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After to post this question: Help with a proof in Hartshorne's book I realized that I have another doubt in this proof:

What I know is

$\varphi_p(s_p)=t_p\implies (\varphi_{V_p}(s(P))_p=t_p$

I don't understand why $(\varphi_{V_p}(s(P))_p=(t_{|V_P})_p$

NOTATION:

Instead of $\varphi(V_P)$, to simplify notation I write $\varphi_{V_P}$

Sorry to post the same proof again

Thanks a lot.

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4  
Recall the definition of the stalk - it answers your question directly. My advice is not to read this Proposition step my step as if it is was a purely algebraic argument. Instead, do it on your own and use pictures. Really, you can prove this Proposition just using pictures, and thereby gain a lot of intuition for the subject. I would explain it to you if we were in front of a blackboard. :) –  Martin Brandenburg Sep 18 '13 at 12:14
    
@MartinBrandenburg Unfortunately I don't have maturity in this field yet to use pictures. –  user42912 Sep 19 '13 at 8:08
    
@MartinBrandenburg maybe there is some typo or simplification in the notation I don't see? –  user42912 Sep 19 '13 at 8:09
    
@MartinBrandenburg please can you see my answer? –  user42912 Sep 19 '13 at 12:08
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2 Answers

Consider the morphisms

  • $\rho_{X/V_P}:\mathcal{G}(X)\to\mathcal{G}(V_P)$
  • $\sigma_{V_P/P}:\mathcal{G}(V_P) \to \mathcal{G}_P$
  • $\sigma_{X/P}:\mathcal{G}(X)\to\mathcal{G}_P$

It follows from the definition of a stalk that $\sigma_{X/P}=\sigma_{V_P/P}\circ\rho_{X/V_P}$ because $V_P$ and $X$ are both open neighborhoods of $P$. Look up the definition of a direct limit to confirm this, if you like.

In other words, the above statement means $t_P=(t|_{V_P})_P$. Now given what you know, this yields what you want to understand.

Edit I would write it down as follows: We may replace $V_P$ by $V_P\cap U$, then $V_P$ is an open subset contained in both $V_P$ and $U$. Hence, we have $t_P=(t|_{V_P})_P$. Thus, $\varphi_P(s_P)=t_P$ implies $(\varphi_{V_P}(s(P)))_P=t_P=(t|_{V_P})_P$.

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Thank you for your help, please can you see my answer? –  user42912 Sep 19 '13 at 12:08
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Lema 1

Let $V\subset U$ and $t\in \mathcal F(U)$, we have $(t_{|V})_p=t_p$

Proof

$(t_{|V})_p=t_p \Leftrightarrow \exists \ W\subset V\cap U=V;\ t_{|W}=(t_{|V})_{|W}$, but

$(t_{|V})_{|W}=t_{|W}$

Proof of the question:

$\varphi_P(s(P))=t_P\implies (\text{by definition of}\ \varphi_P)\\ (\varphi_{V_P}(s(P)))_P=t_P\implies\\ \exists\ W\subset V_P\cap U=V_P; (\varphi_{V_P}(s(P)))_{|W}=t_{|W}\implies \ (\text{since} \ t_{|W}=(t_{|V_P})_{|W})\\(\varphi_{V_P}(s(P)))_{|W}=(t_{|V_P})_{|W}\implies\\ ((\varphi_{V_P}(s(P)))_{|W})_P=((t_{|V_P})_{|W})_P\implies\text{(Lemma 1)}\\ (\varphi_{V_P}(s(P)))_P=((t_{|V_P}))_P$

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1  
This looks correct, but it does use quite a large amount of brackets. I think this can be seen very easily in diagram language, while the above is hard to read. –  Jesko Hüttenhain Sep 20 '13 at 6:28
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