Lemma 71.4.4. 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

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

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

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

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

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

$x_0$ 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 71.3.7. We may pick $x''$ mapping to $x'$ (Properties of Spaces, Lemma 66.4.3). Then Lemma 71.3.5 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.
$\square$

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