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25.5. Affine schemes

Let $R$ be a ring. Consider the topological space $\mathop{\mathrm{Spec}}(R)$ associated to $R$, see Algebra, Section 10.16. We will endow this space with a sheaf of rings $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$ and the resulting pair $(\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ will be an affine scheme.

Recall that $\mathop{\mathrm{Spec}}(R)$ has a basis of open sets $D(f)$, $f \in R$ which we call standard opens, see Algebra, Definition 10.16.3. In addition, the intersection of two standard opens is another: $D(f) \cap D(g) = D(fg)$, $f, g\in R$.

Lemma 25.5.1. Let $R$ be a ring. Let $f \in R$.

  1. If $g\in R$ and $D(g) \subset D(f)$, then
    1. $f$ is invertible in $R_g$,
    2. $g^e = af$ for some $e \geq 1$ and $a \in R$,
    3. there is a canonical ring map $R_f \to R_g$, and
    4. there is a canonical $R_f$-module map $M_f \to M_g$ for any $R$-module $M$.
  2. Any open covering of $D(f)$ can be refined to a finite open covering of the form $D(f) = \bigcup_{i = 1}^n D(g_i)$.
  3. If $g_1, \ldots, g_n \in R$, then $D(f) \subset \bigcup D(g_i)$ if and only if $g_1, \ldots, g_n$ generate the unit ideal in $R_f$.

Proof. Recall that $D(g) = \mathop{\mathrm{Spec}}(R_g)$ (see Algebra, Lemma 10.16.6). Thus (a) holds because $f$ maps to an element of $R_g$ which is not contained in any prime ideal, and hence invertible, see Algebra, Lemma 10.16.2. Write the inverse of $f$ in $R_g$ as $a/g^d$. This means $g^d - af$ is annihilated by a power of $g$, whence (b). For (c), the map $R_f \to R_g$ exists by (a) from the universal property of localization, or we can define it by mapping $b/f^n$ to $a^nb/g^{ne}$. The equality $M_f = M \otimes_R R_f$ can be used to obtain the map on modules, or we can define $M_f \to M_g$ by mapping $x/f^n$ to $a^nx/g^{ne}$.

Recall that $D(f)$ is quasi-compact, see Algebra, Lemma 10.28.1. Hence the second statement follows directly from the fact that the standard opens form a basis for the topology.

The third statement follows directly from Algebra, Lemma 10.16.2. $\square$

In Sheaves, Section 6.30 we defined the notion of a sheaf on a basis, and we showed that it is essentially equivalent to the notion of a sheaf on the space, see Sheaves, Lemmas 6.30.6 and 6.30.9. Moreover, we showed in Sheaves, Lemma 6.30.4 that it is sufficient to check the sheaf condition on a cofinal system of open coverings for each standard open. By the lemma above it suffices to check on the finite coverings by standard opens.

Definition 25.5.2. Let $R$ be a ring.

  1. A standard open covering of $\mathop{\mathrm{Spec}}(R)$ is a covering $\mathop{\mathrm{Spec}}(R) = \bigcup_{i = 1}^n D(f_i)$, where $f_1, \ldots, f_n \in R$.
  2. Suppose that $D(f) \subset \mathop{\mathrm{Spec}}(R)$ is a standard open. A standard open covering of $D(f)$ is a covering $D(f) = \bigcup_{i = 1}^n D(g_i)$, where $g_1, \ldots, g_n \in R$.

Let $R$ be a ring. Let $M$ be an $R$-module. We will define a presheaf $\widetilde M$ on the basis of standard opens. Suppose that $U \subset \mathop{\mathrm{Spec}}(R)$ is a standard open. If $f, g \in R$ are such that $D(f) = D(g)$, then by Lemma 25.5.1 above there are canonical maps $M_f \to M_g$ and $M_g \to M_f$ which are mutually inverse. Hence we may choose any $f$ such that $U = D(f)$ and define $$ \widetilde M(U) = M_f. $$ Note that if $D(g) \subset D(f)$, then by Lemma 25.5.1 above we have a canonical map $$ \widetilde M(D(f)) = M_f \longrightarrow M_g = \widetilde M(D(g)). $$ Clearly, this defines a presheaf of abelian groups on the basis of standard opens. If $M = R$, then $\widetilde R$ is a presheaf of rings on the basis of standard opens.

