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Let $k$ be an algebraically closed field and let $\mathfrak m$ be a maximal ideal in $R=k[x_1,\dots,x_r]$. Show that there is a $k$-algebra automorphism of $R$ taking $\mathfrak m$ to $(x_1,\dots,x_r)$.

Since $k$ is algebraically closed, every maximal ideal $\mathfrak m$ is of the form $(x_1-a_1,\dots,x_n-a_n)$. So would the map be defined as follows? Let $\phi: R\to R$ be defined on $x_i$ as $x_i\mapsto x_i+a_i$ and let $\phi(a_nx^n+\dots a_1x+a_0)=a_n\phi(x_1)^n+\dots+a_1\phi(x)+a_0$. Then by construction, $\phi$ is a $k$-algebra automorphism. Does this look OK?

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  • $\begingroup$ Elements of $k[x_1,...,x_r]$ are of the form $\sum_\alpha a_\alpha x^\alpha$ for $\alpha=(\alpha_1,...,\alpha_r)$ and $x^\alpha=x_1^{\alpha_1}...x_r^{\alpha_r}$. Then $\phi(\sum_\alpha a_\alpha x^\alpha)=\sum_\alpha a_\alpha (x+a)^\alpha$, where $a=(a_1,...,a_r)$ and $x+a=(x_1+a_1,...,x_r+a_r)$. $\endgroup$ – user647486 Apr 26 at 15:12
  • $\begingroup$ The meaty part of the proof is that $\mathfrak m$ is of the form $(x_1-a_1,...,x_r-a_r)$. You need to judge if your readers would be convinced that that is true, and therefore if you should leave that part out or not. $\endgroup$ – user647486 Apr 26 at 15:14
  • $\begingroup$ Oops, I assumed without realizing it that $r=1$. The form of $\mathfrak m$ is a corollary of the Nullstellensatz. $\endgroup$ – user419669 Apr 26 at 15:17

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