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I would like to show that $\mathbb{P}^n$ is a variety, given that it is a prevariety. This is how I've started...

To show that $\mathbb{P}^n$ is in fact a variety, we must show that the diagonal of $\mathbb{P}^n$,$\Delta(\mathbb{P}^n) = \{(x,x): x \in \mathbb{P}^n\}$ is a closed set in the product prevariety $\mathbb{P}^n \times \mathbb{P}^n$. We can use the Segre embedding, i.e. the map,

$\phi: \mathbb{P}^n \times \mathbb{P}^m \rightarrow \mathbb{P}^{(n+1)(m+1)-1}: ([x_0,\ldots,x_n],[y_0,\ldots,y_m]) \mapsto [x_0y_0, \ldots,x_iy_j,\ldots,x_ny_m],$

(where $0 \leq i \leq n$ and $0 \leq j \leq m$), to embed the elements of $\Delta(\mathbb{P}^n)$ in the prevariety $\mathbb{P}^{(n+1)(m+1)-1}$.

Not too sure where to go from here though. Any help would be greatly appreciated!

sc636
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    You could try taking $m=n$ and finding a closed subset $X$ of $\mathbb{P}^{(n+1)(n+1)-1}$ such that $\phi^{-1}(X)$ is the diagonal. – Exit path Feb 11 '18 at 18:02

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