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Let $A$ be a ring and $\mathfrak{p} \subset R$ be a prime ideal. Set $A_\mathfrak{p}=R[U^{-1}]$, where $U= A-\mathfrak{p}$. What is the relation between $A/\mathfrak{p}$ and $A_\mathfrak{p}$?

My motivation for this question comes from the definition of affine scheme

This is from the Notes of Gathmann

Remark 5.1.3. Any element $f \in A$ can be considered to be a “function” on $Spec(A)$ in the following sense: for $\mathfrak{p} \in Spec(A)$, denote by $f(\mathfrak{p})$ the image of f under the composite map $A → A/\mathfrak{p} → k(\mathfrak{p})$.

Now the structure sheaf of an affine scheme is defined as follows

For an open set $U \subset Spec(A)$, we define $O(U)$ to be set of functions $s:U \rightarrow \coprod A_\mathfrak{p}$ such that $s(\mathfrak{p}) \in A_\mathfrak{p} $ and require that for each $\mathfrak{p} \in U$ there is a neighbourhood $V$ of $\mathfrak{p}$, contained in $U$ and elements $a,f \in A$ such that for each $q \in V$ and $s(q)=a/f$.

Now my question is can $S(q)$ be interpreted as $a(q)/f(q)$ using the Remark 5.1.3? Does this even make sense? Any ring $A$ as a ring of functions on the prime spectrum $Spec(A)$. Can this interpretation be expanded to the localizations the ring $A$? Can they be thought of as rings of some sort of "rational functions"?

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Why do you expect there to be any relation between them? –  Tobias Kildetoft Feb 12 '13 at 14:06
    
I've expanted my question to include my motivation for this question. –  Mohan Feb 12 '13 at 17:58
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3 Answers 3

I can't see any direct, straightforward general relation. For example, take

$$R=\Bbb Z\;\;,\;\;\mathfrak p:=2\Bbb Z\;\;,\;\;\Bbb Z_\mathfrak{p}:=\left\{\frac{a}{b}\in\Bbb Q\;;\;b\notin\mathfrak p\right\}\;\;,\;\;R/\mathfrak p=\Bbb Z/2\Bbb Z$$

The localization gives us an infinite subring of the rationals, whereas the quotient ring is the smallest possible (non-zero) field...

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The difference between the two notions you describe is sort of like the difference between, in ordinary real-valued calculus, "the value of $f$ at $P$" and "the Taylor series of $f$ at $P$"; while the latter is a value of some sort at $P$, it contains enough information to reconstruct $f$ in an entire open neighborhood of $P$. (I'm assuming $f$ is analytic in the analogy)

EDIT: for your new question, the answer is yes. Let $f \in A$. For the open set

$$ U = D(f) = \{ \mathfrak{p} \mid f \notin \mathfrak{p} \} = \{ \mathfrak{p} \mid f(\mathfrak{p}) \neq 0 \} $$

you have $\mathcal{O}(U) \cong A[1/f]$. In fact, the open subscheme of $\mathop{\text{Spec}} A$ defined by $U$ is isomorphic to $\mathop{\text{Spec}} A[1/f] $.

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thanks for the reply. But can we interprete $S(q)$ as $a(q)/f(q)$ using the Remark 5.1.3. In the above definition $S(q)=a/f \in A_q$ but $a(q)/f(q) \in k(q)$? So, can by some modification can they be made equal? –  Mohan Feb 12 '13 at 20:09
    
Well, there's the obvious "evaluate $s(\mathfrak{q})$ at $\mathfrak{q}$".... –  Hurkyl Feb 12 '13 at 20:14
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At least, there is an injective map $\mathrm R / \mathfrak{p} \to \mathrm{R}_{\mathfrak p}/\mathfrak p \mathrm{R}_{\mathfrak p}$.

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Right, so $R / \mathfrak p$ embeds in the residue field of the localization. –  JSchlather Feb 12 '13 at 14:52
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Actually the RHS is the field of fractions of the LHS. –  user18119 Feb 12 '13 at 18:02
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