proof in English of Proposition on morphisms into affine schemes












0














Does anybody know where to find an English proof of the following proposition




Let $(X, mathcal{O}_X)$ be a locally ringed space, $Y = operatorname{Spec} A$ an affine scheme. Then the natural map
$$begin{align}operatorname{Hom}(X, Y ) &to operatorname{Hom}(A, Γ(X, mathcal{O}_X)),\
(f, f^flat) &mapsto f^flat_Y ,end{align}$$

is a bijection.




This is Prop. 3.4 in Görtz's and Wedhorn's Algebraic Geometry I (there the proof is only given for the case where $X$ is a scheme) and Proposition 1.6.3 in the 1971 edition of Grothendieck's Eléments de Géométrie Algébrique (full proof given there, but it's in French and AFAIK there is no translation).










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  • I imagine you ask for a proof in english, because the proofs have no nationally (as far as I know).
    – xarles
    Dec 26 '18 at 18:42






  • 3




    You can find one in the answer to mathoverflow.net/questions/228137/…
    – xarles
    Dec 26 '18 at 18:52








  • 3




    Also in stacks.math.columbia.edu/tag/01I1
    – xarles
    Dec 26 '18 at 18:54






  • 1




    @jgon The problem is most textbooks only seem to prove it in the case where $X$ is a scheme
    – 0x539
    Dec 26 '18 at 20:08






  • 1




    I explain this in great detail in my answer here: math.stackexchange.com/questions/56854/…
    – Keenan Kidwell
    Dec 26 '18 at 21:31
















0














Does anybody know where to find an English proof of the following proposition




Let $(X, mathcal{O}_X)$ be a locally ringed space, $Y = operatorname{Spec} A$ an affine scheme. Then the natural map
$$begin{align}operatorname{Hom}(X, Y ) &to operatorname{Hom}(A, Γ(X, mathcal{O}_X)),\
(f, f^flat) &mapsto f^flat_Y ,end{align}$$

is a bijection.




This is Prop. 3.4 in Görtz's and Wedhorn's Algebraic Geometry I (there the proof is only given for the case where $X$ is a scheme) and Proposition 1.6.3 in the 1971 edition of Grothendieck's Eléments de Géométrie Algébrique (full proof given there, but it's in French and AFAIK there is no translation).










share|cite|improve this question
























  • I imagine you ask for a proof in english, because the proofs have no nationally (as far as I know).
    – xarles
    Dec 26 '18 at 18:42






  • 3




    You can find one in the answer to mathoverflow.net/questions/228137/…
    – xarles
    Dec 26 '18 at 18:52








  • 3




    Also in stacks.math.columbia.edu/tag/01I1
    – xarles
    Dec 26 '18 at 18:54






  • 1




    @jgon The problem is most textbooks only seem to prove it in the case where $X$ is a scheme
    – 0x539
    Dec 26 '18 at 20:08






  • 1




    I explain this in great detail in my answer here: math.stackexchange.com/questions/56854/…
    – Keenan Kidwell
    Dec 26 '18 at 21:31














0












0








0


1





Does anybody know where to find an English proof of the following proposition




Let $(X, mathcal{O}_X)$ be a locally ringed space, $Y = operatorname{Spec} A$ an affine scheme. Then the natural map
$$begin{align}operatorname{Hom}(X, Y ) &to operatorname{Hom}(A, Γ(X, mathcal{O}_X)),\
(f, f^flat) &mapsto f^flat_Y ,end{align}$$

is a bijection.




This is Prop. 3.4 in Görtz's and Wedhorn's Algebraic Geometry I (there the proof is only given for the case where $X$ is a scheme) and Proposition 1.6.3 in the 1971 edition of Grothendieck's Eléments de Géométrie Algébrique (full proof given there, but it's in French and AFAIK there is no translation).










share|cite|improve this question















Does anybody know where to find an English proof of the following proposition




Let $(X, mathcal{O}_X)$ be a locally ringed space, $Y = operatorname{Spec} A$ an affine scheme. Then the natural map
$$begin{align}operatorname{Hom}(X, Y ) &to operatorname{Hom}(A, Γ(X, mathcal{O}_X)),\
(f, f^flat) &mapsto f^flat_Y ,end{align}$$

is a bijection.