Let us compute the stalk of $\widetilde M$ at a point $x \in \mathop{\mathrm{Spec}}(R)$. Suppose that $x$ corresponds to the prime $\mathfrak p \subset R$. By definition of the stalk we see that $$ \widetilde M_x = \mathop{\mathrm{colim}}\nolimits_{f\in R, f\not\in \mathfrak p} M_f $$ Here the set $\{f \in R, f \not \in \mathfrak p\}$ is preordered by the rule $f \geq f' \Leftrightarrow D(f) \subset D(f')$. If $f_1, f_2 \in R \setminus \mathfrak p$, then we have $f_1f_2 \geq f_1$ in this ordering. Hence by Algebra, Lemma 10.9.9 we conclude that $$ \widetilde M_x = M_{\mathfrak p}. $$

Next, we check the sheaf condition for the standard open coverings. If $D(f) = \bigcup_{i = 1}^n D(g_i)$, then the sheaf condition for this covering is equivalent with the exactness of the sequence $$ 0 \to M_f \to \bigoplus M_{g_i} \to \bigoplus M_{g_ig_j}. $$ Note that $D(g_i) = D(fg_i)$, and hence we can rewrite this sequence as the sequence $$ 0 \to M_f \to \bigoplus M_{fg_i} \to \bigoplus M_{fg_ig_j}. $$ In addition, by Lemma 25.5.1 above we see that $g_1, \ldots, g_n$ generate the unit ideal in $R_f$. Thus we may apply Algebra, Lemma 10.22.2 to the module $M_f$ over $R_f$ and the elements $g_1, \ldots, g_n$. We conclude that the sequence is exact. By the remarks made above, we see that $\widetilde M$ is a sheaf on the basis of standard opens.

Thus we conclude from the material in Sheaves, Section 6.30 that there exists a unique sheaf of rings $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$ which agrees with $\widetilde R$ on the standard opens. Note that by our computation of stalks above, the stalks of this sheaf of rings are all local rings.

Similarly, for any $R$-module $M$ there exists a unique sheaf of $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$-modules $\mathcal{F}$ which agrees with $\widetilde M$ on the standard opens, see Sheaves, Lemma 6.30.12.

Definition 25.5.3. Let $R$ be a ring.

  1. The structure sheaf $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$ of the spectrum of $R$ is the unique sheaf of rings $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$ which agrees with $\widetilde R$ on the basis of standard opens.
  2. The locally ringed space $(\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ is called the spectrum of $R$ and denoted $\mathop{\mathrm{Spec}}(R)$.
  3. The sheaf of $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$-modules extending $\widetilde M$ to all opens of $\mathop{\mathrm{Spec}}(R)$ is called the sheaf of $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$-modules associated to $M$. This sheaf is denoted $\widetilde M$ as well.

We summarize the results obtained so far.

Lemma 25.5.4. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\widetilde M$ be the sheaf of $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$-modules associated to $M$.

  1. We have $\Gamma(\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)}) = R$.
  2. We have $\Gamma(\mathop{\mathrm{Spec}}(R), \widetilde M) = M$ as an $R$-module.
  3. For every $f \in R$ we have $\Gamma(D(f), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)}) = R_f$.
  4. For every $f\in R$ we have $\Gamma(D(f), \widetilde M) = M_f$ as an $R_f$-module.
  5. Whenever $D(g) \subset D(f)$ the restriction mappings on $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$ and $\widetilde M$ are the maps $R_f \to R_g$ and $M_f \to M_g$ from Lemma 25.5.1.
  6. Let $\mathfrak p$ be a prime of $R$, and let $x \in \mathop{\mathrm{Spec}}(R)$ be the corresponding point. We have $\mathcal{O}_{\mathop{\mathrm{Spec}}(R), x} = R_{\mathfrak p}$.
  7. Let $\mathfrak p$ be a prime of $R$, and let $x \in \mathop{\mathrm{Spec}}(R)$ be the corresponding point. We have $\mathcal{F}_x = M_{\mathfrak p}$ as an $R_{\mathfrak p}$-module.

