# The existence of a $k$-algebra automorphism

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?

• 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 $x+a=(x_1+a_1,...,x_r+a_r)$. – user647486 Apr 26 at 15:12
• 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. – user647486 Apr 26 at 15:14
• Oops, I assumed without realizing it that $r=1$. The form of $\mathfrak m$ is a corollary of the Nullstellensatz. – user419669 Apr 26 at 15:17