This is Prop. 3.4 in Görtz's and Wedhorn's Algebraic Geometry I (there the proof is only given for the case where $X$ is a scheme) and Proposition 1.6.3 in the 1971 edition of Grothendieck's Eléments de Géométrie Algébrique (full proof given there, but it's in French and AFAIK there is no translation).







algebraic-geometry reference-request schemes mathematical-french






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share|cite|improve this question













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edited Dec 26 '18 at 22:35

























asked Dec 26 '18 at 17:24









0x539

731314




731314












  • I imagine you ask for a proof in english, because the proofs have no nationally (as far as I know).
    – xarles
    Dec 26 '18 at 18:42






  • 3




    You can find one in the answer to mathoverflow.net/questions/228137/…
    – xarles
    Dec 26 '18 at 18:52








  • 3




    Also in stacks.math.columbia.edu/tag/01I1
    – xarles
    Dec 26 '18 at 18:54






  • 1




    @jgon The problem is most textbooks only seem to prove it in the case where $X$ is a scheme
    – 0x539
    Dec 26 '18 at 20:08






  • 1




    I explain this in great detail in my answer here: math.stackexchange.com/questions/56854/…
    – Keenan Kidwell
    Dec 26 '18 at 21:31


















  • I imagine you ask for a proof in english, because the proofs have no nationally (as far as I know).
    – xarles
    Dec 26 '18 at 18:42






  • 3




    You can find one in the answer to mathoverflow.net/questions/228137/…
    – xarles
    Dec 26 '18 at 18:52








  • 3




    Also in stacks.math.columbia.edu/tag/01I1
    – xarles
    Dec 26 '18 at 18:54






  • 1




    @jgon The problem is most textbooks only seem to prove it in the case where $X$ is a scheme
    – 0x539
    Dec 26 '18 at 20:08






  • 1




    I explain this in great detail in my answer here: math.stackexchange.com/questions/56854/…
    – Keenan Kidwell
    Dec 26 '18 at 21:31
















I imagine you ask for a proof in english, because the proofs have no nationally (as far as I know).
– xarles
Dec 26 '18 at 18:42




I imagine you ask for a proof in english, because the proofs have no nationally (as far as I know).
– xarles
Dec 26 '18 at 18:42




3




3




You can find one in the answer to mathoverflow.net/questions/228137/…
– xarles
Dec 26 '18 at 18:52






You can find one in the answer to mathoverflow.net/questions/228137/…
– xarles
Dec 26 '18 at 18:52






3




3




Also in stacks.math.columbia.edu/tag/01I1
– xarles
Dec 26 '18 at 18:54




Also in stacks.math.columbia.edu/tag/01I1
– xarles
Dec 26 '18 at 18:54




1




1




@jgon The problem is most textbooks only seem to prove it in the case where $X$ is a scheme
– 0x539
Dec 26 '18 at 20:08




@jgon The problem is most textbooks only seem to prove it in the case where $X$ is a scheme
– 0x539
Dec 26 '18 at 20:08




1




1




I explain this in great detail in my answer here: math.stackexchange.com/questions/56854/…
– Keenan Kidwell
Dec 26 '18 at 21:31




I explain this in great detail in my answer here: math.stackexchange.com/questions/56854/…
– Keenan Kidwell
Dec 26 '18 at 21:31










1 Answer
1






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2














Here's my effort at a translation. The original text uses $S = Y$, as I will.




Let $DeclareMathOperator{Hom}{Hom} DeclareMathOperator{O}{mathcal{O}} A = Gamma(S, O_S)$, and consider a ring homomorphism $varphi: A to Gamma(X, O_X)$. For each $x in X$, the set of $f in A$ such that $varphi(f)(x) = 0$ (0, 4.1.9) is a prime ideal of $A$, because $DeclareMathOperator{m}{mathfrak{m}} O_x/m_x = kappa(x)$ is a field; it is thus an element of $DeclareMathOperator{Spec}{Spec} S = Spec(A)$, that we will again denote ${^a varphi}(x)$. Moreover, for each $f in A$, we have by definition (0, 4.1.13) $newcommand{varphia}{{^a varphi}} varphia^{-1}(D(f)) = X_{varphi(f)}$, which shows that $varphia$ is a continuous map from $X$ to $S$. Define next a homomorphism $newcommand{varphitilde}{tilde{varphi}} varphitilde: O_S to varphia_*(O_X)$ of $O_S$-modules: for each $f in A$, we have $Gamma(D(f), O_S) = A_f$ (1.3.6); for each $s in A$, we will make correspond to $s/f in A_f$ the element
$$
(varphi(s)|X_{varphi(f)})(varphi(f)|X_{varphi(f)})^{-1}
$$

