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Let $R$ be an integral domain and $P,Q$ be proper prime ideals of $R$. Let $R_P,R_Q$ be localizations of $R$ at $P,Q$. If $R_P\subseteq R_Q$, is $Q\subseteq P$?

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Presumably, you mean $\subseteq$ when considered as subsets of the field of fractions of $R$? –  Thomas Andrews Jan 14 '13 at 19:03
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up vote 2 down vote accepted

More generally, if $R$ is a commutative ring, then $R$-algebra homomorphisms $R_P \to R_Q$ exist and are unique - by the universal property of localizations - iff $R \to R_Q$ maps $R \setminus P$ to $(R_Q)^* = R_Q \setminus Q R_Q$ iff (since $Q R_Q \cap R = Q$) $R \setminus P \subseteq R \setminus Q$ iff $Q \subseteq P$.

More generally, if $X$ is a scheme with points $x,y$, then $x$ is a specialization of $y$ iff there is a morphism of $X$-schemes $\mathrm{Spec}(\mathcal{O}_{X,y}) \to \mathrm{Spec}(\mathcal{O}_{X,x})$.

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thank you very much Martin –  Azadeh Jan 20 '13 at 9:49
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Take any element $q\in Q$. If it was not it $P$, then it would be invertible in $R_P$, but not in $R_Q$, which would contradict the fact that $R_P\subseteq R_Q$.

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Thank you very much –  Azadeh Jan 14 '13 at 19:20
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