Moreover, all these identifications are functorial in the $R$ module $M$. In particular, the functor $M \mapsto \widetilde M$ is an exact functor from the category of $R$-modules to the category of $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$-modules.

Proof. Assertions (1) - (7) are clear from the discussion above. The exactness of the functor $M \mapsto \widetilde M$ follows from the fact that the functor $M \mapsto M_{\mathfrak p}$ is exact and the fact that exactness of short exact sequences may be checked on stalks, see Modules, Lemma 17.3.1. $\square$

Definition 25.5.5. An affine scheme is a locally ringed space isomorphic as a locally ringed space to $\mathop{\mathrm{Spec}}(R)$ for some ring $R$. A morphism of affine schemes is a morphism in the category of locally ringed spaces.

It turns out that affine schemes play a special role among all locally ringed spaces, which is what the next section is about.

    The code snippet corresponding to this tag is a part of the file schemes.tex and is located in lines 502–759 (see updates for more information).

    \section{Affine schemes}
    \label{section-affine-schemes}
    
    \noindent
    Let $R$ be a ring. Consider the topological space $\Spec(R)$
    associated to $R$, see Algebra, Section \ref{algebra-section-spectrum-ring}.
    We will endow this space with a sheaf of rings $\mathcal{O}_{\Spec(R)}$
    and the resulting pair $(\Spec(R), \mathcal{O}_{\Spec(R)})$
    will be an affine scheme.
    
    \medskip\noindent
    Recall that $\Spec(R)$ has a basis of open sets $D(f)$,
    $f \in R$ which we call standard opens, see Algebra,
    Definition \ref{algebra-definition-Zariski-topology}.
    In addition, the intersection of two standard opens is another:
    $D(f) \cap D(g) = D(fg)$, $f, g\in R$.
    
    \begin{lemma}
    \label{lemma-standard-open}
    Let $R$ be a ring. Let $f \in R$.
    \begin{enumerate}
    \item If $g\in R$ and $D(g) \subset D(f)$, then
    \begin{enumerate}
    \item $f$ is invertible in $R_g$,
    \item $g^e = af$ for some $e \geq 1$ and $a \in R$,
    \item there is a canonical ring map $R_f \to R_g$, and
    \item there is a canonical $R_f$-module map
    $M_f \to M_g$ for any $R$-module $M$.
    \end{enumerate}
    \item Any open covering of $D(f)$ can be refined to a finite
    open covering of the form $D(f) = \bigcup_{i = 1}^n D(g_i)$.
    \item If $g_1, \ldots, g_n \in R$, then $D(f) \subset \bigcup D(g_i)$
    if and only if $g_1, \ldots, g_n$ generate the unit ideal in $R_f$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Recall that $D(g) = \Spec(R_g)$ (see
    Algebra, Lemma \ref{algebra-lemma-standard-open}).
    Thus (a) holds because $f$
    maps to an element of $R_g$ which is not
    contained in any prime ideal, and hence invertible,
    see Algebra, Lemma \ref{algebra-lemma-Zariski-topology}.
    Write the inverse of $f$ in $R_g$ as $a/g^d$.
    This means $g^d - af$ is annihilated by a power of $g$, whence (b).
    For (c), the map $R_f \to R_g$ exists by (a) from the universal property
    of localization, or we can define it by mapping $b/f^n$
    to $a^nb/g^{ne}$. The equality $M_f = M \otimes_R R_f$
    can be used to obtain the map on modules, or
    we can define $M_f \to M_g$ by mapping
    $x/f^n$ to $a^nx/g^{ne}$.
    
    \medskip\noindent
    Recall that $D(f)$ is quasi-compact, see
    Algebra, Lemma \ref{algebra-lemma-qc-open}.
    Hence the second statement follows directly
    from the fact that the standard opens form
    a basis for the topology.
    
