If $R = k[x]_{(x)}$, why is $R_{(0)}\cong k(x)$?

Question

I'm having some problems understanding why this should be true. I don't have a strong background in ring theory and I'm not even sure if $(0)$ is a prime ideal of $R$. I would like to know how to prove the isomorphism in the title.

Answer 1

Notice that $R$ is an integral domain, since it is a subring of the field $k(x)$. Therefore, your question follows from the following result.

Claim: if $R$ is an integral domain, then $(0)$ is a prime ideal of $R$ and $R_{(0)} \cong \mathrm{Frac}(R)$.

Indeed, $(0) \subset R$ is a prime ideal, because if $xy \in (0)$, then either $x=0$ or $y=0$ (since $R$ is an integral domain). Now, $R_{(0)} = (R \setminus \{0\})^{-1}R$, by definition. This is precisely the fraction field of $R$, which is $k(x)$ in your case.

Answer 2

$0$ is a prime ideal since $k[x]$ (and hence $R$) is an integral domain. By definition of integral domain, $ab = 0 \implies a=0$ or $b= 0$, that is $a \in (0)$ or $b \in (0)$. Localizing at $0$ is formally inverting all nonzero polynomials, that is, elements of $R_{(0)}$ are of the form $$\frac{f(x)}{g(x)}$$ For $f, \ g \in R$, both nonzero. Then, note that the composition $k [ x] \to R_{(0)}$ has the property that all nonzero elements are units in the image. Recall that we have a universal property of fraction fields in this case, and this should be enough to conclude the result.