Lemma 29.55.4. Let $f : Y \to X$ be a quasi-finite morphism of schemes. Let $T \subset Y$ be a closed nowhere dense subset of $Y$. Then $f(T) \subset X$ is a nowhere dense subset of $X$.
Proof. As in the proof of Lemma 29.48.7 this reduces immediately to the case where the base $X$ is affine. In this case $Y = \bigcup _{i = 1, \ldots , n} Y_ i$ is a finite union of affine opens (as $f$ is quasi-compact). Since each $T \cap Y_ i$ is nowhere dense, and since a finite union of nowhere dense sets is nowhere dense (see Topology, Lemma 5.21.2), it suffices to prove that the image $f(T \cap Y_ i)$ is nowhere dense in $X$. This reduces us to the case where both $X$ and $Y$ are affine. At this point we apply Lemma 29.55.3 above to get a diagram
\[ \xymatrix{ Y \ar[rd]_ f \ar[rr]_ j & & Z \ar[ld]^\pi \\ & X & } \]
with $Z$ affine, $\pi $ finite and $j$ an open immersion. Set $\overline{T} = \overline{j(T)} \subset Z$. By Topology, Lemma 5.21.3 we see $\overline{T}$ is nowhere dense in $Z$. Since $f(T) \subset \pi (\overline{T})$ the lemma follows from the corresponding result in the finite case, see Lemma 29.48.7. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: