## 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 $Y$.

In Examples, Section 108.23 the reader finds an example for both phenomena. These phenomena can arise even for immersions, see Examples, Section 108.24. 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. $\square$

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

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 $Y$.

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 $R_{\mathfrak p} \to (A_1)_{\mathfrak p} \times \ldots \times (A_1)_{\mathfrak p}$ 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: $s:V\to Y$ shall be $s:V\to X$.

Comment #5947 by Bjorn Poonen on

In response to Comment #2119 by David Hansen on July 16, 2016 at 18:45 Comment #2140 by Johan on July 21, 2016 at 19:47

I wonder if the search behavior has changed since then. For me a search for "scheme-theoretic image" does find this section.

In any case, I agree with David Hansen that "scheme theoretic" should be "scheme-theoretic". Likewise, "set theoretic" should be "set-theoretic", not just in this section, but throughout.

Comment #5948 by Bjorn Poonen on

Regarding (1) and (2) after Definition 01R7, there is a counterexample simpler than Example 01QW: the disjoint union $X = \coprod_{n \ge 1} \operatorname{Spec} k[t]/(t^n)$ mapping to $Y=\operatorname{Spec} k[t]$.

(1) Here $\overline{f(X)}$ is a point, but the scheme-theoretic image $Z$ is all of $Y$.

(2) The formation of the scheme-theoretic image does not commute with restriction to $U = \operatorname{Spec} k[t,1/t] \subset Y$ (the preimage of $U$ in $X$ is empty).

Comment #6133 by on

Thanks for the superior example! I have added it in this commit. I have not yet addressed the issue about naming "scheme-theoretic" things. Does everybody agree with this?

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