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This is example I-22 in Eisenbud's Geometry of Schemes.

Let $K$ be a field and $R=K[x]_{(x)}$ the localization of $K[x]$ at the maximal ideal (x). So $R$ is basically $K(x)$ except the elements where the denominator has a non-zero constant term. In the book it is written that $Spec(R)=\{(0), (x)\}$, but how can $(x)$ be a prime ideal of $R$, since it clearly contains $1$ and hence is equal to $R$, which is by definition not a prime ideal?

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" can $(x)$ be a prime ideal of $R$, since it clearly contains $1$..." I think you're confused about what $K[x]_{(x)}$ is. It consists of the fractions in $K(x)$ of the form $f/g$ where $g\notin(x)$, i.e., where $g$ has a non-zero constant term. So, in particular, $x$ does not get inverted in $K[x]_{(x)}$. The multiplicative set at which you're localizing is the complement of the prime ideal $xK[x]$. – Keenan Kidwell Dec 10 '11 at 19:51
up vote 3 down vote accepted

The ideal $xR$ is proper, and you can prove it directly: If $xf(x)/g(x) = 1$ with $f(x), g(x) \in K[x]$, then $xf(x) = g(x)$; however, as you said, in $R$ our denominators must have non-zero constant terms.

For algebraic geometry, it's important to know that $\mathfrak{p} \mapsto S^{-1}\mathfrak{p} = \mathfrak{p}(S^{-1}A)$ is a bijection between the primes of $A$ not meeting $S$ and the primes of $S^{-1}A$. This is Proposition 6.4 of Milne's notes and should be done in any commutative algebra textbook. [I think Atiyah-Macdonald is still horribly expensive, but that's the one I'd recommend.]

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