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I'm trying to justify to myself the assertion (used here) that given a field $K$ and elements $b_1,\dots,b_n\in K$, the map $\varphi(X_i) = X_i + b_i$ is a $K$-automorphism of $K[X_1,\dots,X_n]$.

First of all, $\varphi$ is by definition a $K$-endomorphism of $K[X_1,\dots,X_n]$, so I'm not really worried about that. What I am trying to see is why $\varphi$ is bijective.

For surjectivity, I can suppose $f(X_1,\dots,X_n)\in K[X_1,\dots,X_n]$, then note that $f(X_1 - b_1,\dots,X_n - b_n)$ is a polynomial in $K[X_1,\dots,X_n]$ and $$ f(X_1 - b_1,\dots,X_n - b_n) \overset\varphi\mapsto f(X_1,\dots,X_n) \text. $$ For injectivity, I can try to show $\ker\varphi$ is trivial. So, suppose $g(X_1,\dots,X_n) \in \ker\varphi$, that is, $g(X_1 - b_1, \dots, X_n - b_n) = 0$. But it's not jumping out at me why $g(X_1,\dots,X_n) = 0$. How should I see this?

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If for $\mathbf b=(b_1,\ldots, b_n)\in K^n$ we denote with $\phi_{\mathbf b}$ the endomorphism of $K[X_1,\ldots,X_n]$ given by $\phi_{\mathbf b}(X_i)=X_i+b_i$, then we see that $\phi_{-\mathbf b}\circ\phi_{\mathbf b}=\phi_{\mathbf b}\circ\phi_{-\mathbf b}=\operatorname{id}$ because these compositions map $X_i\mapsto X_i+b_i\mapsto X_i$ (resp. $X_i\mapsto X_i-b_i\mapsto X_i$).

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The most important property of polynomial rings is the following: given a ring homomorphism $\varphi\colon K\to R$ and $r_1,\dots,r_n\in R$, there exists a unique ring homomorphism $\hat{h}\colon K[X_1,\dots,X_n]$ such that $\hat{\varphi}(k)=\varphi(k)$, for $k\in K$, and $\hat{\varphi}(X_i)=r_i$, for $i=1,\dots,n$.

Here $R$ is any commutative ring.

Now apply this to the homomorphism $f$ defined to be the identity on $K$ and mapping $X_i$ to $X_i+b_i$, but also to the homomorphism $g$ defined to be the identity on $K$ and mapping $X_i$ to $X_i-b_i$.

Since $f\circ g$ and $g\circ f$ map $X_i$ to $X_i$, they are both the identity. The uniqueness in the property above is the key.

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