    \medskip\noindent
    The third statement follows directly from
    Algebra, Lemma \ref{algebra-lemma-Zariski-topology}.
    \end{proof}
    
    \noindent
    In Sheaves, Section \ref{sheaves-section-bases} we defined
    the notion of a sheaf on a basis, and we showed that it is
    essentially equivalent to the notion of a sheaf on the space,
    see Sheaves, Lemmas \ref{sheaves-lemma-extend-off-basis} and
    \ref{sheaves-lemma-extend-off-basis-structures}. Moreover,
    we showed in
    Sheaves, Lemma \ref{sheaves-lemma-cofinal-systems-coverings-standard-case}
    that it is sufficient to check the sheaf
    condition on a cofinal system of open coverings for each
    standard open. By the lemma above it suffices to check
    on the finite coverings by standard opens.
    
    \begin{definition}
    \label{definition-standard-covering}
    Let $R$ be a ring.
    \begin{enumerate}
    \item A {\it standard open covering} of $\Spec(R)$
    is a covering $\Spec(R) = \bigcup_{i = 1}^n D(f_i)$,
    where $f_1, \ldots, f_n \in R$.
    \item Suppose that $D(f) \subset \Spec(R)$ is a standard
    open. A {\it standard open covering} of $D(f)$
    is a covering $D(f) = \bigcup_{i = 1}^n D(g_i)$,
    where $g_1, \ldots, g_n \in R$.
    \end{enumerate}
    \end{definition}
    
    \noindent
    Let $R$ be a ring. Let $M$ be an $R$-module. We will define
    a presheaf $\widetilde M$ on the basis of standard opens.
    Suppose that $U \subset \Spec(R)$ is a standard open.
    If $f, g \in R$ are such that $D(f) = D(g)$, then
    by Lemma \ref{lemma-standard-open} above there are canonical
    maps $M_f \to M_g$ and $M_g \to M_f$ which are mutually inverse.
    Hence we may choose any $f$ such that $U = D(f)$
    and define
    $$
    \widetilde M(U) = M_f.
    $$
    Note that if $D(g) \subset D(f)$, then by
    Lemma \ref{lemma-standard-open} above we have
    a canonical map
    $$
    \widetilde M(D(f)) = M_f \longrightarrow M_g = \widetilde M(D(g)).
    $$
    Clearly, this defines a presheaf of abelian groups on the basis
    of standard opens. If $M = R$, then $\widetilde R$ is a presheaf
    of rings on the basis of standard opens.
    
    \medskip\noindent
    Let us compute the stalk of $\widetilde M$ at a point $x \in \Spec(R)$.
    Suppose that $x$ corresponds to the prime $\mathfrak p \subset R$.
    By definition of the stalk we see that
    $$
    \widetilde M_x = \colim_{f\in R, f\not\in \mathfrak p} M_f
    $$
    Here the set $\{f \in R, f \not \in \mathfrak p\}$ is preordered by
    the rule $f \geq f' \Leftrightarrow D(f) \subset D(f')$.
    If $f_1, f_2 \in R \setminus \mathfrak p$, then we have
    $f_1f_2 \geq f_1$ in this ordering. Hence by
    Algebra, Lemma \ref{algebra-lemma-localization-colimit}
    we conclude that
    $$
    \widetilde M_x = M_{\mathfrak p}.
    $$
    
    \medskip\noindent
    Next, we check the sheaf condition for the standard open coverings.
    If $D(f) = \bigcup_{i = 1}^n D(g_i)$, then the sheaf condition
    for this covering is equivalent with the exactness of the
    sequence
    $$
    0 \to M_f \to \bigoplus M_{g_i} \to \bigoplus M_{g_ig_j}.
    $$
    Note that $D(g_i) = D(fg_i)$, and hence we can rewrite this
    sequence as the sequence
    $$
    0 \to M_f \to \bigoplus M_{fg_i} \to \bigoplus M_{fg_ig_j}.
    $$
    In addition, by Lemma \ref{lemma-standard-open} above
    we see that $g_1, \ldots, g_n$ generate the unit ideal
    in $R_f$. Thus we may apply
    Algebra, Lemma \ref{algebra-lemma-cover-module}
    to the module $M_f$ over $R_f$ and the elements $g_1, \ldots, g_n$.
    We conclude that the sequence is exact. By the remarks
    made above, we see that $\widetilde M$ is a sheaf
    on the basis of standard opens.
    
