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Denote by $\Psi : k^n\to k^n$ the map of affine spaces corresponding to $\Phi$, and without loss of generality assume $\Psi(0) = 0$. Putting $M = (x_1,\ldots,x_n)$ and $N = (y_1,\ldots,y_n)$, this means that $\Phi^{-1}(N) = M$, so $\Phi(M) = N$ since $\Phi$ is surjective. We then get an induced map $\Phi_\ast:M/M^2 \Phi_a:M^a/M^{a+1} \to N/N^2$N^{a}/N^{a+1}$for any$a\geq 1$. Here both$M/M^2$M^a/M^{a+1}$ and $N/N^2$ N^{a}/N^{a+1}$are$n$-dimensional$k$-vector spaces of the same dimension, and$\Phi_\ast$\Phi_a$ is clearly thus an isomorphism since it is clearly surjective. But now if $\Phi(f) = 0$ for some $f$, then $\Phi(f) \in N$ and hence $f\in M$. We then have Choose $\Phi_\ast(f) = 0$, a contradiction. Thus a$so that$\Phi$is injective. Geometrically, saying f$ lies in $\Phi$ is surjective means M^a$but not in$\Psi$is M^{a+1}$ (such an immersion, and so its differential $a$ clearly exists: it is an isomorphism. But if the degree of the lowest degree homogeneous piece of $\Phi$ fails to be injective, f$). We then$\Psi$maps$k^n$onto a lower dimensional subvariety of have$k^n$, \Phi_a(f) = 0$ and so its differential isn't an isomorphism.

I feel obliged to add that we are somehow still using "dimension theory" here$f\notin M^{a+1}$, in the sense contradicting that the dimension of affine space is encoded in the dimension of its tangent space. However, knowledge of chains of primes etc. $\Phi_a$ is clearly overkillan isomorphism.

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Denote by $\Psi : k^n\to k^n$ the map of affine spaces corresponding to $\Phi$, and without loss of generality assume $\Psi(0) = 0$. Putting $M = (x_1,\ldots,x_n)$ and $N = (y_1,\ldots,y_n)$, this means that $\Phi^{-1}(N) = M$, so $\Phi(M) = N$ since $\Phi$ is surjective. We then get an induced map $\Phi_\ast:M/M^2 \to N/N^2$. Here both $M/M^2$ and $N/N^2$ are $n$-dimensional $k$-vector spaces, and $\Phi_\ast$ is clearly an isomorphism. But now if $\Phi(f) = 0$ for some $f$, then $\Phi(f) \in N$ and hence $f\in M$. We then have $\Phi^\ast(f) \Phi_\ast(f) = 0$, a contradiction. Thus $\Phi$ is injective.

Geometrically, saying $\Phi$ is surjective means $\Psi$ is an immersion, and so its differential is an isomorphism. But if $\Phi$ fails to be injective, then $\Psi$ maps $k^n$ onto a lower dimensional subvariety of $k^n$, and so its differential isn't an isomorphism.

I feel obliged to add that we are somehow still using "dimension theory" here, in the sense that the dimension of affine space is encoded in the dimension of its tangent space. However, knowledge of chains of primes etc. is clearly overkill.

3 added 246 characters in body

Denote by $\Psi : k^n\to k^n$ the map of affine spaces corresponding to $\Phi$, and without loss of generality assume $\Psi(0) = 0$. Putting $M = (x_1,\ldots,x_n)$ and $N = (y_1,\ldots,y_n)$, this means that $\Phi^{-1}(N) = M$, so $\Phi(M) = N$ since $\Phi$ is surjective. We then get an induced map $\Phi_\ast:M/M^2 \to N/N^2$. Here both $M/M^2$ and $N/N^2$ are $n$-dimensional $k$-vector spaces, and $\Phi_\ast$ is clearly an isomorphism. But now if $\Phi(f) = 0$ for some $f$, then $\Phi(f) \in N$ and hence $f\in M$. We then have $\Phi^\ast(f) = 0$, a contradiction. Thus $\Phi$ is injective.

Geometrically, saying $\Phi$ is surjective means $\Psi$ is an immersion, and so its differential is an isomorphism. But if $\Phi$ fails to be injective, then $\Psi$ maps $k^n$ onto a lower dimensional subvariety of $k^n$, and so its differential isn't an isomorphism.

I feel obliged to add that we are somehow still using "dimension theory" here, in the sense that the dimension of affine space is encoded in the dimension of its tangent space. However, knowledge of chains of primes etc. is clearly overkill.

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