of $Gamma(X_{varphi(f)}, O_X) = Gamma(D(f), varphia_*(O_X))$, and we prove immediately, by passing from $D(f)$ to $D(fg)$ for each $g in A$, that this defines a homomorphism of $O_S$-modules; thus we have obtained a morphism $(varphia, varphitilde)$ of ringed spaces. Furthermore, with the same notation, and setting $y = varphia(x)$, we see immediately (0, 3.7.1) that we have $varphitilde_x^#(s_y/f_y) = (varphi(s)_x)(varphi(f)_x)^{-1}$; as the relation $s_y in m_y$ is by definition equivalent to $varphi(s)_x in m_x$, we see that $varphitilde_x^#$ is a local homomorphism $O_y to O_x$, in other words $(varphia, varphitilde)$ is a morphism of locally ringed spaces. Thus we have defined a canonical map
begin{equation}
sigma : Hom(Gamma(S, O_S), Gamma(X, O_X)) to Hom_{text{loc}}(X,S) , . tag{1.6.3.2}
end{equation}



It remains to prove that $rho_{text{loc}}$ and $sigma$ are mutually inverse. However, the definition given above for $varphitilde$ shows immediately that $Gamma(varphitilde) = varphi$, and hence $rho_{text{loc}} circ sigma$ is the identity. To see that $sigma circ rho_text{loc}$ is the identity, we begin with a morphism $(psi, theta): X to S$ of locally ringed spaces, and set $varphi: Gamma(theta)$; the hypothesis that $theta_x^#$ is local allows us to deduce that this homomorphism, by passing to quotients, is a monomorphism of fields $theta^x : kappa(theta(psi(x)) to kappa(x)$ such that, for each section $f in A = Gamma(S, O_S)$, we have $theta^x(f(psi(x)) = varphi(f)(x)$; the relation $f(psi(x)) = 0$ is thus equivalent to $varphi(f)(x) = 0$, which shows that $varphia = psi$ by virtue of the definition of $varphia$. On the other hand, the definitions imply that the diagram [see diagram in text] is commutative, and the same is true of the analogous diagram where $theta_x^#$ is replaced by $varphitilde_x^#$, hence $varphitilde_x^# = theta_x^#$ (Bourbaki, Alg. comm., chap. II, $S 2$, no. 1, prop. 1), which implies that $varphitilde = theta$.







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    Here's my effort at a translation. The original text uses $S = Y$, as I will.




    Let $DeclareMathOperator{Hom}{Hom} DeclareMathOperator{O}{mathcal{O}} A = Gamma(S, O_S)$, and consider a ring homomorphism $varphi: A to Gamma(X, O_X)$. For each $x in X$, the set of $f in A$ such that $varphi(f)(x) = 0$ (0, 4.1.9) is a prime ideal of $A$, because $DeclareMathOperator{m}{mathfrak{m}} O_x/m_x = kappa(x)$ is a field; it is thus an element of $DeclareMathOperator{Spec}{Spec} S = Spec(A)$, that we will again denote ${^a varphi}(x)$. Moreover, for each $f in A$, we have by definition (0, 4.1.13) $newcommand{varphia}{{^a varphi}} varphia^{-1}(D(f)) = X_{varphi(f)}$, which shows that $varphia$ is a continuous map from $X$ to $S$. Define next a homomorphism $newcommand{varphitilde}{tilde{varphi}} varphitilde: O_S to varphia_*(O_X)$ of $O_S$-modules: for each $f in A$, we have $Gamma(D(f), O_S) = A_f$ (1.3.6); for each $s in A$, we will make correspond to $s/f in A_f$ the element
    $$
    (varphi(s)|X_{varphi(f)})(varphi(f)|X_{varphi(f)})^{-1}
    $$

    of $Gamma(X_{varphi(f)}, O_X) = Gamma(D(f), varphia_*(O_X))$, and we prove immediately, by passing from $D(f)$ to $D(fg)$ for each $g in A$, that this defines a homomorphism of $O_S$-modules; thus we have obtained a morphism $(varphia, varphitilde)$ of ringed spaces. Furthermore, with the same notation, and setting $y = varphia(x)$, we see immediately (0, 3.7.1) that we have $varphitilde_x^#(s_y/f_y) = (varphi(s)_x)(varphi(f)_x)^{-1}$; as the relation $s_y in m_y$ is by definition equivalent to $varphi(s)_x in m_x$, we see that $varphitilde_x^#$ is a local homomorphism $O_y to O_x$, in other words $(varphia, varphitilde)$ is a morphism of locally ringed spaces. Thus we have defined a canonical map
    begin{equation}
    sigma : Hom(Gamma(S, O_S), Gamma(X, O_X)) to Hom_{text{loc}}(X,S) , . tag{1.6.3.2}
    end{equation}



