The Stacks project

Lemma 70.4.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$ and $y = f(x) \in |Y|$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Consider commutative diagrams

\[ \xymatrix{ X \ar[d] & X \times _ Y V \ar[d] \ar[l] & X_ v \ar[d] \ar[l] \\ Y & V \ar[l] & v \ar[l] } \quad \xymatrix{ X \ar[d] & U \ar[d] \ar[l] & U_ v \ar[d] \ar[l] \\ Y & V \ar[l] & v \ar[l] } \quad \xymatrix{ x \ar@{|->}[d] & x' \ar@{|->}[d] \ar@{|->}[l] & u \ar@{|->}[ld] \ar@{|->}[l] \\ y & v \ar@{|->}[l] } \]

where $V$ and $U$ are schemes, $V \to Y$ and $U \to X \times _ Y V$ are étale, $v \in V$, $x' \in |X_ v|$, $u \in U$ are points related as in the last diagram. Denote $\mathcal{F}|_{X_ v}$ and $\mathcal{F}|_{U_ v}$ the pullbacks of $\mathcal{F}$. The following are equivalent

  1. for some $V, v, x'$ as above $x'$ is a weakly associated point of $\mathcal{F}|_{X_ v}$,

  2. for every $V \to Y, v, x'$ as above $x'$ is a weakly associated point of $\mathcal{F}|_{X_ v}$,

  3. for some $U, V, u, v$ as above $u$ is a weakly associated point of $\mathcal{F}|_{U_ v}$,

  4. for every $U, V, u, v$ as above $u$ is a weakly associated point of $\mathcal{F}|_{U_ v}$,

  5. for some field $k$ and morphism $\mathop{\mathrm{Spec}}(k) \to Y$ representing $y$ and some $t \in |X_ k|$ mapping to $x$, the point $t$ is a weakly associated point of $\mathcal{F}|_{X_ k}$.

If there exists a field $k_0$ and a monomorphism $\mathop{\mathrm{Spec}}(k_0) \to Y$ representing $y$, then these are also equivalent to

  1. $x_0$ is a weakly associated point of $\mathcal{F}|_{X_{k_0}}$ where $x_0 \in |X_{k_0}|$ is the unique point mapping to $x$.

If the fibre of $f$ over $y$ is locally Noetherian, then in conditions (1), (2), (3), (4), and (6) we may replace “weakly associated” with “associated”.

Proof. Observe that given $V, v, x'$ as in the lemma we can find $U \to X \times _ Y V$ and $u \in U$ mapping to $x'$ and then the morphism $U_ v \to X_ v$ is étale. Thus it is clear that (1) and (3) are equivalent as well as (2) and (4). Each of these implies (5). We will show that (5) implies (2). Suppose given $V, v, x'$ as well as $\mathop{\mathrm{Spec}}(k) \to X$ and $t \in |X_ k|$ such that the point $t$ is a weakly associated point of $\mathcal{F}|_{X_ k}$. We can choose a point $w \in v \times _ Y \mathop{\mathrm{Spec}}(k)$. Then we obtain the morphisms

\[ X_ v \longleftarrow X_ w \longrightarrow X_ k \]

Since $V \to Y$ is étale and since $w$ may be viewed as a point of $V \times _ Y \mathop{\mathrm{Spec}}(k)$, we see that $\kappa (w)/k$ is a finite separable extension of fields (Morphisms, Lemma 29.36.7). Thus $X_ w \to X_ k$ is a finite étale morphism as a base change of $w \to \mathop{\mathrm{Spec}}(k)$. Thus any point $x''$ of $X_ w$ lying over $t$ is a weakly associated point of $\mathcal{F}|_{X_ w}$ by Lemma 70.3.7. We may pick $x''$ mapping to $x'$ (Properties of Spaces, Lemma 65.4.3). Then Lemma 70.3.5 implies that $x'$ is a weakly associated point of $\mathcal{F}|_{X_ v}$.

To finish the proof it suffices to show that the equivalent conditions (1) – (5) imply (6) if we are given $\mathop{\mathrm{Spec}}(k_0) \to Y$ as in (6). In this case the morphism $\mathop{\mathrm{Spec}}(k) \to Y$ of (5) factors uniquely as $\mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k_0) \to Y$. Then $x_0$ is the image of $t$ under the morphism $X_ k \to X_{k_0}$. Hence the same lemma as above shows that (6) is true. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0CUZ. Beware of the difference between the letter 'O' and the digit '0'.