Let $\mathbb{K}$ be a field. The domain of $f = \frac{g}h\in \mathbb{K}(x), $ where $g,h\in \mathbb{K}[x]$ is defined to be the set $Dom(f) := \{c\in \mathbb{K} : h(c) \neq 0\}.$ Let $S(c) := \{f\in \mathbb{K}(x) : c\in Dom(f)\}$ for $c\in \mathbb{K}.$ Show that for every $c\in \mathbb{K},$ $\mathbb{K}[x]_{(x-c)}\cong S(c),$ where $(x-c)$ is the smallest ideal of $\mathbb{K}[x]$ containing $x-c$ and $\mathbb{K}[x]_{(x-c)}$ is the localization of $\mathbb{K}[x]$ with respect to $(x-c)$ and its elements can be thought of as fractions with numerators in $\mathbb{K}[x]$ and denominators in $\mathbb{K}[x]\backslash (x-c).$
I know that localizations are unique in the sense that if $R$ is a commutative ring, $S$ is a multiplicatively closed subset of $R$, and $Q$ is a commutative ring so that there exists an injective homomorphism $\phi : R\to Q$ so that
- $\phi(x)$ is a unit of $Q$ for every $x\in S$ and every element of $Q$ is of the form $\phi(x)\phi(y)^{-1}$ for $x\in R, y\in S$
- If $\psi : R\to T$ is a morphism so that $\psi(x)$ is a unit of $T$ for all $x\in S,$ there exists a morphism $\tilde{\psi} : Q\to T$ so that $\tilde{\psi}\circ \phi = \psi,$
then $Q$ is isomorphic to the localization of $R$ at $S.$ So I think one way to prove part $3$ is to show these two conditions are satisfied for $S(c).$
Alternatively, is there a more "direct" approach to this problem? I was thinking of defining the map $\phi : \mathbb{K}[x]_{(x-c)}\to S(c), \phi(\frac{a}b) = \frac{a}b$. However, it seems that showing that this map is an isomorphism would also show that $S(c) = \mathbb{K}[x]_{(x-c)}.$ I know $S(c)$ is a subring of $\mathbb{K}(x)$ and hence is commutative. Let $R = \mathbb{K}[x], S = \mathbb{K}[x]\backslash (x-c).$ Define $\phi : R\to S(c)$ as follows. $\phi(x) = \frac{x}1.$ Observe that this map is clearly injective. Since for $f=g/h\in S(c), h(c)\neq 0, h\not\in (x-c)\Rightarrow h\in S.$ So $f = \phi(g)\phi(h)^{-1}.$ Then define $\tilde{\psi} : S(c)\to T$ so that $\tilde{\psi}(\frac{a}b) = \psi(a)\psi(b)^{-1}.$ Then one can verify that $\tilde{\psi}$ has the required properties.