Let $V$ be a closed subvariety of $\mathbf{P}^n$. (We work over an algebraically closed field.) Define $\pi:(\mathbf{P}^n\setminus P_0)\to \mathbf{P}^{n-1}$ by $\pi(x_0:x_1:...:x_n) = (x_0:x_1,...:x_{n-1})$, where $P_0$ is the point $(0,0,...,0,*)$ in $\mathbf{P}^n$.

If only $\pi$ were defined in all of $\mathbf{P}^n$, $\pi(V)$ would be a closed subvariety of $\mathbf{P}^{n-1}$. It isn't, and $V$ need not be the a closed subvariety of $\mathbf{P}^{n-1}$. (Easy example: $V:x_0^2 = x_1 x_2$.) Can one still say that $\pi(V)$ contains $\overline{\pi(V)}\setminus W$, where $W$ is a closed subvariety of positive codimension in $\overline{\pi(V)}$ and degree $\leq \deg(V)$, say? How?

  • $\begingroup$ It seems natural to guess that one should start by blowing up $P_0$, but I don't know where to go from there. $\endgroup$ Commented Nov 30, 2020 at 10:06
  • $\begingroup$ In the projective space $(x_0:x_1:\dots:x_n)$ is a usual practical notation for elements. $\endgroup$
    – YCor
    Commented Nov 30, 2020 at 11:50

1 Answer 1


Blow up to get a morphism $\Pi: Bl_{P_0}\mathbf P^n \rightarrow \mathbf P^{n-1}$. Let $\widetilde{V}$ be the proper transform of $V$ in $Bl_{P_0}\mathbf P^n$. Then $\overline{\pi(V)}=\Pi(\widetilde{V})$.

Now we can write $\widetilde{V}=V \setminus \{P_0\} \ \cup \mathbf P(C_{P_O}V)$ where $C_{P_0}V$ is the tangent cone of $V$ at $P_0$.

So $\pi(V \setminus \{P_0\})$ (which in your notation is $\pi(V)$) contains $\Pi(\widetilde{V}) \setminus \Pi (\mathbf P(C_{P_O}V))$.

As noted above, $\Pi(\widetilde{V})$ equals $\overline{\pi(V)}$. Moreover, $\mathbf P(C_{P_O}V))$ is a closed subset of the exceptional divisor $E$, and $\Pi_{|E} \colon E \rightarrow \mathbf P^{n-1}$ is an isomorphism.

So we get that $\pi(V)$ (in your notation) contains $\overline{\pi(V)} \setminus W$ where $W \subset \mathbf P^{n-1}$ is a closed subset isomorphic to the projectivisation of the tangent cone of $V$ at $P_0$.

The closed set $W$ has dimension $\operatorname{dim}(V)-1$. On the other hand, $\pi(V)$ has the same dimension as $V$ unless $V$ is a cone whose vertex contains $P_0$, but in that case $\pi(V)$ is a closed set .

As for degree, the degree of $\mathbf P(C_{P_O}V))$ as a subscheme of $E$ equals the multiplicity of $V$ at $P_0$, hence is bounded above by $\operatorname{deg}(V)$. Since $W$ is (isomorphic to) the underlying closed subset of this scheme, its degree is not greater than that of the scheme. So we have $\operatorname{deg}(W) \leq \operatorname{deg}(V)$ as required.

  • $\begingroup$ Ah yes (and two colleagues just explained the same to me). Thanks! I'm a bit unsure of what happens if $V$ is singular at $P_0$ - I take you are addressing that at the end? $\endgroup$ Commented Nov 30, 2020 at 12:07
  • $\begingroup$ Never mind - everything works out. $\endgroup$ Commented Nov 30, 2020 at 12:19

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