    \medskip\noindent
    Thus we conclude from the material in
    Sheaves, Section \ref{sheaves-section-bases}
    that there exists a
    unique sheaf of rings $\mathcal{O}_{\Spec(R)}$
    which agrees with $\widetilde R$ on the standard opens.
    Note that by our computation of stalks above, the
    stalks of this sheaf of rings are all local rings.
    
    \medskip\noindent
    Similarly, for any $R$-module $M$ there exists
    a unique sheaf of $\mathcal{O}_{\Spec(R)}$-modules
    $\mathcal{F}$ which agrees with $\widetilde M$ on the
    standard opens, see
    Sheaves, Lemma \ref{sheaves-lemma-extend-off-basis-module}.
    
    \begin{definition}
    \label{definition-structure-sheaf}
    Let $R$ be a ring.
    \begin{enumerate}
    \item The {\it structure sheaf $\mathcal{O}_{\Spec(R)}$ of the
    spectrum of $R$} is the unique sheaf of rings $\mathcal{O}_{\Spec(R)}$
    which agrees with $\widetilde R$ on the basis of standard opens.
    \item The locally ringed space
    $(\Spec(R), \mathcal{O}_{\Spec(R)})$ is called
    the {\it spectrum} of $R$ and denoted $\Spec(R)$.
    \item The sheaf of $\mathcal{O}_{\Spec(R)}$-modules
    extending $\widetilde M$ to all opens of $\Spec(R)$
    is called the sheaf of $\mathcal{O}_{\Spec(R)}$-modules
    associated to $M$. This sheaf is denoted $\widetilde M$ as
    well.
    \end{enumerate}
    \end{definition}
    
    \noindent
    We summarize the results obtained so far.
    
    \begin{lemma}
    \label{lemma-spec-sheaves}
    Let $R$ be a ring. Let $M$ be an $R$-module. Let $\widetilde M$
    be the sheaf of $\mathcal{O}_{\Spec(R)}$-modules
    associated to $M$.
    \begin{enumerate}
    \item We have $\Gamma(\Spec(R), \mathcal{O}_{\Spec(R)}) = R$.
    \item We have $\Gamma(\Spec(R), \widetilde M) = M$ as an $R$-module.
    \item For every $f \in R$ we have
    $\Gamma(D(f), \mathcal{O}_{\Spec(R)}) = R_f$.
    \item For every $f\in R$ we have $\Gamma(D(f), \widetilde M) = M_f$
    as an $R_f$-module.
    \item Whenever $D(g) \subset D(f)$ the restriction mappings
    on $\mathcal{O}_{\Spec(R)}$ and $\widetilde M$
    are the maps
    $R_f \to R_g$ and $M_f \to M_g$ from Lemma
    \ref{lemma-standard-open}.
    \item Let $\mathfrak p$ be a prime of $R$, and let $x \in \Spec(R)$
    be the corresponding point. We have
    $\mathcal{O}_{\Spec(R), x} = R_{\mathfrak p}$.
    \item Let $\mathfrak p$ be a prime of $R$, and let $x \in \Spec(R)$
    be the corresponding point. We have $\mathcal{F}_x = M_{\mathfrak p}$
    as an $R_{\mathfrak p}$-module.
    \end{enumerate}
    Moreover, all these identifications are functorial in the $R$
    module $M$. In particular, the functor $M \mapsto \widetilde M$
    is an exact functor from the category of $R$-modules
    to the category of $\mathcal{O}_{\Spec(R)}$-modules.
    \end{lemma}
    
    \begin{proof}
    Assertions (1) - (7) are clear from the discussion above.
    The exactness of the functor $M \mapsto \widetilde M$
    follows from the fact that the functor $M \mapsto M_{\mathfrak p}$
    is exact and the fact that exactness of short exact sequences
    may be checked on stalks, see
    Modules, Lemma \ref{modules-lemma-abelian}.
    \end{proof}
    
    \begin{definition}
    \label{definition-affine-scheme}
    An {\it affine scheme} is a locally ringed space isomorphic
    as a locally ringed space to $\Spec(R)$ for some ring $R$.
    A {\it morphism of affine schemes} is a morphism in the category
    of locally ringed spaces.
    \end{definition}
    
    \noindent
    It turns out that affine schemes play a special role
    among all locally ringed spaces, which is what the next
    section is about.

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