    It remains to prove that $rho_{text{loc}}$ and $sigma$ are mutually inverse. However, the definition given above for $varphitilde$ shows immediately that $Gamma(varphitilde) = varphi$, and hence $rho_{text{loc}} circ sigma$ is the identity. To see that $sigma circ rho_text{loc}$ is the identity, we begin with a morphism $(psi, theta): X to S$ of locally ringed spaces, and set $varphi: Gamma(theta)$; the hypothesis that $theta_x^#$ is local allows us to deduce that this homomorphism, by passing to quotients, is a monomorphism of fields $theta^x : kappa(theta(psi(x)) to kappa(x)$ such that, for each section $f in A = Gamma(S, O_S)$, we have $theta^x(f(psi(x)) = varphi(f)(x)$; the relation $f(psi(x)) = 0$ is thus equivalent to $varphi(f)(x) = 0$, which shows that $varphia = psi$ by virtue of the definition of $varphia$. On the other hand, the definitions imply that the diagram [see diagram in text] is commutative, and the same is true of the analogous diagram where $theta_x^#$ is replaced by $varphitilde_x^#$, hence $varphitilde_x^# = theta_x^#$ (Bourbaki, Alg. comm., chap. II, $S 2$, no. 1, prop. 1), which implies that $varphitilde = theta$.







    share|cite|improve this answer




























      2














      Here's my effort at a translation. The original text uses $S = Y$, as I will.




      Let $DeclareMathOperator{Hom}{Hom} DeclareMathOperator{O}{mathcal{O}} A = Gamma(S, O_S)$, and consider a ring homomorphism $varphi: A to Gamma(X, O_X)$. For each $x in X$, the set of $f in A$ such that $varphi(f)(x) = 0$ (0, 4.1.9) is a prime ideal of $A$, because $DeclareMathOperator{m}{mathfrak{m}} O_x/m_x = kappa(x)$ is a field; it is thus an element of $DeclareMathOperator{Spec}{Spec} S = Spec(A)$, that we will again denote ${^a varphi}(x)$. Moreover, for each $f in A$, we have by definition (0, 4.1.13) $newcommand{varphia}{{^a varphi}} varphia^{-1}(D(f)) = X_{varphi(f)}$, which shows that $varphia$ is a continuous map from $X$ to $S$. Define next a homomorphism $newcommand{varphitilde}{tilde{varphi}} varphitilde: O_S to varphia_*(O_X)$ of $O_S$-modules: for each $f in A$, we have $Gamma(D(f), O_S) = A_f$ (1.3.6); for each $s in A$, we will make correspond to $s/f in A_f$ the element
      $$
      (varphi(s)|X_{varphi(f)})(varphi(f)|X_{varphi(f)})^{-1}
      $$

      of $Gamma(X_{varphi(f)}, O_X) = Gamma(D(f), varphia_*(O_X))$, and we prove immediately, by passing from $D(f)$ to $D(fg)$ for each $g in A$, that this defines a homomorphism of $O_S$-modules; thus we have obtained a morphism $(varphia, varphitilde)$ of ringed spaces. Furthermore, with the same notation, and setting $y = varphia(x)$, we see immediately (0, 3.7.1) that we have $varphitilde_x^#(s_y/f_y) = (varphi(s)_x)(varphi(f)_x)^{-1}$; as the relation $s_y in m_y$ is by definition equivalent to $varphi(s)_x in m_x$, we see that $varphitilde_x^#$ is a local homomorphism $O_y to O_x$, in other words $(varphia, varphitilde)$ is a morphism of locally ringed spaces. Thus we have defined a canonical map
      begin{equation}
      sigma : Hom(Gamma(S, O_S), Gamma(X, O_X)) to Hom_{text{loc}}(X,S) , . tag{1.6.3.2}
      end{equation}



