where $R$ is a ring. Somehow I we did not discussed much how this is used in localizations or even more with respect to integral elements/integral closure ect. of a ring. Somehow in the statements about integral elements or integral closure they never speak about a field of fractions even though they use it in each proof.
Since this is new for me I would like to see why they use it in all the proofs, so I want to understand it by heart. Could someone help me mabye?
I would be very thankful.
The guidelines for the construction of the ring of fractions are those for the field of fractions.
Let $R$ be a (commutative) integral domain (with unit) and consider the (regular) set $S=R - \{0\}$. Let $s_1,s_2 \in S$. Since $R$ is an integral domain, $s_1s_2=0$ gives $s_1=0$ or $s_2=0.$ Thus, $S$ is closed under multiplication. Define an equivalence relation on $R \times S$ by $(r_1,s_1) \sim (r_2,s_2)$ if and only if $s_3(r_1s_2-r_2s_1)=0$ for some $s_3 \in S$. For this particular pair $(R,S)$, the latter condition can simply be stated as $r_1s_2=r_2s_1$. Together with the two operations $(r_1,s_1)+(r_2,s_2)=(r_1s_2+r_2s_1,s_1s_2)$ and $(r_1,s_1) \times (r_2,s_2)=(r_1r_2,s_1s_2)$, you obtain a field $((R,S),+,\times)$ called the field of fractions.
This procedure generalizes to more general commutative rings $R$ and multiplicative sets $S$. You retrieve the former when $R$ is an integral domain and $S=R-P$ where $P=(0)$ is the zero ideal. The ideal $P$ as defined is a prime ideal: If $r_1r_2 \in P$ then $r_1 \in P$ or $r_2 \in P$, as $R$ is an integral domain.
So the localization of a commutative ring $R$ with respect to the multiplicative set $S$ is the set $R \times S / \sim$ under these two operations. It is usually denoted $S^{-1}R$. The ring $S^{-1}R$ is a subring of the field of fractions when $S$ excludes $\{0\}$ and $R$ is an integral domain.
If $P$ is a prime ideal and $R$ a commutative ring, then $S=R-P$ is a multiplicative set, since $s_1s_2 \in P$ implies $s_1 \in P$ or $s_2 \in P$, $s_1,s_2 \in S$. Denoting the non-units in $S^{-1}R$ by $(S^{-1}R)^{*}$, you have the following:
Indeed, if $r \in S$ then $(r,s)\times(s,r)=(1,1)$ and $(r,s) \in S^{-1}R$ is a unit. Conversely, if $(r_1,s_1) \in S^{-1}R$ is a unit, then $(r_1,s_1) \times (r_2,s_2)=(1,1)$ gives $s(r_1r_2-s_1s_2)=0$ some $s \in S$. Hence, $r_1sr_2=ss_1s_2$ and $r_1 \in S$.
As such, we get that $(r_1s_2-r_2s_1,s_1s_2), (r_1r_3,s_1s_3) \in (S^{-1}R)^{*}$ whenever $(r_1,s_1),(r_2,s_2) \in (S^{-1}R)^{*}$ and $(r_3,s_3) \in S^{-1}R$. This means that $(S^{-1}R)^{*}$ form an ideal in $S^{-1}R$ and so is a local ring.