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I know the group of automorphisms of $\mathbb{P}^n$ is equal to the degree $1$ birational transformations of $\mathbb{P}^n$ and every birational transformation of $\mathbb{P}^n$ can be written as $[f_1:...:f_{n+1}]$, where the $f_i$ are homogeneous polynomials.

Now I am considering this birational map $\phi:\mathbb{P}^2\to \mathbb{P}^2: [x:y:z]\mapsto [z:x-y:x-y]$, it's defined everywhere except $[1:1:0]$ as this formula. But this is indeed an automorphism of $\mathbb{P}^2$. I wonder how to get a formula at $[1:1:0]$ so that $\phi$ determines the automorphism?

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    $\begingroup$ The map you wrote down has the wrong dimensions, there should be two colons in the target (otherwise it is mapping into $\mathbb P^1$). $\endgroup$ – pre-kidney Jan 2 at 1:05
  • $\begingroup$ @pre-kidney sorry about that typo, I have fixed $\endgroup$ – 6666 Jan 2 at 1:24
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The map is not an automorphism of $\mathbb P^2$, because its image has dimension $1$: it is supported in the diagonal copy of $\mathbb P^1$ given by $[s\colon t\colon t]$ as $s,t$ range over the ground field.

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