      It remains to prove that $rho_{text{loc}}$ and $sigma$ are mutually inverse. However, the definition given above for $varphitilde$ shows immediately that $Gamma(varphitilde) = varphi$, and hence $rho_{text{loc}} circ sigma$ is the identity. To see that $sigma circ rho_text{loc}$ is the identity, we begin with a morphism $(psi, theta): X to S$ of locally ringed spaces, and set $varphi: Gamma(theta)$; the hypothesis that $theta_x^#$ is local allows us to deduce that this homomorphism, by passing to quotients, is a monomorphism of fields $theta^x : kappa(theta(psi(x)) to kappa(x)$ such that, for each section $f in A = Gamma(S, O_S)$, we have $theta^x(f(psi(x)) = varphi(f)(x)$; the relation $f(psi(x)) = 0$ is thus equivalent to $varphi(f)(x) = 0$, which shows that $varphia = psi$ by virtue of the definition of $varphia$. On the other hand, the definitions imply that the diagram [see diagram in text] is commutative, and the same is true of the analogous diagram where $theta_x^#$ is replaced by $varphitilde_x^#$, hence $varphitilde_x^# = theta_x^#$ (Bourbaki, Alg. comm., chap. II, $S 2$, no. 1, prop. 1), which implies that $varphitilde = theta$.







      share|cite|improve this answer


























        2












        2








        2






        Here's my effort at a translation. The original text uses $S = Y$, as I will.




        Let $DeclareMathOperator{Hom}{Hom} DeclareMathOperator{O}{mathcal{O}} A = Gamma(S, O_S)$, and consider a ring homomorphism $varphi: A to Gamma(X, O_X)$. For each $x in X$, the set of $f in A$ such that $varphi(f)(x) = 0$ (0, 4.1.9) is a prime ideal of $A$, because $DeclareMathOperator{m}{mathfrak{m}} O_x/m_x = kappa(x)$ is a field; it is thus an element of $DeclareMathOperator{Spec}{Spec} S = Spec(A)$, that we will again denote ${^a varphi}(x)$. Moreover, for each $f in A$, we have by definition (0, 4.1.13) $newcommand{varphia}{{^a varphi}} varphia^{-1}(D(f)) = X_{varphi(f)}$, which shows that $varphia$ is a continuous map from $X$ to $S$. Define next a homomorphism $newcommand{varphitilde}{tilde{varphi}} varphitilde: O_S to varphia_*(O_X)$ of $O_S$-modules: for each $f in A$, we have $Gamma(D(f), O_S) = A_f$ (1.3.6); for each $s in A$, we will make correspond to $s/f in A_f$ the element
        $$
        (varphi(s)|X_{varphi(f)})(varphi(f)|X_{varphi(f)})^{-1}
        $$

        of $Gamma(X_{varphi(f)}, O_X) = Gamma(D(f), varphia_*(O_X))$, and we prove immediately, by passing from $D(f)$ to $D(fg)$ for each $g in A$, that this defines a homomorphism of $O_S$-modules; thus we have obtained a morphism $(varphia, varphitilde)$ of ringed spaces. Furthermore, with the same notation, and setting $y = varphia(x)$, we see immediately (0, 3.7.1) that we have $varphitilde_x^#(s_y/f_y) = (varphi(s)_x)(varphi(f)_x)^{-1}$; as the relation $s_y in m_y$ is by definition equivalent to $varphi(s)_x in m_x$, we see that $varphitilde_x^#$ is a local homomorphism $O_y to O_x$, in other words $(varphia, varphitilde)$ is a morphism of locally ringed spaces. Thus we have defined a canonical map
        begin{equation}
        sigma : Hom(Gamma(S, O_S), Gamma(X, O_X)) to Hom_{text{loc}}(X,S) , . tag{1.6.3.2}
        end{equation}



        It remains to prove that $rho_{text{loc}}$ and $sigma$ are mutually inverse. However, the definition given above for $varphitilde$ shows immediately that $Gamma(varphitilde) = varphi$, and hence $rho_{text{loc}} circ sigma$ is the identity. To see that $sigma circ rho_text{loc}$ is the identity, we begin with a morphism $(psi, theta): X to S$ of locally ringed spaces, and set $varphi: Gamma(theta)$; the hypothesis that $theta_x^#$ is local allows us to deduce that this homomorphism, by passing to quotients, is a monomorphism of fields $theta^x : kappa(theta(psi(x)) to kappa(x)$ such that, for each section $f in A = Gamma(S, O_S)$, we have $theta^x(f(psi(x)) = varphi(f)(x)$; the relation $f(psi(x)) = 0$ is thus equivalent to $varphi(f)(x) = 0$, which shows that $varphia = psi$ by virtue of the definition of $varphia$. On the other hand, the definitions imply that the diagram [see diagram in text] is commutative, and the same is true of the analogous diagram where $theta_x^#$ is replaced by $varphitilde_x^#$, hence $varphitilde_x^# = theta_x^#$ (Bourbaki, Alg. comm., chap. II, $S 2$, no. 1, prop. 1), which implies that $varphitilde = theta$.







        share|cite|improve this answer














        Here's my effort at a translation. The original text uses $S = Y$, as I will.




