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Suppose $ V $ is a variety in $ \mathbb{P}^n $ and $ V \supset H_\infty $. Show that $ V= \mathbb{P}^n $ or $ V = H_\infty $. If $ V= \mathbb{P}^n, V_* = \mathbb {A}^n $ while if $ V = H_\infty, V_* = \emptyset $.

My attempt: $ H_\infty = \mathbb{P}^n $\ $ U_{n+1} \subset V \subset \mathbb{P}^n $. Suppose $ v \in V $. I am trying to prove that $ v \notin U_{n+1} $.

Any help is much appreciated. Thanks in advance for any replies.

Notation: $ \mathbb{P}^n = $ projective $ n $ space. $ H_\infty = $ Hyperplace at infinity. $ V_* = V(I^*) $ where $ I^* = \{ (F_*)|F \in I \} $. $ F_* = F(X_1,...,X_n,1) $. $ U_{n+1} = \{ [x_1:...:x_{n+1}] \in \mathbb{P}^n | x_{n+1} \neq 0 \}. $

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    $\begingroup$ Please explain your notations. $\endgroup$ – user379195 May 9 '17 at 4:23
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  1. Helpful fact: If $X$ and $Y$ are varieties (in particular, irreducible algebraic sets cut out by a radical ideal) of the same dimension, then $X \subset Y$ implies that $X = Y$.

  2. There are two options, $dim V = n$ or $dim V = n -1$. In the first case, 1. implies that $P^n = V$. In the second case, $1.$ implies that $H_{\infty} = V$.

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