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Let $K$ be an algebraically closed field and let $\mathfrak{m}$ be a maximal ideal of $K[x_{1},..,x_{n}]$. How to show there is a ring automorphism $f$ of $K[x_{1},..,x_{n}]$ such that:

$f(\mathfrak{m}) = (x_1, x_{2},..,x_{n})$ here $()$ denotes the ideal generated by.

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Since $K$ is algebraically closed, $\mathfrak{m}$ has a very simple form. Look at the "weak Nullstellensatz" in Hochster's handout. –  Dylan Moreland Jul 18 '11 at 16:22
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I think, since $K$ is algebraically closed, all maximal ideals are of the form $m=(X_1-a_1,\dots,X_n-a_n)$. To specify a ring homomorphism from $K[X_1,\dots,X_n]$ to any other (commutative) ring $A$, you only need to specify a ring homomorphism $\rho:K\rightarrow A$ and $n$ points in $A$ that the $X_i$ will be mapped to. So in your case you can take the following ring homomorphism : $\rho:K\hookrightarrow K[X_1,\dots,X_n]$ and send $X_i$ to $X_i+a_i$. This defines a unique ring homomorphism $K[X_1,\dots,X_n]\rightarrow K[X_1,\dots,X_n]$, and it's an isomorphism (you can construct its inverse in the same manner). This sends your maximal ideal $m=(X_1-a_1,\dots,X_n-a_n)$ to $(X_1,\dots,X_n)$.

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It is a consequence of Hilbert Nullstellensatz that if $K$ is algebraically closed then ${\bf m}=(x_1-a_1,\ldots,x_n-a_n)$ for some $(a_i)\in K^n$. Then $f(x_i)=x_i+a_i$ does the trick.

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