The Stacks project

29.6 Scheme theoretic image

Caution: Some of the material in this section is ultra-general and behaves differently from what you might expect.

Lemma 29.6.1. Let $f : X \to Y$ be a morphism of schemes. There exists a closed subscheme $Z \subset Y$ such that $f$ factors through $Z$ and such that for any other closed subscheme $Z' \subset Y$ such that $f$ factors through $Z'$ we have $Z \subset Z'$.

Proof. Let $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to f_*\mathcal{O}_ X)$. If $\mathcal{I}$ is quasi-coherent then we just take $Z$ to be the closed subscheme determined by $\mathcal{I}$, see Lemma 29.2.3. This works by Schemes, Lemma 26.4.6. In general the same lemma requires us to show that there exists a largest quasi-coherent sheaf of ideals $\mathcal{I}'$ contained in $\mathcal{I}$. This follows from Lemma 29.4.2. $\square$

Definition 29.6.2. Let $f : X \to Y$ be a morphism of schemes. The scheme theoretic image of $f$ is the smallest closed subscheme $Z \subset Y$ through which $f$ factors, see Lemma 29.6.1 above.

For a morphism $f : X \to Y$ of schemes with scheme theoretic image $Z$ we often denote $f : X \to Z$ the factorization of $f$ through its scheme theoretic image. If the morphism $f$ is not quasi-compact, then (in general)

  1. the set theoretic inclusion $\overline{f(X)} \subset Z$ is not an equality, i.e., $f(X) \subset Z$ is not a dense subset, and

  2. the construction of the scheme theoretic image does not commute with restriction to open subschemes to $Y$.

Namely, the immersion of Example 29.3.4 gives an example for both phenomena. (If $Z \to U \to X$ is as in Example 29.3.4, then the scheme theoretic image of $Z \to X$ is $X$ and $Z$ is not topologically dense in $X$. Also, the scheme theoretic image of $Z = Z \cap U \to U$ is just $Z$ which is not equal to $U \cap X = U$.) However, the next lemma shows that both disasters are avoided when the morphism is quasi-compact.

Lemma 29.6.3. Let $f : X \to Y$ be a morphism of schemes. Let $Z \subset Y$ be the scheme theoretic image of $f$. If $f$ is quasi-compact then

  1. the sheaf of ideals $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to f_*\mathcal{O}_ X)$ is quasi-coherent,

  2. the scheme theoretic image $Z$ is the closed subscheme determined by $\mathcal{I}$,

  3. for any open $U \subset Y$ the scheme theoretic image of $f|_{f^{-1}(U)} : f^{-1}(U) \to U$ is equal to $Z \cap U$, and

  4. the image $f(X) \subset Z$ is a dense subset of $Z$, in other words the morphism $X \to Z$ is dominant (see Definition 29.8.1).

Proof. Part (4) follows from part (3). To show (3) it suffices to prove (1) since the formation of $\mathcal{I}$ commutes with restriction to open subschemes of $Y$. And if (1) holds then in the proof of Lemma 29.6.1 we showed (2). Thus it suffices to prove that $\mathcal{I}$ is quasi-coherent. Since the property of being quasi-coherent is local we may assume $Y$ is affine. As $f$ is quasi-compact, we can find a finite affine open covering $X = \bigcup _{i = 1, \ldots , n} U_ i$. Denote $f'$ the composition

\[ X' = \coprod U_ i \longrightarrow X \longrightarrow Y. \]

Then $f_*\mathcal{O}_ X$ is a subsheaf of $f'_*\mathcal{O}_{X'}$, and hence $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to f'_*\mathcal{O}_{X'})$. By Schemes, Lemma 26.24.1 the sheaf $f'_*\mathcal{O}_{X'}$ is quasi-coherent on $Y$. Hence we win. $\square$

Example 29.6.4. If $A \to B$ is a ring map with kernel $I$, then the scheme theoretic image of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is the closed subscheme $\mathop{\mathrm{Spec}}(A/I)$ of $\mathop{\mathrm{Spec}}(A)$. This follows from Lemma 29.6.3.

If the morphism is quasi-compact, then the scheme theoretic image only adds points which are specializations of points in the image.

Lemma 29.6.5. Let $f : X \to Y$ be a quasi-compact morphism. Let $Z$ be the scheme theoretic image of $f$. Let $z \in Z$1. There exists a valuation ring $A$ with fraction field $K$ and a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[rr] \ar[d] & & X \ar[d] \ar[ld] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Z \ar[r] & Y } \]

such that the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to $z$. In particular any point of $Z$ is the specialization of a point of $f(X)$.

Proof. Let $z \in \mathop{\mathrm{Spec}}(R) = V \subset Y$ be an affine open neighbourhood of $z$. By Lemma 29.6.3 the intersection $Z \cap V$ is the scheme theoretic image of $f^{-1}(V) \to V$. Hence we may replace $Y$ by $V$ and assume $Y = \mathop{\mathrm{Spec}}(R)$ is affine. In this case $X$ is quasi-compact as $f$ is quasi-compact. Say $X = U_1 \cup \ldots \cup U_ n$ is a finite affine open covering. Write $U_ i = \mathop{\mathrm{Spec}}(A_ i)$. Let $I = \mathop{\mathrm{Ker}}(R \to A_1 \times \ldots \times A_ n)$. By Lemma 29.6.3 again we see that $Z$ corresponds to the closed subscheme $\mathop{\mathrm{Spec}}(R/I)$ of $Y$. If $\mathfrak p \subset R$ is the prime corresponding to $z$, then we see that $I_{\mathfrak p} \subset R_{\mathfrak p}$ is not an equality. Hence (as localization is exact, see Algebra, Proposition 10.9.12) we see that $R_{\mathfrak p} \to (A_1)_{\mathfrak p} \times \ldots \times (A_ n)_{\mathfrak p}$ is not zero. Hence one of the rings $(A_ i)_{\mathfrak p}$ is not zero. Hence there exists an $i$ and a prime $\mathfrak q_ i \subset A_ i$ lying over a prime $\mathfrak p_ i \subset \mathfrak p$. By Algebra, Lemma 10.50.2 we can choose a valuation ring $A \subset K = \kappa (\mathfrak q_ i)$ dominating the local ring $R_{\mathfrak p}/\mathfrak p_ iR_{\mathfrak p} \subset \kappa (\mathfrak q_ i)$. This gives the desired diagram. Some details omitted. $\square$

Lemma 29.6.6. Let

\[ \xymatrix{ X_1 \ar[d] \ar[r]_{f_1} & Y_1 \ar[d] \\ X_2 \ar[r]^{f_2} & Y_2 } \]

be a commutative diagram of schemes. Let $Z_ i \subset Y_ i$, $i = 1, 2$ be the scheme theoretic image of $f_ i$. Then the morphism $Y_1 \to Y_2$ induces a morphism $Z_1 \to Z_2$ and a commutative diagram

\[ \xymatrix{ X_1 \ar[r] \ar[d] & Z_1 \ar[d] \ar[r] & Y_1 \ar[d] \\ X_2 \ar[r] & Z_2 \ar[r] & Y_2 } \]

Proof. The scheme theoretic inverse image of $Z_2$ in $Y_1$ is a closed subscheme of $Y_1$ through which $f_1$ factors. Hence $Z_1$ is contained in this. This proves the lemma. $\square$

Lemma 29.6.7. Let $f : X \to Y$ be a morphism of schemes. If $X$ is reduced, then the scheme theoretic image of $f$ is the reduced induced scheme structure on $\overline{f(X)}$.

Proof. This is true because the reduced induced scheme structure on $\overline{f(X)}$ is clearly the smallest closed subscheme of $Y$ through which $f$ factors, see Schemes, Lemma 26.12.7. $\square$

Lemma 29.6.8. Let $f : X \to Y$ be a separated morphism of schemes. Let $V \subset Y$ be a retrocompact open. Let $s : V \to X$ be a morphism such that $f \circ s = \text{id}_ V$. Let $Y'$ be the scheme theoretic image of $s$. Then $Y' \to Y$ is an isomorphism over $V$.

Proof. The assumption that $V$ is retrocompact in $Y$ (Topology, Definition 5.12.1) means that $V \to Y$ is a quasi-compact morphism. By Schemes, Lemma 26.21.14 the morphism $s : V \to X$ is quasi-compact. Hence the construction of the scheme theoretic image $Y'$ of $s$ commutes with restriction to opens by Lemma 29.6.3. In particular, we see that $Y' \cap f^{-1}(V)$ is the scheme theoretic image of a section of the separated morphism $f^{-1}(V) \to V$. Since a section of a separated morphism is a closed immersion (Schemes, Lemma 26.21.11), we conclude that $Y' \cap f^{-1}(V) \to V$ is an isomorphism as desired. $\square$

[1] By Lemma 29.6.3 set-theoretically $Z$ agrees with the closure of $f(X)$ in $Y$.

Comments (6)

Comment #2119 by David Hansen on

A minor complaint: if you search the Stacks project for "scheme-theoretic image", this section does NOT show up in the search results! This is bad, since "scheme-theoretic" is more grammatically correct than "scheme theoretic".

Comment #2140 by on

Please read the explanation about search terms on the search page. For example putting in David-Hansen searches for text containing the word David but not containing the word Hansen. So I don't think we can fix this right now due to the behaviour of search.

Comment #4288 by Xuande Liu on

Hence (as localization is exact, see Algebra, Proposition 10.9.12) we see that is not zero. The last subscript should be n but not 1.

Comment #4734 by Damie Calaque on

There's a typo in the proof of the last Lemma: shall be .


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