        Let $DeclareMathOperator{Hom}{Hom} DeclareMathOperator{O}{mathcal{O}} A = Gamma(S, O_S)$, and consider a ring homomorphism $varphi: A to Gamma(X, O_X)$. For each $x in X$, the set of $f in A$ such that $varphi(f)(x) = 0$ (0, 4.1.9) is a prime ideal of $A$, because $DeclareMathOperator{m}{mathfrak{m}} O_x/m_x = kappa(x)$ is a field; it is thus an element of $DeclareMathOperator{Spec}{Spec} S = Spec(A)$, that we will again denote ${^a varphi}(x)$. Moreover, for each $f in A$, we have by definition (0, 4.1.13) $newcommand{varphia}{{^a varphi}} varphia^{-1}(D(f)) = X_{varphi(f)}$, which shows that $varphia$ is a continuous map from $X$ to $S$. Define next a homomorphism $newcommand{varphitilde}{tilde{varphi}} varphitilde: O_S to varphia_*(O_X)$ of $O_S$-modules: for each $f in A$, we have $Gamma(D(f), O_S) = A_f$ (1.3.6); for each $s in A$, we will make correspond to $s/f in A_f$ the element
        $$
        (varphi(s)|X_{varphi(f)})(varphi(f)|X_{varphi(f)})^{-1}
        $$

        of $Gamma(X_{varphi(f)}, O_X) = Gamma(D(f), varphia_*(O_X))$, and we prove immediately, by passing from $D(f)$ to $D(fg)$ for each $g in A$, that this defines a homomorphism of $O_S$-modules; thus we have obtained a morphism $(varphia, varphitilde)$ of ringed spaces. Furthermore, with the same notation, and setting $y = varphia(x)$, we see immediately (0, 3.7.1) that we have $varphitilde_x^#(s_y/f_y) = (varphi(s)_x)(varphi(f)_x)^{-1}$; as the relation $s_y in m_y$ is by definition equivalent to $varphi(s)_x in m_x$, we see that $varphitilde_x^#$ is a local homomorphism $O_y to O_x$, in other words $(varphia, varphitilde)$ is a morphism of locally ringed spaces. Thus we have defined a canonical map
        begin{equation}
        sigma : Hom(Gamma(S, O_S), Gamma(X, O_X)) to Hom_{text{loc}}(X,S) , . tag{1.6.3.2}
        end{equation}



        It remains to prove that $rho_{text{loc}}$ and $sigma$ are mutually inverse. However, the definition given above for $varphitilde$ shows immediately that $Gamma(varphitilde) = varphi$, and hence $rho_{text{loc}} circ sigma$ is the identity. To see that $sigma circ rho_text{loc}$ is the identity, we begin with a morphism $(psi, theta): X to S$ of locally ringed spaces, and set $varphi: Gamma(theta)$; the hypothesis that $theta_x^#$ is local allows us to deduce that this homomorphism, by passing to quotients, is a monomorphism of fields $theta^x : kappa(theta(psi(x)) to kappa(x)$ such that, for each section $f in A = Gamma(S, O_S)$, we have $theta^x(f(psi(x)) = varphi(f)(x)$; the relation $f(psi(x)) = 0$ is thus equivalent to $varphi(f)(x) = 0$, which shows that $varphia = psi$ by virtue of the definition of $varphia$. On the other hand, the definitions imply that the diagram [see diagram in text] is commutative, and the same is true of the analogous diagram where $theta_x^#$ is replaced by $varphitilde_x^#$, hence $varphitilde_x^# = theta_x^#$ (Bourbaki, Alg. comm., chap. II, $S 2$, no. 1, prop. 1), which implies that $varphitilde = theta$.








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        edited Dec 26 '18 at 20:06

























        answered Dec 26 '18 at 19:13









        André 3000

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