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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 $\Phi_\ast(f) = 0$, 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.

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 $\Phi_a:M^a/M^{a+1} \to N^{a}/N^{a+1}$ for any $a\geq 1$. Here both $M^a/M^{a+1}$ and $N^{a}/N^{a+1}$ are $k$-vector spaces of the same dimension, and $\Phi_a$ is thus an isomorphism since it is clearly surjective. But now if $\Phi(f) = 0$ for some $f$, then $\Phi(f) \in N$ and hence $f\in M$. Choose $a$ so that $f$ lies in $M^a$ but not in $M^{a+1}$ (such an $a$ clearly exists: it is the degree of the lowest degree homogeneous piece of $f$). We then have $\Phi_a(f) = 0$ and $f\notin M^{a+1}$, contradicting that $\Phi_a$ is an isomorphism.

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.

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.

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.

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.